Let $f\in C^1([0,1],\mathbb{R})$ such that $f(0) = 0$. I am trying to show that $\sup_{x\in[0,1]}|f(x)|\leq \sqrt{\int_0^1(f'(x))^2dx}$. I have a suspicion that we ought to use the mean value theorem for the integral and the standard MVT, for
$$\forall r \in (0, 1]:\exists (c_1, c_2)\in (0, r)\times [0, r]:\left|\frac{f(r)}{r}\right| = |f'(c_1)| \land (f'(c_2))^2 = \frac{1}{r}\int_0^r(f'(x))^2dx$$
since $(f'(x))^2$ is continuous on $[0,1]$. But I don't see how I could have power over the $c_1, c_2$ in the sense that I could bound $|f(x)|$ uniformly.
Best Answer
\begin{eqnarray}|f(x)| & = & |f(x) - f(0)| \\ &=& \left|\int_0^x f^\prime(t)dt\right| \\ &\le& \int_0^x|f^\prime(t)|dt \\ &\le& \int_0^1|f^\prime(t)|dt \\ & = & \int_0^11\cdot|f^\prime(t)|dt \\ &\le& \left(\int_0^1 1^2 dt\right)^{\frac{1}{2}}\left(\int_0^1 f^\prime(t)^2 dt\right)^{\frac{1}{2}} \end{eqnarray} Now take the $\sup$ with respect to $x$.