How to show that $\int_0^1 f(x)^2 dx \le \int_0^1 f'(x)^2 dx$ if $f(0) = f(1) = 0$? I am working in $$H_0^1([0, 1]) = \{u:[0, 1] \to \mathbb{R}| \int_0^1 u'(x)^2 dx < \infty, \text{ and }u(0) = u(1) = 0 \}$$ with inner product $\langle u, v \rangle = \int_0^1 u'(x) v'(x) dx.$ I am trying to show that for $f \in H_0^1,$ $||f||_{\mathcal{L}^2} \le ||f||_{H_0^1}$, which comes out to
$$\int_0^1 f(x)^2 \le \int_0^1 f'(x)^2 dx.$$
Clearly the $f(0) = f(1) = 0$ hypothesis is crucial here ($f(x) = 1$ violates the inequality for example) but I am not sure how to use it. I tried to play around with integration by parts, and with Fubini's Theorem (by writing $f(x) = \int_0^x f'(t) dt$) but no luck.
I believe it's true that for $x \in [0, 1]$ we have $f(x)^2 \le \int_0^1 f'(t)^2 dt$, and if I could prove this, we would be done.

Using Holder inequality we can similarly
prove $\int_0^1 f^p dx\leq \int_0^1 (f'(x))^pdx$ for any $p>1$.
