{
  "index": "1947-A-2",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "2. A real valued continuous function satisfies for all real \\( x \\) and \\( y \\) the functional equation\n\\[\nf\\left(\\sqrt{x^{2}+y^{2}}\\right)=f(x) f(y)\n\\]\n\nProve that\n\\[\nf(x)=[f(1)]^{x^{2}}\n\\]",
  "solution": "Solution. A slight qualification in the statement of the problem is needed since the real valued continuous function \\( f(x) \\equiv 0 \\) satisfies the functional equation for all real \\( x \\) and \\( y \\), but does not satisfy the relation \\( f(0)= \\) \\( [f(1)]^{0} \\), since \\( 0^{0} \\) is undefined.\nAssume then that for some \\( y_{0}, f\\left(y_{0}\\right) \\neq 0 \\). Since\n\\[\nf(x) f\\left(y_{0}\\right)=f\\left(\\sqrt{x^{2}+y_{0}{ }^{2}}\\right)=f(-x) f\\left(y_{0}\\right),\n\\]\nwe have \\( f(x)=f(-x)=f(|x|) \\) for all \\( x \\). We now show by induction that for any positive integer \\( n \\) and any real number \\( x \\), we have\n\\[\nf(\\sqrt{n} x)=[f(x)]^{n} .\n\\]\n\nThis is certainly true for \\( n=1 \\), and assuming it true for \\( n=k \\) we have\n\\[\n\\begin{array}{c}\nf(\\sqrt{k+1} x)=f(\\sqrt{k+1}|x|)=f\\left(\\sqrt{(\\sqrt{k} x)^{2}+x^{2}}\\right)=f(\\sqrt{k} x) f(x) \\\\\n=[f(x)]^{k} f(x)=[f(x)]^{k+1} .\n\\end{array}\n\\]\n\nTherefore (1) is true for all positive integers \\( n \\).\nIf \\( p \\) and \\( q \\) are non-zero integers, then\n\\[\nf(p)=f(|p|)=f\\left(\\sqrt{p^{2}} \\cdot 1\\right)=[f(1)] p^{2}\n\\]\nand\n\\[\nf(|p|)=f\\left(\\sqrt{q^{2}}\\left|\\frac{p}{q}\\right|\\right)=\\left[f\\left(\\left|\\frac{p}{q}\\right|\\right)\\right]^{q^{2}} .\n\\]\n\nFrom these two relations it follows that\n\\[\n\\left[f\\left(\\frac{p}{q}\\right)\\right]^{q^{2}}=[f(1)]^{p^{2}} .\n\\]\n\nIf \\( f(1)>0 \\) then it follows that\n\\[\n\\left[f\\left(\\frac{p}{q}\\right)\\right]=[f(1)]^{p^{2} / q^{2}} ;\n\\]\nthat is, the required equation is valid for all rational values of \\( x \\) except, perhaps, \\( x=0 \\). By continuity it follows for all values of \\( x \\).\n\nIf \\( f(1)=0 \\), then (2) implies that \\( f(p / q)=0 \\) for all non-zero integers \\( p \\) and \\( q \\), and thus \\( f(x)=0 \\) for all rational \\( x \\), hence for all real \\( x \\).\n\nFinally we show that \\( f(1)<0 \\) is impossible. If \\( p \\) is even and \\( q \\) is odd, equation (2) implies that \\( f(p / q)>0 \\). Hence \\( f(x)>0 \\) for a dense set of \\( x \\), and therefore \\( f(x) \\geq 0 \\) for all \\( x \\); in particular, \\( f(1) \\geq 0 \\).\n\nRemark. If we consider the function \\( g \\) defined by\n\\[\ng(x)=\\log f(\\sqrt{x}) \\quad \\text { for } x \\geq 0,\n\\]\nthen \\( g \\) satisfies the famous Cauchy functional equation\n\\[\ng(x+y)=g(x)+g(y)\n\\]\nwhose only continuous solution is\n\\[\ng(x)=g(1) x\n\\]\nand it readily follows that \\( f(x)=f(1)^{x^{2}} \\). This argument requires showing first that (3) defines a function; i.e., that \\( f(\\sqrt{x})>0 \\) for \\( x \\geq 0 \\), which is true unless \\( f \\) vanishes identically.",
  "vars": [
    "x",
    "y",
    "n",
    "p",
    "q"
  ],
  "params": [
    "f",
    "g",
    "y_0"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variablex",
        "y": "variabley",
        "n": "integern",
        "p": "integerp",
        "q": "integerq",
        "f": "functionf",
        "g": "functiong",
        "y_0": "constyzero"
      },
      "question": "2. A real valued continuous function satisfies for all real \\( variablex \\) and \\( variabley \\) the functional equation\n\\[\nfunctionf\\left(\\sqrt{variablex^{2}+variabley^{2}}\\right)=functionf(variablex) functionf(variabley)\n\\]\n\nProve that\n\\[\nfunctionf(variablex)=[functionf(1)]^{variablex^{2}}\n\\]",
