An inequality of integrals Let $f\in C^1([a,b])$ with $f(a)=0$. How can I show that there exists a positive constant $M$ independent of $f$ such that $\int^b_a|f(x)|^2dx\leq M\int^b_a|f^\prime(x)|^2dx$?

For any $x\in[a,b]$, you have
$$
\begin{align*}
|f(x)|^2
&=
\left|\int_a^xf'(t)\,dt\right|^2\\
&\leq
\int_a^x|f'(t)|^2\,dt\int_a^x1\,dt\\
&=
(x-a)\int_a^x|f'(t)|^2\,dt\\
&\leq
(x-a)\int_a^b|f'(t)|^2\,dt,
\end{align*}
$$
where the first inequality is the Cauchy-Schwarz inequality. Now integrate both sides over $[a,b]$ with respect to $x$ to get your desired inequality, with $M=(b-a)^2/2$.
