Upper bound for $\Vert f \Vert^{2}$, where $f: [0,1] \to \mathbb{R}$ continuously differentiable. Let $f: [0,1] \to \mathbb{R}$ be continuously differentiable with $f(0)=0$. Prove that $$\Vert f \Vert^{2} \leq \int_{0}^{1} (f'(x))^{2}dx$$
Here $\Vert f \Vert$ is given by $\sup\{|f(t)|: t \in [0,1]\}$.
I'm just a bit unclear how to proceed.

Let x = argsup f. Because f(0) = 0, f(x) is the integral up to x of f', which is less than or equal to the L2 norm of f' up to x by convexity of the squaring function, which is less than or equal to the L2 norm of f' on the entire interval. Now square both sides.