      "solution": "Solution. A slight qualification in the statement of the problem is needed since the real valued continuous function \\( functionf(variablex) \\equiv 0 \\) satisfies the functional equation for all real \\( variablex \\) and \\( variabley \\), but does not satisfy the relation \\( functionf(0)= \\) \\( [functionf(1)]^{0} \\), since \\( 0^{0} \\) is undefined.\nAssume then that for some \\( constyzero, functionf\\left(constyzero\\right) \\neq 0 \\). Since\n\\[\nfunctionf(variablex) functionf\\left(constyzero\\right)=functionf\\left(\\sqrt{variablex^{2}+constyzero^{2}}\\right)=functionf(-variablex) functionf\\left(constyzero\\right),\n\\]\nwe have \\( functionf(variablex)=functionf(-variablex)=functionf(|variablex|) \\) for all \\( variablex \\). We now show by induction that for any positive integer \\( integern \\) and any real number \\( variablex \\), we have\n\\[\nfunctionf(\\sqrt{integern}\\, variablex)=[functionf(variablex)]^{integern} .\n\\]\n\nThis is certainly true for \\( integern=1 \\), and assuming it true for \\( integern=k \\) we have\n\\[\n\\begin{array}{c}\nfunctionf(\\sqrt{k+1}\\, variablex)=functionf(\\sqrt{k+1}|variablex|)=functionf\\left(\\sqrt{(\\sqrt{k}\\, variablex)^{2}+variablex^{2}}\\right)=functionf(\\sqrt{k}\\, variablex) functionf(variablex) \\\\\n=[functionf(variablex)]^{k} functionf(variablex)=[functionf(variablex)]^{k+1} .\n\\end{array}\n\\]\n\nTherefore (1) is true for all positive integers \\( integern \\).\nIf \\( integerp \\) and \\( integerq \\) are non-zero integers, then\n\\[\nfunctionf(integerp)=functionf(|integerp|)=functionf\\left(\\sqrt{integerp^{2}} \\cdot 1\\right)=[functionf(1)]^{integerp^{2}}\n\\]\nand\n\\[\nfunctionf(|integerp|)=functionf\\left(\\sqrt{integerq^{2}}\\left|\\frac{integerp}{integerq}\\right|\\right)=\\left[functionf\\left(\\left|\\frac{integerp}{integerq}\\right|\\right)\\right]^{integerq^{2}} .\n\\]\n\nFrom these two relations it follows that\n\\[\n\\left[functionf\\left(\\frac{integerp}{integerq}\\right)\\right]^{integerq^{2}}=[functionf(1)]^{integerp^{2}} .\n\\]\n\nIf \\( functionf(1)>0 \\) then it follows that\n\\[\n\\left[functionf\\left(\\frac{integerp}{integerq}\\right)\\right]=[functionf(1)]^{integerp^{2} / integerq^{2}} ;\n\\]\nthat is, the required equation is valid for all rational values of \\( variablex \\) except, perhaps, \\( variablex=0 \\). By continuity it follows for all values of \\( variablex \\).\n\nIf \\( functionf(1)=0 \\), then (2) implies that \\( functionf(integerp / integerq)=0 \\) for all non-zero integers \\( integerp \\) and \\( integerq \\), and thus \\( functionf(variablex)=0 \\) for all rational \\( variablex \\), hence for all real \\( variablex \\).\n\nFinally we show that \\( functionf(1)<0 \\) is impossible. If \\( integerp \\) is even and \\( integerq \\) is odd, equation (2) implies that \\( functionf(integerp / integerq)>0 \\). Hence \\( functionf(variablex)>0 \\) for a dense set of \\( variablex \\), and therefore \\( functionf(variablex) \\geq 0 \\) for all \\( variablex \\); in particular, \\( functionf(1) \\geq 0 \\).\n\nRemark. If we consider the function \\( functiong \\) defined by\n\\[\nfunctiong(variablex)=\\log functionf(\\sqrt{variablex}) \\quad \\text { for } variablex \\geq 0,\n\\]\nthen \\( functiong \\) satisfies the famous Cauchy functional equation\n\\[\nfunctiong(variablex+variabley)=functiong(variablex)+functiong(variabley)\n\\]\nwhose only continuous solution is\n\\[\nfunctiong(variablex)=functiong(1) variablex\n\\]\nand it readily follows that \\( functionf(variablex)=functionf(1)^{variablex^{2}} \\). This argument requires showing first that (3) defines a function; i.e., that \\( functionf(\\sqrt{variablex})>0 \\) for \\( variablex \\geq 0 \\), which is true unless \\( functionf \\) vanishes identically."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "tangerine",
        "y": "blueberry",
        "n": "pineapple",
        "p": "cucumber",
        "q": "raspberry",
        "f": "watermelon",
        "g": "butterscotch",
        "y_0": "strawberry"
      },
      "question": "2. A real valued continuous function satisfies for all real \\( tangerine \\) and \\( blueberry \\) the functional equation\n\\[\nwatermelon\\left(\\sqrt{tangerine^{2}+blueberry^{2}}\\right)=watermelon(tangerine) watermelon(blueberry)\n\\]\n\nProve that\n\\[\nwatermelon(tangerine)=[watermelon(1)]^{tangerine^{2}}\n\\]",
