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


*

*Show that $f$ is differentiable

*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. 
 A: 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.
A: 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.
