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Functional equations such as this one appear only once every several years on exams, so I feel it's hard to have a sure-fire way to approach the problem, unlike, say, solving a series convergence problem, multiple variable integration, or proving some results using basic Fourier series.

So, when I do see a solution offered for one of these problems and study the solution for a substantial amount of time, I still cannot remember how to solve these types of problems, when I come across another one.

But the question is:

Find all the real-valued continuous functions $f$ on $\mathbb R$ which satisfy $$f(x)f(y)=f(x_1)f(y_1)$$

for all $x$, $y$, $x_1$, $y_1$ such that $x^2+y^2=x_1^2+y_1^2$.

Ideally, besides offering a solution, I would love to hear about your intuition on how to solve these functional equations.

Thanks,

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    $\begingroup$ +1. I wanna know how the experts solve functional equations too.... $\endgroup$ – Jack's wasted life Jun 26 '15 at 22:56
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You can linearize the problem by introducing $$g(u):=\ln\left(f(\sqrt u)\right).$$

Then with $u=x^2,v=y^2$, $$u+v=u'+v'\implies g(u)+g(v)=g(u')+g(v').$$

Setting $u'=0,v'=u+v$,

$$g(u)+g(v)=g(0)+g(u+v)$$

shows that the function must be affine,

$$g(u)=au+b,$$ and $$f(x)=e^{ax^2+b}=F_0\left(\frac{F_1}{F_0}\right)^{x^2}.$$


The intuitions/tricks behind this:

  • it is often advantageous to linearize to benefit of what we know from linear algebra and make the equations look more familiar;
  • products can be linearized by means of logarithms;
  • non-linear functions can be linearized by means of a change of variable with the function inverse;
  • when you have a property involving several variables, try to exploit it by assigning particular values to some of them.
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  • $\begingroup$ Hi @YvesDaoust - I really like this approach. I just have one follow-up question: how do you get the line g(u) = au+b, hence showing the function, g, is affine? $\endgroup$ – User001 Jun 29 '15 at 2:03
  • $\begingroup$ From your previous line, it follows that g(u) = g(0) + g(u+v) - g(v)... $\endgroup$ – User001 Jun 29 '15 at 2:06
  • $\begingroup$ @LebronJames: let $h(u)=g(u)-g(0)$, then $h(u)+h(v)=h(u+v)$. Assuming $h$ continuous, this is enough to say that $h$ is linear. $\endgroup$ – Yves Daoust Jun 29 '15 at 6:19
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All solutions are the functions $f(x) = \alpha e^{\beta x^2}$, $\alpha,\beta \in \mathbb{R}$. Any of this functions satisfies the OP query: $$f(x)f(y) = \alpha e^{\beta x^2} \alpha e^{\beta y^2} = \alpha^2 e^{\beta(x^2 + y^2)} = \alpha^2 e^{\beta(x_1^2 + y_1^2)} = \alpha e^{\beta x_1^2} \alpha e^{\beta y_1^2} = f(x_1)f(y_1) \, .$$

Here is why these are the only ones. Observe that the hypothesis implies that there is a function $\psi$ such that $$f(x)f(y) = \psi(x^2 + y^2) \, .$$ If $f(0) = 0$ then $0.f(y) = \psi(y^2) = 0$ hence $f(x).f(y) \equiv 0$. So $f$ must be identically zero. Assume that $f(0) = \alpha \neq 0$. Then $\tilde{f}(x) := \frac{f(x)}{\alpha}$ also satisfies the OP hypotesis. So we can assume w.l.o.g. that $\alpha = 1$. From this we get that $f(x) = \psi(x^2)$ and $\psi(r)$ is a continuous function for $r \geq 0$. Moreover the function $\psi$ satisfies $$f(x) f(y) = \psi(x^2) \psi(y^2) = \psi(x^2 + y^2) \, .$$ By taking $x=y$ we see that $\psi \geq 0$. Actually, $\psi(x) > 0$. Indeed, if $\psi(x_0^2) = 0$ then $\psi(x_0^2 + r) = 0$ for $r \geq 0$. W.l.o.g. we can assume $x_0>0$. Observe that also $f(x_0)=0$. So there are values $y_0$ such that $f(y_0) = 0$ and $0 \leq y_0 < x_0$. But then also $\psi(y_0^2) = 0$. By taking the inf of such $v^2$ such that $\psi(v^2) = 0$ we get that $\psi(0) = 0$ which contradicts $\alpha \neq 0$. Finally, we can take logarithms. Namely, we define the function $\lambda(x) := \log(\psi(x))$, for $x \geq 0$. Then $$\lambda(x) + \lambda(y) = \lambda(x+y)$$ for all $x,y \geq 0$. Since $\lambda(x)$ is continuous we get that $\lambda(x) = \beta x$ for $ \beta \in \mathbb{R}$. Then $\psi(x^2) = e^{\beta x^2}$. So $f(x) = e^{\beta x^2}$. But we had assumed that $f(0)=1$. So the general solution is as I claimed : $f(x) = \alpha e^{\beta x^2}$.