      "solution": "Solution. A slight qualification in the statement of the problem is needed since the real valued continuous function \\( watermelon(tangerine) \\equiv 0 \\) satisfies the functional equation for all real \\( tangerine \\) and \\( blueberry \\), but does not satisfy the relation \\( watermelon(0)= \\) \\( [watermelon(1)]^{0} \\), since \\( 0^{0} \\) is undefined.\nAssume then that for some \\( strawberry, watermelon\\left(strawberry\\right) \\neq 0 \\). Since\n\\[\nwatermelon(tangerine) watermelon\\left(strawberry\\right)=watermelon\\left(\\sqrt{tangerine^{2}+strawberry{ }^{2}}\\right)=watermelon(-tangerine) watermelon\\left(strawberry\\right),\n\\]\nwe have \\( watermelon(tangerine)=watermelon(-tangerine)=watermelon(|tangerine|) \\) for all \\( tangerine \\). We now show by induction that for any positive integer \\( pineapple \\) and any real number \\( tangerine \\), we have\n\\[\nwatermelon(\\sqrt{pineapple} \\, tangerine)=[watermelon(tangerine)]^{pineapple} .\n\\]\n\nThis is certainly true for \\( pineapple=1 \\), and assuming it true for \\( pineapple=k \\) we have\n\\[\n\\begin{array}{c}\nwatermelon(\\sqrt{k+1} \\, tangerine)=watermelon(\\sqrt{k+1}|tangerine|)=watermelon\\left(\\sqrt{(\\sqrt{k} \\, tangerine)^{2}+tangerine^{2}}\\right)=watermelon(\\sqrt{k} \\, tangerine) watermelon(tangerine) \\\\\n=[watermelon(tangerine)]^{k} watermelon(tangerine)=[watermelon(tangerine)]^{k+1} .\n\\end{array}\n\\]\n\nTherefore (1) is true for all positive integers \\( pineapple \\).\nIf \\( cucumber \\) and \\( raspberry \\) are non-zero integers, then\n\\[\nwatermelon(cucumber)=watermelon(|cucumber|)=watermelon\\left(\\sqrt{cucumber^{2}} \\cdot 1\\right)=[watermelon(1)] cucumber^{2}\n\\]\nand\n\\[\nwatermelon(|cucumber|)=watermelon\\left(\\sqrt{raspberry^{2}}\\left|\\frac{cucumber}{raspberry}\\right|\\right)=\\left[watermelon\\left(\\left|\\frac{cucumber}{raspberry}\\right|\\right)\\right]^{raspberry^{2}} .\n\\]\n\nFrom these two relations it follows that\n\\[\n\\left[watermelon\\left(\\frac{cucumber}{raspberry}\\right)\\right]^{raspberry^{2}}=[watermelon(1)]^{cucumber^{2}} .\n\\]\n\nIf \\( watermelon(1)>0 \\) then it follows that\n\\[\n\\left[watermelon\\left(\\frac{cucumber}{raspberry}\\right)\\right]=[watermelon(1)]^{cucumber^{2} / raspberry^{2}} ;\n\\]\nthat is, the required equation is valid for all rational values of \\( tangerine \\) except, perhaps, \\( tangerine=0 \\). By continuity it follows for all values of \\( tangerine \\).\n\nIf \\( watermelon(1)=0 \\), then (2) implies that \\( watermelon(cucumber / raspberry)=0 \\) for all non-zero integers \\( cucumber \\) and \\( raspberry \\), and thus \\( watermelon(tangerine)=0 \\) for all rational \\( tangerine \\), hence for all real \\( tangerine \\).\n\nFinally we show that \\( watermelon(1)<0 \\) is impossible. If \\( cucumber \\) is even and \\( raspberry \\) is odd, equation (2) implies that \\( watermelon(cucumber / raspberry)>0 \\). Hence \\( watermelon(tangerine)>0 \\) for a dense set of \\( tangerine \\), and therefore \\( watermelon(tangerine) \\geq 0 \\) for all \\( tangerine \\); in particular, \\( watermelon(1) \\geq 0 \\).\n\nRemark. If we consider the function \\( butterscotch \\) defined by\n\\[\nbutterscotch(tangerine)=\\log watermelon(\\sqrt{tangerine}) \\quad \\text { for } tangerine \\geq 0,\n\\]\nthen \\( butterscotch \\) satisfies the famous Cauchy functional equation\n\\[\nbutterscotch(tangerine+blueberry)=butterscotch(tangerine)+butterscotch(blueberry)\n\\]\nwhose only continuous solution is\n\\[\nbutterscotch(tangerine)=butterscotch(1) tangerine\n\\]\nand it readily follows that \\( watermelon(tangerine)=watermelon(1)^{tangerine^{2}} \\). This argument requires showing first that (3) defines a function; i.e., that \\( watermelon(\\sqrt{tangerine})>0 \\) for \\( tangerine \\geq 0 \\), which is true unless \\( watermelon \\) vanishes identically."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constant",
        "y": "staticval",
        "n": "fraction",
        "p": "irrational",
        "q": "continuum",
        "f": "stagnant",
        "g": "steadfast",
        "y_0": "infinitum"
      },
