Let $f:[0,1]\rightarrow \mathbb{R} $ be a continuous function so that $\int_{0}^{1}f(x)\cdot g(x)dx=\int_{0}^{1}f(x)dx\cdot \int_{0}^{1}g(x)dx$, for any $g:[0,1]\rightarrow \mathbb{R}$, continuous and not differentiable.
Prove that $f$ is constant.

I have no idea on what to do.

  • $\begingroup$ so $f$ could or could not be differentiable? $\endgroup$ – tired Mar 2 '17 at 20:21
  • $\begingroup$ Yes, I think so. $\endgroup$ – ztefelina Mar 2 '17 at 20:24
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    $\begingroup$ "Not derivable" - I assume you mean not differentiable. Everywhere non-differentiable, or not everywhere differentiable? $\endgroup$ – Robert Israel Mar 2 '17 at 20:43
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    $\begingroup$ In any case, non-differentiable functions are uniformly dense in the continuous functions on $[0,1]$, so by continuity the equation will be true for all continuous $g$. $\endgroup$ – Robert Israel Mar 2 '17 at 20:48

As @RobertIsrael pointed out, the non-differentiable continuous functions are dense in $C[0,1]$ with respect to the uniform norm, and therefore

$$\int_0^1 f(x) g(x) \, dx = \int_0^1 f(x) \, dx \int_0^1 g(x) \, dx$$

holds for any continuous function $g$. If we choose $f=g$, then we find

$$\int_0^1 f(x)^2 \, dx = \left( \int_0^1 f(x) \, dx \right)^2.$$

This means that

$$\int_0^1 ( f(x) - a)^2 \, dx = 0$$ for $a:= \int_0^1 f(y) \, dy$. Since the integrand $x \mapsto (f(x)-a)^2$ is continuous and non-negative, this implies

$$f(x)-a=0 \qquad \text{for all $x \in [0,1]$},$$


$$f(x) = a = \int_0^1 f(y) \, dy \qquad \text{for all $x \in [0,1]$}.$$

Remark: More generally, if $f$ is a continuous function and $V$ strictly convex, then $$\int_0^1 V(f(x)) \, dx = V \left( \int_0^1 f(x) \, dx \right)$$ if and only if $f$ is constant, see this question; here, we have proved this statement for $V(x) = x^2$.


Suppose $\{g(x)\}$ is a sequence of functions $g_n(x)=\frac{1}{n\sqrt {2\pi}} e^{-\frac {\mu}{2n^2}x^2}$

The sequence converges to the Dirac delta function $\delta(x-\mu)$

$\int_0^1 f(x)g(x) = f(\mu)$ if $\mu \in [0,1]$

$\int_0^1 f(x)\int_0^1 g(x) = \int_0^1 f(x) = f(\mu)$

and if we consider all $g.$

$f(\mu) = \int_0^1 f(x)$ for all $\mu \in [0,1]$

$f(\mu)$ is constant over the interval.

  • $\begingroup$ this is sloppy. those integrals are not equal to $1$ and they do not converge to the delta function you indicate. $\endgroup$ – zhw. Mar 2 '17 at 21:11
  • $\begingroup$ You intend $e^{-\frac{(x-\mu)^2}{2n^2}}$. It also might be good to mention the idea here, which is essentially to estimate $f(y)$ by an integral against an appropriate function $g_y$. (You also dropped the whole nondifferentiability thing, though that turned out to be more or less irrelevant.) $\endgroup$ – Ian Mar 2 '17 at 21:14
  • $\begingroup$ @Ian And even then it only works for $\mu\in (0,1).$ $\endgroup$ – zhw. Mar 2 '17 at 21:19
  • $\begingroup$ @zhw. To be fair continuity then gives the right endpoint values. Still, saz's argument is better. $\endgroup$ – Ian Mar 2 '17 at 22:09

Fro the first mean value theorem for definite integrals follows that the exists $c \in (0, 1)$ such that $$\int_{0}^{1}f(x)dx = f(c)$$

With the assumption above

$$\int_{0}^{1}(f(x)-f(c))^2dx = \int_{0}^{1}f^2(x)dx - \int_{0}^{1}f^2(c)dx = (\int_{0}^{1}f(x)dx)^2 - \int_{0}^{1}f^2(c)dx = 0$$

follows $f(x)-f(c) = 0$ on $(0,1)$.


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