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Let $f$ be a continuous function and $f(x)=\frac{2}{t^2}\int^t_0 f(x+s)sds$, $\forall t>0.$

  1. Show that $f$ is differentiable
  2. Show that $f$ is a constant

Solution:

We can observe that $f(x)=h(t)$.

$f'(x)=h_x'(t)=0$

$f'(x)=h_t'(t)=-\frac{4}{t^3}\int^t_0 f(x+s)sds+\frac{2}{t^2}f(x+t)t$

Then

$-\frac{4}{t^3}\int^t_0 f(x+s)sds+\frac{2}{t^2}f(x+t)t=0$

$\Leftrightarrow \frac{2}{t^2}\int_0^t f(x+s)sds=f(x+t)$

$\Leftrightarrow f(x)=f(x+t), \forall t>0 $

From here, we can obtain $f$ is a constant. This implies $(1)$.

But I am not sure about $f'_x(x)=f'_t(x)$. Thanks for your help.

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  • $\begingroup$ What is $h(t)$ ? To prove $f$ is differentiable, utilize definition of differentiability on $x$ and $f$ is bounded on closed interval $[0,t]$ due to continuity of $f$. $\endgroup$
    – Belive
    Oct 14, 2017 at 6:25
  • $\begingroup$ The right-hand side would be a function that depends on $t$, let say it $h(t)$. $\endgroup$
    – Ross
    Oct 14, 2017 at 6:29
  • $\begingroup$ @Belive Why $f$ is bounded on a closed interval? I can not get it $\endgroup$
    – Ross
    Oct 14, 2017 at 6:31
  • $\begingroup$ en.wikipedia.org/wiki/Extreme_value_theorem , you can find the boundedness theorem in the first few lines. $h$ you suggested is independent of t, consider a fixed $x_0$, then Left hand side is $f(x_0)$, while right hand is side equals to $f(x_0)$, for all positive $t$. $\endgroup$
    – Belive
    Oct 14, 2017 at 6:39

2 Answers 2

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While your point is taken, I think some rephrasing will do this the world of good.

Fix $x$. Define $h(t) = \frac 2{t^2} \int_{0}^t f(x+s)sds$. By definition, we know that $h(t)$ is a constant, for all $t > 0$. Furthermore, since $h$ is the product of two differentiable functions by the fundamental theorem of calculus, it is differentiable, and $h'(t) = \frac{-4}{t^3}\int_{0}^t f(x+s)sds + \frac{2}{t^2}f(x+t)t = 0 \forall t$.

This, upon simplification, gives $0 = \frac{2}{t^2}f(x+t)t - \frac{2f(x)}{t}$ for all $t$, which simplifies to $f(x+t) = f(x)$ for all $t$, giving that $f$ is a constant.

Nowhere did we use the fact that $f$ is differentiable. But if we fix $t$, say $t = 1$, then $f(x) = 2\int_{0}^1 f(x+s)sds$, and hence $f$ is defined as the integral of a continuous function $g(s) = f(x+s)s$, and is therefore differentiable. Infact repeating this argument will tell you that $f$ is infinitely differentiable, which is obvious once you figure out it's a constant.

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    $\begingroup$ The differentiability of $f$ with respect to $x$ is not assumed, so should it be continuity of $f$ in the 4th paragraph instead? $\endgroup$
    – Belive
    Oct 14, 2017 at 6:50
  • $\begingroup$ The continuity of $f$ implies the integrability of $g(s) = f(x+s)s$, and note that $h = \frac{2}{t^2} \int_{0}^t g ds$, so the continuity of $f$ was used, not the differentiability. $\endgroup$ Oct 14, 2017 at 6:53
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    $\begingroup$ you mentioned that " $f$ is differentiable "is a fact, so I am wondering should it be " $f$ is continuous " , btw, nice use of differentiating $h$ $\endgroup$
    – Belive
    Oct 14, 2017 at 6:57
  • $\begingroup$ Thank you! Actually, I should have answered it in the reverse order, but I realized that one can prove that $f$ is a constant without needing it's differentiability at all, so that's why I reversed the answers. $\endgroup$ Oct 14, 2017 at 7:19
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    $\begingroup$ I think proofing $h$ is sufficient for the original question, as constant function is always differentiable. $\endgroup$
    – Belive
    Oct 14, 2017 at 7:56
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Since $\int_0^t s \, ds = t^2/2$, the assumption can be rewritten as $$ \frac{2}{t^2} \int_0^t [f(x+s) - f(x)] s\, ds = 0 \qquad \forall t > 0, $$ i.e. $$ \int_0^t [f(x+s) - f(x)] s\, ds = 0 \qquad \forall t > 0. $$ Since the integrand is a continuous function, by the FTC we have that $$ [f(x+t) - f(x)] t = 0 \qquad \forall t > 0, $$ i.e. $$ f(x+t) - f(x) = 0 \qquad \forall t > 0. $$ Hence $f$ in constant on $[x, +\infty)$ for every $x\in\mathbb{R}$, so that it is a constant function.

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