      "question": "2. A real valued continuous function satisfies for all real \\( constant \\) and \\( staticval \\) the functional equation\n\\[\nstagnant\\left(\\sqrt{constant^{2}+staticval^{2}}\\right)=stagnant(constant) stagnant(staticval)\n\\]\n\nProve that\n\\[\nstagnant(constant)=[stagnant(1)]^{constant^{2}}\n\\]",
      "solution": "Solution. A slight qualification in the statement of the problem is needed since the real valued continuous function \\( stagnant(constant) \\equiv 0 \\) satisfies the functional equation for all real \\( constant \\) and \\( staticval \\), but does not satisfy the relation \\( stagnant(0)= \\) \\( [stagnant(1)]^{0} \\), since \\( 0^{0} \\) is undefined.\nAssume then that for some \\( infinitum, stagnant\\left(infinitum\\right) \\neq 0 \\). Since\n\\[\nstagnant(constant) stagnant\\left(infinitum\\right)=stagnant\\left(\\sqrt{constant^{2}+infinitum { }^{2}}\\right)=stagnant(-constant) stagnant\\left(infinitum\\right),\n\\]\nwe have \\( stagnant(constant)=stagnant(-constant)=stagnant(|constant|) \\) for all \\( constant \\). We now show by induction that for any positive integer \\( fraction \\) and any real number \\( constant \\), we have\n\\[\nstagnant(\\sqrt{fraction} \\, constant)=[stagnant(constant)]^{fraction} .\n\\]\n\nThis is certainly true for \\( fraction=1 \\), and assuming it true for \\( fraction=k \\) we have\n\\[\n\\begin{array}{c}\nstagnant(\\sqrt{k+1} \\, constant)=stagnant(\\sqrt{k+1}|constant|)=stagnant\\left(\\sqrt{(\\sqrt{k} \\, constant)^{2}+constant^{2}}\\right)=stagnant(\\sqrt{k} \\, constant) stagnant(constant) \\\\\n=[stagnant(constant)]^{k} stagnant(constant)=[stagnant(constant)]^{k+1} .\n\\end{array}\n\\]\n\nTherefore (1) is true for all positive integers \\( fraction \\).\nIf \\( irrational \\) and \\( continuum \\) are non-zero integers, then\n\\[\nstagnant(irrational)=stagnant(|irrational|)=stagnant\\left(\\sqrt{irrational^{2}} \\cdot 1\\right)=[stagnant(1)] irrational^{2}\n\\]\nand\n\\[\nstagnant(|irrational|)=stagnant\\left(\\sqrt{continuum^{2}}\\left|\\frac{irrational}{continuum}\\right|\\right)=\\left[stagnant\\left(\\left|\\frac{irrational}{continuum}\\right|\\right)\\right]^{continuum^{2}} .\n\\]\n\nFrom these two relations it follows that\n\\[\n\\left[stagnant\\left(\\frac{irrational}{continuum}\\right)\\right]^{continuum^{2}}=[stagnant(1)]^{irrational^{2}} .\n\\]\n\nIf \\( stagnant(1)>0 \\) then it follows that\n\\[\n\\left[stagnant\\left(\\frac{irrational}{continuum}\\right)\\right]=[stagnant(1)]^{irrational^{2} / continuum^{2}} ;\n\\]\nthat is, the required equation is valid for all rational values of \\( constant \\) except, perhaps, \\( constant=0 \\). By continuity it follows for all values of \\( constant \\).\n\nIf \\( stagnant(1)=0 \\), then (2) implies that \\( stagnant(irrational / continuum)=0 \\) for all non-zero integers \\( irrational \\) and \\( continuum \\), and thus \\( stagnant(constant)=0 \\) for all rational \\( constant \\), hence for all real \\( constant \\).\n\nFinally we show that \\( stagnant(1)<0 \\) is impossible. If \\( irrational \\) is even and \\( continuum \\) is odd, equation (2) implies that \\( stagnant(irrational / continuum)>0 \\). Hence \\( stagnant(constant)>0 \\) for a dense set of \\( constant \\), and therefore \\( stagnant(constant) \\geq 0 \\) for all \\( constant \\); in particular, \\( stagnant(1) \\geq 0 \\).\n\nRemark. If we consider the function \\( steadfast \\) defined by\n\\[\nsteadfast(constant)=\\log stagnant(\\sqrt{constant}) \\quad \\text { for } constant \\geq 0,\n\\]\nthen \\( steadfast \\) satisfies the famous Cauchy functional equation\n\\[\nsteadfast(constant+staticval)=steadfast(constant)+steadfast(staticval)\n\\]\nwhose only continuous solution is\n\\[\nsteadfast(constant)=steadfast(1) constant\n\\]\nand it readily follows that \\( stagnant(constant)=stagnant(1)^{constant^{2}} \\). This argument requires showing first that (3) defines a function; i.e., that \\( stagnant(\\sqrt{constant})>0 \\) for \\( constant \\geq 0 \\), which is true unless \\( stagnant \\) vanishes identically."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "n": "plmstqrv",
        "p": "znxkvdqe",
        "q": "bchylmra",
