Let $f:[0,1]\to \mathbb{R}$ be continuously differentiable, and $f(0)=0$. Prove that $$\sup_{0\le x \le 1}|f(x)| \le \sqrt{\int_0^1 (f'(x))^2dx}$$
Can you please verify my proof below? I'm especially interested in seeing if it can be simplified, or the claim proved more directly; my proof looks clunky to me. Thanks!
Proof. Let $S$ be the supremum on the left side. We aim to prove that $$S^2 = (\sup_{0\le x \le 1}|f(x)|)^2 \le \int_0^1 (f'(x))^2dx$$
Since $[0,1]$ is compact, $S$ is realized at some point $x_S$: $f(x_S)=S$. We want to show that we can assume $x_S=1$, and so restrict ourselves to the case when $f(1)=S$. One way to see this is to redefine $f(x)=S$ for all $x\ge x_s$. This leaves the left-hand side of the inequality without change, and right-hand side can only decrease or stay the same since $f'(x)=0$ for $x\ge x_S$ now. Therefore the new inequality implies the original one, so we can restrict ourselves to the case $f(1)=S$. However, this leaves $f'$ with a possible discontinuity at $x_S$. The rest of the proof works fine with $f'$ piecewise continuous.
Alternatively, if we want to leave $f'$ continuous, we can replace the upper limit $1$ in the supremum and the integral by $x_S$, and add another $x_S$ factor on the right: $$(\sup_{0\le x \le x_S}|f(x)|)^2 \le x_S\int_0^{x_S} (f'(x))^2dx$$ As before, this inequality clearly implies the original one; if we now change variables $y=x_S x$, we get back the upper limit $1$, the new factor $x_S$ cancels out and the value at $y=1$ is $S$ as desired.
Therefore we assume now that $f(1)=S$, and we recall that $f(0)=0$ is given.
Next, we consider a Riemann sum for the right-hand side integral by choosing a tagged partition $x_0=0<x_1<x_2<\ldots<x_n=1$, $\delta_i=x_{i+1}-x_i$, $t_i\in(x_i,x_{i+1})$. We use the Mean Value Theorem to choose each $t_i$, which turns the sum into $$\sum_{0\le i \le {n-1}} f'(t_i)^2\delta_i =\sum_{0\le i \le {n-1}} \left(\frac{f(x_{i+1})-f(x_i)}{\delta_i}\right)^2\delta_i=\sum_{0\le i \le {n-1}} \frac{(f(x_{i+1})-f(x_i))^2}{\delta_i}$$ If we now prove that $S^2$ is less or equal to this sum, the main claim will follow, because the integral is the limit of such sums.
We simplify this sum by repeatedly removing some inner point $x_i$: we remove the two terms that use $f(x_i)$ from the sum and replace them by a new term that connects $f(x_{i-1})$ and $f(x_{i+1})$ over the combined interval. Let's prove that doing that will not increase the sum, that is the loss is greater or equal to the gain: $$\frac{(f(x_{i})-f(x_{i-1}))^2}{\delta_{i-1}} + \frac{(f(x_{i+1})-f(x_i))^2}{\delta_i} \ge \frac{(f(x_{i+1})-f(x_{i-1}))^2}{\delta_{i-1}+\delta_{i}}$$ After renaming, this becomes the inequality $$\frac{A^2}{\delta_1}+\frac{B^2}{\delta_2} \ge \frac{(A+B)^2}{\delta_1+\delta_2}$$. By scaling the equation we can assume $\delta_1=1$ and simplify to $$A^2+\frac{B^2}{\delta} \ge \frac{{A+B}^2}{\delta+1} $$This becomes $(\delta+1)(\delta A^2+B^2) \ge \delta(A^2+B^2)$, and after cancelling common terms the whole thing turns out to be $(\delta A - B)^2 \ge 0$, Q.E.D.
After removing all inner points, we are left with just one term, and by what we just proved it's not larger than the original Riemann sum: $$\frac{(f(1)-f(0))^2}{1} \le \sum_{0\le i \le {n-1}} \frac{(f(x_{i+1})-f(x_i))^2}{\delta_i}$$ As $f(0)=0$, and we showed we can assume $f(1)=S$, the left-hand side is $S^2$, which is indeed less than or equal to the Riemann sum, and so to the integral, Q.E.D.
Since you are interested in a simpler proof I will provide one: $f(x)=\int_0^{x} f'(t)\, dt$ so $|f(x)| \leq \int_0^{x} |f'(t)|\, dt \leq \int_0^{1} |f'(t)|\, dt\leq \sqrt {\int_0^{1} |f'(t)|^{2}\, dt}$ by Cauchy - Schwarz inequality. Square this and take supremum.