Let \(f:\mathbb R\to\mathbb R\) be a nonzero \(C^\infty\) function satisfying \[ f(x)f(y)=f\!\left(\sqrt{x^2+y^2}\right) \] for all \(x,y\in\mathbb R\), and suppose \[ f(x)\longrightarrow0 \qquad\text{as }|x|\to\infty. \]
Prove that \(f\) is even and \(f(0)=1\).
Prove that \[ f'(x)=f''(0)\,x f(x), \] and find the most general function satisfying the given conditions.
Solution
1. One has\[ f(0)\in\{0,1\}. \]
Set \(x=y=0\) in the functional equation: \[ f(0)^2=f(0). \] Thus \(f(0)\) is an idempotent real number, hence is \(0\) or \(1\).
2. The case\[ f(0)=0 \]would force\[ f\equiv0. \]
If \(f(0)=0\), set \(y=0\) in the functional equation. For every real \(x\), \[ 0 = f(x)f(0) = f(\abs{x}), \] so \(f(t)=0\) for every \(t\geq0\).
Now for arbitrary \(x\in\RR\), set \(y=x\). Then \[ f(x)^2 = f(\sqrt2\,\abs{x}). \] The argument on the right is nonnegative, so the right side is \(0\). Therefore \(f(x)=0\) for every \(x\), contradicting the assumption that \(f\) is nonzero.
3. One has\[ \boxed{f(0)=1} \]and \(f\) is even.
Steps 1 and 2 rule out \(f(0)=0\), so \(f(0)=1\). Setting \(y=0\) in the functional equation now gives \[ f(x) = f(x)f(0) = f(\abs{x}). \] Hence \[ f(-x)=f(x) \] for every \(x\), so \(f\) is even.
4. Fix \(x>0\) and define\[ r(y)\coloneqq\sqrt{x^2+y^2}. \]Then\[ r'(0)=0 \qquad\text{and}\qquad r''(0)=\frac1x. \]
Direct differentiation gives \[ r'(y) = \frac{y}{\sqrt{x^2+y^2}}, \] so \(r'(0)=0\). Differentiating once more, \[ r''(y) = \frac{x^2}{(x^2+y^2)^{3/2}}, \] and therefore \[ r''(0)=\frac{x^2}{x^3}=\frac1x. \]
5. For every \(x>0\),\[ f'(x)=f''(0)\,x f(x). \]
For fixed \(x>0\), the functional equation reads \[ f(x)f(y)=f(r(y)). \] Differentiate twice with respect to \(y\). At \(y=0\), the left side becomes \[ f(x)f''(0). \] For the right side, the chain rule gives \[ \frac{d^2}{dy^2}f(r(y)) = f''(r(y))(r'(y))^2 +f'(r(y))r''(y). \] Using step 4 at \(y=0\) gives \[ f(x)f''(0) = \frac{f'(x)}x. \] Multiplying by \(x\) gives the claim.
6. For every real \(x\),\[ \boxed{ f'(x)=f''(0)\,x f(x) }. \]
Step 5 proves the identity for \(x>0\). Since \(f\) is even by step 3, \(f'\) is odd. Both sides of the displayed identity are therefore odd functions of \(x\), so the identity extends to \(x<0\). At \(x=0\), evenness gives \(f'(0)=0\), while the right side also vanishes.
7. Set\[ c\coloneqq f''(0). \]Then\[ f(x) = \exp\left(\frac c2x^2\right) \]for every real \(x\).
The differential equation in step 6 is \[ f'=cxf. \] The function \[ F(x)\coloneqq \exp\left(-\frac c2x^2\right)f(x) \] satisfies \[ F'(x) = \exp\left(-\frac c2x^2\right) \bigl(f'(x)-cxf(x)\bigr) =0. \] Thus \(F\) is constant. By step 3, \[ F(0)=f(0)=1, \] so \(F\equiv1\), which gives the displayed formula.
8. The decay hypothesis is equivalent to\[ c<0. \]
If \(c<0\), then \[ \exp\left(\frac c2x^2\right)\longrightarrow0 \] as \(\abs{x}\to\infty\). If \(c=0\), the function is identically \(1\). If \(c>0\), the function tends to \(+\infty\). Hence decay occurs exactly when \(c<0\).
9. The most general function satisfying all the hypotheses is\[ \boxed{ f(x)=e^{-a x^2}, \qquad a>0 }. \]
By steps 7 and 8, write \[ a=-\frac c2>0. \] This gives every possible solution. Conversely, if \(a>0\), then \[ f(x)=e^{-ax^2} \] is nonzero and smooth, tends to zero at infinity, and satisfies \[ f(x)f(y) = e^{-a(x^2+y^2)} = f\left(\sqrt{x^2+y^2}\right). \]
10. Q.E.D.
Step 3 proves part 1, step 6 proves the requested differential identity, and step 9 gives the complete classification.