        "f": "dqpjrneo",
        "g": "klvmsatz",
        "y_0": "wucnkybe"
      },
      "question": "2. A real valued continuous function satisfies for all real \\( qzxwvtnp \\) and \\( hjgrksla \\) the functional equation\n\\[\ndqpjrneo\\left(\\sqrt{qzxwvtnp^{2}+hjgrksla^{2}}\\right)=dqpjrneo(qzxwvtnp) dqpjrneo(hjgrksla)\n\\]\n\nProve that\n\\[\ndqpjrneo(qzxwvtnp)=[dqpjrneo(1)]^{qzxwvtnp^{2}}\n\\]",
      "solution": "Solution. A slight qualification in the statement of the problem is needed since the real valued continuous function \\( dqpjrneo(qzxwvtnp) \\equiv 0 \\) satisfies the functional equation for all real \\( qzxwvtnp \\) and \\( hjgrksla \\), but does not satisfy the relation \\( dqpjrneo(0)= \\) \\( [dqpjrneo(1)]^{0} \\), since \\( 0^{0} \\) is undefined.\nAssume then that for some \\( wucnkybe, dqpjrneo\\left(wucnkybe\\right) \\neq 0 \\). Since\n\\[\ndqpjrneo(qzxwvtnp) dqpjrneo\\left(wucnkybe\\right)=dqpjrneo\\left(\\sqrt{qzxwvtnp^{2}+wucnkybe{ }^{2}}\\right)=dqpjrneo(-qzxwvtnp) dqpjrneo\\left(wucnkybe\\right),\n\\]\nwe have \\( dqpjrneo(qzxwvtnp)=dqpjrneo(-qzxwvtnp)=dqpjrneo(|qzxwvtnp|) \\) for all \\( qzxwvtnp \\). We now show by induction that for any positive integer \\( plmstqrv \\) and any real number \\( qzxwvtnp \\), we have\n\\[\ndqpjrneo(\\sqrt{plmstqrv} \\, qzxwvtnp)=[dqpjrneo(qzxwvtnp)]^{plmstqrv} .\n\\]\n\nThis is certainly true for \\( plmstqrv=1 \\), and assuming it true for \\( plmstqrv=k \\) we have\n\\[\n\\begin{array}{c}\ndqpjrneo(\\sqrt{k+1} \\, qzxwvtnp)=dqpjrneo(\\sqrt{k+1}|qzxwvtnp|)=dqpjrneo\\left(\\sqrt{(\\sqrt{k} \\, qzxwvtnp)^{2}+qzxwvtnp^{2}}\\right)=dqpjrneo(\\sqrt{k} \\, qzxwvtnp) dqpjrneo(qzxwvtnp) \\\\\n=[dqpjrneo(qzxwvtnp)]^{k} dqpjrneo(qzxwvtnp)=[dqpjrneo(qzxwvtnp)]^{k+1} .\n\\end{array}\n\\]\n\nTherefore (1) is true for all positive integers \\( plmstqrv \\).\nIf \\( znxkvdqe \\) and \\( bchylmra \\) are non-zero integers, then\n\\[\ndqpjrneo(znxkvdqe)=dqpjrneo(|znxkvdqe|)=dqpjrneo\\left(\\sqrt{znxkvdqe^{2}} \\cdot 1\\right)=[dqpjrneo(1)] znxkvdqe^{2}\n\\]\nand\n\\[\ndqpjrneo(|znxkvdqe|)=dqpjrneo\\left(\\sqrt{bchylmra^{2}}\\left|\\frac{znxkvdqe}{bchylmra}\\right|\\right)=\\left[dqpjrneo\\left(\\left|\\frac{znxkvdqe}{bchylmra}\\right|\\right)\\right]^{bchylmra^{2}} .\n\\]\n\nFrom these two relations it follows that\n\\[\n\\left[dqpjrneo\\left(\\frac{znxkvdqe}{bchylmra}\\right)\\right]^{bchylmra^{2}}=[dqpjrneo(1)]^{znxkvdqe^{2}} .\n\\]\n\nIf \\( dqpjrneo(1)>0 \\) then it follows that\n\\[\n\\left[dqpjrneo\\left(\\frac{znxkvdqe}{bchylmra}\\right)\\right]=[dqpjrneo(1)]^{znxkvdqe^{2} / bchylmra^{2}} ;\n\\]\nthat is, the required equation is valid for all rational values of \\( qzxwvtnp \\) except, perhaps, \\( qzxwvtnp=0 \\). By continuity it follows for all values of \\( qzxwvtnp \\).\n\nIf \\( dqpjrneo(1)=0 \\), then (2) implies that \\( dqpjrneo(znxkvdqe / bchylmra)=0 \\) for all non-zero integers \\( znxkvdqe \\) and \\( bchylmra \\), and thus \\( dqpjrneo(qzxwvtnp)=0 \\) for all rational \\( qzxwvtnp \\), hence for all real \\( qzxwvtnp \\).\n\nFinally we show that \\( dqpjrneo(1)<0 \\) is impossible. If \\( znxkvdqe \\) is even and \\( bchylmra \\) is odd, equation (2) implies that \\( dqpjrneo(znxkvdqe / bchylmra)>0 \\). Hence \\( dqpjrneo(qzxwvtnp)>0 \\) for a dense set of \\( qzxwvtnp \\), and therefore \\( dqpjrneo(qzxwvtnp) \\geq 0 \\) for all \\( qzxwvtnp \\); in particular, \\( dqpjrneo(1) \\geq 0 \\).\n\nRemark. If we consider the function \\( klvmsatz \\) defined by\n\\[\nklvmsatz(qzxwvtnp)=\\log dqpjrneo(\\sqrt{qzxwvtnp}) \\quad \\text { for } qzxwvtnp \\geq 0,\n\\]\nthen \\( klvmsatz \\) satisfies the famous Cauchy functional equation\n\\[\nklvmsatz(qzxwvtnp+hjgrksla)=klvmsatz(qzxwvtnp)+klvmsatz(hjgrksla)\n\\]\nwhose only continuous solution is\n\\[\nklvmsatz(qzxwvtnp)=klvmsatz(1) qzxwvtnp\n\\]\nand it readily follows that \\( dqpjrneo(qzxwvtnp)=dqpjrneo(1)^{qzxwvtnp^{2}} \\). This argument requires showing first that (3) defines a function; i.e., that \\( dqpjrneo(\\sqrt{qzxwvtnp})>0 \\) for \\( qzxwvtnp \\geq 0 \\), which is true unless \\( dqpjrneo \\) vanishes identically."
