If $f$ is an infinitely differentiable function then does this hold?

Solution 1:

We can assume without loss of generality that $f(1)=0$. Now let $g(x) = x f(x)$. We have $g(0)=g(1)=0$. Hence there is a point $c$ in $(0,1)$ where $g'(c) = 0$. It follows that $c f'(c) + f(c) = 0$, QED.

More generally if $\alpha>0$ and we choose $g(x)=x^\alpha (f(x)-f(1))$, the same argument applies, we have $g(0)=g(1)=0$ and there is a point $c\in (0, 1)$ such that $\alpha c^{\alpha-1}(f(c)-f(1))+ c^\alpha f'(c)=0$, i.e. \begin{equation} f'(c) = \alpha \frac{f(1)-f(c)}{c} \end{equation}

Solution 2:

Counter-example $f(x)=\frac 1x$ when $f$ is only $C^{\infty}$ on $(0,1)$.

If $f$ is also continuous on $[0,1]$ then this is a trivial application of MVT to the function $xf(x)$.