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    $\begingroup$ why are we able to assert that $f(x)f(y) = \psi(x^2 + y^2)$? $\endgroup$ – Matematleta Jun 27 '15 at 1:30
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    $\begingroup$ We know that $f(x)f(y) = f(\sqrt{x^2+y^2})f(0)$ for all $x,y$. So, define $\psi(r) = f(\sqrt{r})f(0)$ for $r \ge 0$. $\endgroup$ – JimmyK4542 Jun 27 '15 at 7:11
  • $\begingroup$ @Chilango. $f(x)f(y)$ has the same value for all $x,y$ such that $x^2 + y^2$ have a fix value. Then $f(x)f(y)$ is a function of $x^2 + y^2$. $\endgroup$ – Holonomia Jun 27 '15 at 8:05
  • $\begingroup$ Thanks so much @Holonomia. $\endgroup$ – User001 Jun 29 '15 at 2:08
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    $\begingroup$ Just a last comment: In the solution you accepted by Daoust it is not justified why $f(x) > 0$. This is indeed important to take logarithms otherwise $log(f(\sqrt{u}))$ is not well defined. $\endgroup$ – Holonomia Jun 29 '15 at 5:33
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The aim of the following is to address the intuition side of the question - I doubt that I have more experience than anybody else, and the following is certainly not rigorous - still... One way to the answer - at least in this case! - is to use calculus, i.e., assume everything in sight is differentiable, and perhaps try "to sweep up the loose ends" afterwards.

As pointed out by Holonomia, the function $g(x,y) = f(x)f(y)$ is constant on circles. This means that the gradient of the differentiable $g$ is parallel to the vector $(2x,2y)$, as the latter is normal to $x^2 +y^2 = c$. So $$ {\rm grad}\ g = \lambda\cdot (2x, 2y), $$ where where $\lambda = \lambda(x,y)$ is a scalar function. Comparing the components, one gets $$f'(x) f(y) = \lambda 2 x,$$ and $$f'(y) f(x) = \lambda 2 y.$$

Doing the algebra (formally), one obtains $$ {f'(x) \over 2x f(x)} = {f'(y) \over 2y f(y)}.$$ Therefore, both sides of the equality are constant, i.e., $$ {f'(x) \over 2x f(x)} = \beta,$$ with $\beta$ some constant. Cross-multiplying by $2x$ and integrating, one ends up with $$ f(x) = \alpha e^{\beta x^2},$$ for some constant $\alpha$ - i.e., Holonomia's answer.

  • Edit * Some "sweeping up," by request, to show that any $f$ satisfying the conditions of the problem (continuity, functional equation) is differentiable at $x=c$, for every $c$. Fix $c$, set $a = |c|$, and consider $$ f(x) \int_{a+10}^{a+20} f(y)\, dy = \int_{a+10}^{a+20} f(x) f(y) \,dy = \int_{a+10}^{a+20} \psi( x^2 +y^2) \, dy,$$ where $\psi$ is Holonomia's $\psi$. The integral multiplying $f(x)$ is not zero if $f$ is not identically zero (using a Holonomia-style argument and $f(\sqrt 2 x)f(0) = f(x)^2 $, for instance, to conclude that the continuous $f$ is nowhere $0$ if not identically $0$). With the change of variables $y= \sqrt{r^2-x^2}$, the integral on the right becomes $$ \int_{\sqrt {(a+10)^2 +x^2}}^{\sqrt {(a+20)^2 +x^2}} \psi ( r^2) {r\over \sqrt{r^2-x^2} }\, dr,$$ which is differentiable at (in a neighborhood of) $x=c$, because $r^2-x^2 \ge (a+10)^2 >0$, $\psi$ is continuous, and the limits of the integral are differentiable. Thus $f(x)$ is differentiable.
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  • $\begingroup$ Hi @peterag, how do we then back out of the differentiability assumption? Really cool intuition... $\endgroup$ – User001 Jun 29 '15 at 1:44
  • $\begingroup$ @LebronJames - actually, I just rolled back the edit to address your comment, as I messed something up - Tomorrow... $\endgroup$ – peter a g Jun 29 '15 at 3:47
  • $\begingroup$ Ok, got it. Thanks @peterag! :-) $\endgroup$ – User001 Jun 29 '15 at 3:52
  • $\begingroup$ @LebronJames see the 'edit'. However, as advertised in the first line of this answer, the point really was not 'rigor' - although the first version of this when you posed the question actually had a <problematic> version of this 'edit', but it seemed pointless given Holonomia's answer. Be that as it may, the argument in the edit is based on the standard proof that a measurable character on the reals is continuous (and differentiable). $\endgroup$ – peter a g Jun 29 '15 at 21:24

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