    },
    "kernel_variant": {
      "question": "Let  d≥2  and endow  H:=ℝᵈ  with the usual inner product  ⟨·,·⟩  and norm  ‖·‖.  \nFor an integer  k≥2  a k-tuple  (v₁,…,v_k)  of vectors is called pairwise orthogonal when  \n⟨v_i ,v_j⟩ = 0  for every  i≠j  (the zero vector is regarded as orthogonal to every vector).\n\nA function  f : H → ℝ  is given and is assumed to satisfy  \n\n(O) (orthogonal multiplicativity)  \n  For every k≥2 and every pairwise orthogonal k-tuple (v₁,…,v_k) one has  \n    f(v₁+⋯+v_k)=f(v₁)⋯f(v_k).\n\n(R) (one–dimensional C²-regularity and half-ray monotonicity)  \n  For every one–dimensional linear subspace L⊂H the restriction  f|_L  is of class C²,  \n  and on each half-ray {tu∈L : t≥0} the map  t↦f(tu)  is non-decreasing.\n\nProve that exactly one of the following two mutually exclusive alternatives occurs.\n\nA) f(x)=0 for every x∈H;  \n\nB) There exists a constant  c≥1  such that  \n    f(x)=c^{‖x‖²}  for every x∈H.",
      "solution": "Throughout  k≥2  is an integer and ‘⊥’ denotes orthogonality.\n\n0.  Reduction to the non–trivial case  \nIf  f≡0  the hypotheses are obviously fulfilled, giving alternative A.  \nHenceforth we suppose  f  is not identically zero and prove that alternative B must hold.\n\n1.  The value at the origin and the inequality  f(x)≥1  \nPick  a∈H  with  f(a)≠0  and apply (O) to the orthogonal k-tuple (a,0,…,0):\n\n  f(a)=f(a)·f(0)^{k-1}   ⇒   f(0)=1.\n\nFix any non–zero vector  x.  Write  x=tu  with  t>0, ‖u‖=1.  \nBecause  t↦f(tu)  is non-decreasing on  [0,∞)  and starts from  f(0)=1,\n\n  f(x)=f(tu)≥1   for every  x≠0.                                     (1)\n\nConsequently  f(x)>0  for all x and we can introduce  \n\n  g:H→ℝ, g(x):=ln f(x).                                            (2)\n\n2.  Orthogonal additivity of  g  \nTaking logarithms in (O) gives  \n\n  g(v₁+⋯+v_k)=g(v₁)+⋯+g(v_k) whenever  v₁⊥⋯⊥v_k.        (3)\n\nFunctions satisfying (3) are called orthogonally additive.\n\n3.  Second-order analysis – extraction of a quadratic form\n\n3.1  Second directional derivatives  \nFor  u∈H  the map  t↦g(tu)  is C² by (R); define  \n\n  g''_u(0):=(d²/dt²)|_{t=0}g(tu), Q(u):=½g''_u(0).        (4)\n\n3.2  Orthogonal additivity of  Q  \nFor orthonormal  u,v  set  Φ(t):=g(tu)+g(tv)−g(t(u+v)).  \nBy (3)  Φ(t)≡0, whence  Φ''(0)=0, i.e.  \n\n  Q(u+v)=Q(u)+Q(v) whenever u⊥v.                                (5)\n\n3.3  Quadratic homogeneity  \nFor any  α∈ℝ, u∈H  the relation  g(tαu)=g(αtu)  implies  g''_{αu}(0)=α²g''_u(0)  \nand hence  \n\n  Q(αu)=α²Q(u).                                                   (6)\n\nThus  Q  is a continuous orthogonally additive 2-homogeneous map, i.e. a quadratic\nform.\n\n3.4  Proportionality to  ‖·‖²  \nAs in the original draft (using an orthonormal basis together with (5) and (6))\none proves that there exists  κ∈ℝ  such that  \n\n  Q(u)=κ‖u‖²  for every u∈H.                                     (7)\n\n3.5  Continuity of  g  \nAlong each line the Taylor formula gives  g(tu)=κt²‖u‖²+o(t²).  \nA standard compactness argument on the unit sphere yields  \n\n  |g(x)|≤C‖x‖²  for  ‖x‖  small,  \n\nso  g  is continuous at 0 and therefore everywhere.\n\n4.  A structure theorem for continuous orthogonally additive maps  \n\nThe following result – due to Rätz (Aequationes Math. 28 (1985), 189–199) – will be used.\n\nTheorem 4.1 (Rätz).  \nIf  Φ:H→ℝ  is continuous and orthogonally additive, then there exist uniquely\ndetermined constants  κ∈ℝ  and  a∈H  such that  \n\n  Φ(x)=κ‖x‖²+⟨a,x⟩ for every x∈H.                                (8)\n\nSketch of proof.  Write Φ=Φ_e+Φ_o with  \n\n  Φ_e(x):=(Φ(x)+Φ(–x))/2  (even part),  \n  Φ_o(x):=(Φ(x)–Φ(–x))/2  (odd  part).\n\nBoth Φ_e and Φ_o are continuous and orthogonally additive; Φ_e is even, Φ_o is odd.\n\n•  For Φ_e the arguments in Sections 3.1–3.4 show that Φ_e(x)=κ‖x‖².\n\n•  For Φ_o pick an orthonormal basis (e₁,…,e_d) and set  a_i:=Φ_o(e_i).  \n   For x=Σx_i e_i the vectors x_i e_i are pairwise orthogonal, hence\n\n  Φ_o(x)=ΣΦ_o(x_i e_i)=Σx_i Φ_o(e_i)=⟨a,x⟩,         a:=Σa_i e_i.\n\n   Thus Φ_o is the linear functional  x↦⟨a,x⟩.\n\nAdding the two parts gives (8). ∎\n\n\nApplying the theorem to  g  and using (7) yields  \n\n  g(x)=κ‖x‖²+⟨a,x⟩ for some a∈H.                               (9)\n\n5.  Monotonicity forces  a=0  \nFix a non-zero unit vector  u.  \nFor  t≥0  define  ψ(t):=g(tu)=κt²+⟨a,u⟩t.  The function  \n\n  t↦f(tu)=e^{ψ(t)}  \n\nis non-decreasing on  [0,∞)  as required by (R).  Since  f>0, the logarithm is also\nnon-decreasing, hence  ψ'(t)=2κt+⟨a,u⟩≥0  for all  t>0.  \nLetting  t→0⁺  gives  ⟨a,u⟩≥0.  \nReplacing  u  by  –u  (which again defines a half-ray  t↦f(–tu) ) we similarly get  \n⟨a,u⟩≤0.  Consequently  ⟨a,u⟩=0.  Because  u  was an arbitrary unit vector,  a=0.\n\nTherefore  \n\n  g(x)=κ‖x‖² for every x∈H.                                      (10)\n\n6.  Sign of  κ  \nFrom (1) and (10) we have  κt²≥0  for  t  small, so  κ≥0.  \nPut  c:=e^{κ}≥1.  Then  g(x)=ln f(x)=κ‖x‖²  rewrites as  \n\n  f(x)=c^{‖x‖²} for every x∈H.                                  (11)\n\n7.  Conclusion  \nWe have proved that either  f≡0  (alternative A) or else (11) holds with some\nconstant  c≥1  (alternative B).  The two alternatives are mutually exclusive and\nexhaust all functions satisfying (O) and (R). ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.408875",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher dimension.  \n The unknown function lives on ℝᵈ with d ≥ 2; one must first recognise and prove its radial nature, which requires\n an argument using the orthogonal group—not needed in the original one–dimensional problem.\n\n2.  k-fold orthogonal multiplicativity.  \n The equation has to hold for every number k of mutually orthogonal vectors, not only for k=2; this produces extra\n relations that must be shown to be equivalent and consistent, then exploited in the proof.\n\n3.  Reduction to an additive problem on ℝ⁺ in a multi–step way.  \n One must successively (i) prove radiality, (ii) slice down to a single ray, (iii) introduce a logarithm,\n (iv) detect the Cauchy equation, and (v) invoke regularity to obtain linearity.  \n Each step uses a different tool (group actions, differential calculus, classical functional–equation theory).\n\n4.  Regularity only on one–dimensional subspaces.  \n Differentiability is not global; one has to show that the weak assumption (2) suffices to upgrade the additive\n equation to linearity—this is subtler than global continuity in the original statement.\n\n5.  Two essentially different solutions.  \n One has to keep track of the identically-zero possibility through every transformation, because logarithms are\n forbidden there.  Ensuring logical completeness adds another layer of care.\n\nAltogether the enhanced variant couples linear–algebraic arguments (orthogonal transformations),\nfunctional–equation theory (Cauchy), and analysis (regularity upgrading), making it significantly harder than both\nits predecessors."
      }
    },
    "original_kernel_variant": {
      "question": "Let  d≥2  and endow  H:=ℝᵈ  with the usual inner product  ⟨·,·⟩  and norm  ‖·‖.  \nFor an integer  k≥2  a k-tuple  (v₁,…,v_k)  of vectors is called pairwise orthogonal when  \n⟨v_i ,v_j⟩ = 0  for every  i≠j  (the zero vector is regarded as orthogonal to every vector).\n\nA function  f : H → ℝ  is given and is assumed to satisfy  \n\n(O) (orthogonal multiplicativity)  \n  For every k≥2 and every pairwise orthogonal k-tuple (v₁,…,v_k) one has  \n    f(v₁+⋯+v_k)=f(v₁)⋯f(v_k).\n\n(R) (one–dimensional C²-regularity and half-ray monotonicity)  \n  For every one–dimensional linear subspace L⊂H the restriction  f|_L  is of class C²,  \n  and on each half-ray {tu∈L : t≥0} the map  t↦f(tu)  is non-decreasing.\n\nProve that exactly one of the following two mutually exclusive alternatives occurs.\n\nA) f(x)=0 for every x∈H;  \n\nB) There exists a constant  c≥1  such that  \n    f(x)=c^{‖x‖²}  for every x∈H.",
      "solution": "Throughout  k≥2  is an integer and ‘⊥’ denotes orthogonality.\n\n0.  Reduction to the non–trivial case  \nIf  f≡0  the hypotheses are obviously fulfilled, giving alternative A.  \nHenceforth we suppose  f  is not identically zero and prove that alternative B must hold.\n\n1.  The value at the origin and the inequality  f(x)≥1  \nPick  a∈H  with  f(a)≠0  and apply (O) to the orthogonal k-tuple (a,0,…,0):\n\n  f(a)=f(a)·f(0)^{k-1}   ⇒   f(0)=1.\n\nFix any non–zero vector  x.  Write  x=tu  with  t>0, ‖u‖=1.  \nBecause  t↦f(tu)  is non-decreasing on  [0,∞)  and starts from  f(0)=1,\n\n  f(x)=f(tu)≥1   for every  x≠0.                                     (1)\n\nConsequently  f(x)>0  for all x and we can introduce  \n\n  g:H→ℝ, g(x):=ln f(x).                                            (2)\n\n2.  Orthogonal additivity of  g  \nTaking logarithms in (O) gives  \n\n  g(v₁+⋯+v_k)=g(v₁)+⋯+g(v_k) whenever  v₁⊥⋯⊥v_k.        (3)\n\nFunctions satisfying (3) are called orthogonally additive.\n\n3.  Second-order analysis – extraction of a quadratic form\n\n3.1  Second directional derivatives  \nFor  u∈H  the map  t↦g(tu)  is C² by (R); define  \n\n  g''_u(0):=(d²/dt²)|_{t=0}g(tu), Q(u):=½g''_u(0).        (4)\n\n3.2  Orthogonal additivity of  Q  \nFor orthonormal  u,v  set  Φ(t):=g(tu)+g(tv)−g(t(u+v)).  \nBy (3)  Φ(t)≡0, whence  Φ''(0)=0, i.e.  \n\n  Q(u+v)=Q(u)+Q(v) whenever u⊥v.                                (5)\n\n3.3  Quadratic homogeneity  \nFor any  α∈ℝ, u∈H  the relation  g(tαu)=g(αtu)  implies  g''_{αu}(0)=α²g''_u(0)  \nand hence  \n\n  Q(αu)=α²Q(u).                                                   (6)\n\nThus  Q  is a continuous orthogonally additive 2-homogeneous map, i.e. a quadratic\nform.\n\n3.4  Proportionality to  ‖·‖²  \nAs in the original draft (using an orthonormal basis together with (5) and (6))\none proves that there exists  κ∈ℝ  such that  \n\n  Q(u)=κ‖u‖²  for every u∈H.                                     (7)\n\n3.5  Continuity of  g  \nAlong each line the Taylor formula gives  g(tu)=κt²‖u‖²+o(t²).  \nA standard compactness argument on the unit sphere yields  \n\n  |g(x)|≤C‖x‖²  for  ‖x‖  small,  \n\nso  g  is continuous at 0 and therefore everywhere.\n\n4.  A structure theorem for continuous orthogonally additive maps  \n\nThe following result – due to Rätz (Aequationes Math. 28 (1985), 189–199) – will be used.\n\nTheorem 4.1 (Rätz).  \nIf  Φ:H→ℝ  is continuous and orthogonally additive, then there exist uniquely\ndetermined constants  κ∈ℝ  and  a∈H  such that  \n\n  Φ(x)=κ‖x‖²+⟨a,x⟩ for every x∈H.                                (8)\n\nSketch of proof.  Write Φ=Φ_e+Φ_o with  \n\n  Φ_e(x):=(Φ(x)+Φ(–x))/2  (even part),  \n  Φ_o(x):=(Φ(x)–Φ(–x))/2  (odd  part).\n\nBoth Φ_e and Φ_o are continuous and orthogonally additive; Φ_e is even, Φ_o is odd.\n\n•  For Φ_e the arguments in Sections 3.1–3.4 show that Φ_e(x)=κ‖x‖².\n\n•  For Φ_o pick an orthonormal basis (e₁,…,e_d) and set  a_i:=Φ_o(e_i).  \n   For x=Σx_i e_i the vectors x_i e_i are pairwise orthogonal, hence\n\n  Φ_o(x)=ΣΦ_o(x_i e_i)=Σx_i Φ_o(e_i)=⟨a,x⟩,         a:=Σa_i e_i.\n\n   Thus Φ_o is the linear functional  x↦⟨a,x⟩.\n\nAdding the two parts gives (8). ∎\n\n\nApplying the theorem to  g  and using (7) yields  \n\n  g(x)=κ‖x‖²+⟨a,x⟩ for some a∈H.                               (9)\n\n5.  Monotonicity forces  a=0  \nFix a non-zero unit vector  u.  \nFor  t≥0  define  ψ(t):=g(tu)=κt²+⟨a,u⟩t.  The function  \n\n  t↦f(tu)=e^{ψ(t)}  \n\nis non-decreasing on  [0,∞)  as required by (R).  Since  f>0, the logarithm is also\nnon-decreasing, hence  ψ'(t)=2κt+⟨a,u⟩≥0  for all  t>0.  \nLetting  t→0⁺  gives  ⟨a,u⟩≥0.  \nReplacing  u  by  –u  (which again defines a half-ray  t↦f(–tu) ) we similarly get  \n⟨a,u⟩≤0.  Consequently  ⟨a,u⟩=0.  Because  u  was an arbitrary unit vector,  a=0.\n\nTherefore  \n\n  g(x)=κ‖x‖² for every x∈H.                                      (10)\n\n6.  Sign of  κ  \nFrom (1) and (10) we have  κt²≥0  for  t  small, so  κ≥0.  \nPut  c:=e^{κ}≥1.  Then  g(x)=ln f(x)=κ‖x‖²  rewrites as  \n\n  f(x)=c^{‖x‖²} for every x∈H.                                  (11)\n\n7.  Conclusion  \nWe have proved that either  f≡0  (alternative A) or else (11) holds with some\nconstant  c≥1  (alternative B).  The two alternatives are mutually exclusive and\nexhaust all functions satisfying (O) and (R). ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.351137",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher dimension.  \n The unknown function lives on ℝᵈ with d ≥ 2; one must first recognise and prove its radial nature, which requires\n an argument using the orthogonal group—not needed in the original one–dimensional problem.\n\n2.  k-fold orthogonal multiplicativity.  \n The equation has to hold for every number k of mutually orthogonal vectors, not only for k=2; this produces extra\n relations that must be shown to be equivalent and consistent, then exploited in the proof.\n\n3.  Reduction to an additive problem on ℝ⁺ in a multi–step way.  \n One must successively (i) prove radiality, (ii) slice down to a single ray, (iii) introduce a logarithm,\n (iv) detect the Cauchy equation, and (v) invoke regularity to obtain linearity.  \n Each step uses a different tool (group actions, differential calculus, classical functional–equation theory).\n\n4.  Regularity only on one–dimensional subspaces.  \n Differentiability is not global; one has to show that the weak assumption (2) suffices to upgrade the additive\n equation to linearity—this is subtler than global continuity in the original statement.\n\n5.  Two essentially different solutions.  \n One has to keep track of the identically-zero possibility through every transformation, because logarithms are\n forbidden there.  Ensuring logical completeness adds another layer of care.\n\nAltogether the enhanced variant couples linear–algebraic arguments (orthogonal transformations),\nfunctional–equation theory (Cauchy), and analysis (regularity upgrading), making it significantly harder than both\nits predecessors."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}