# 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 $\int^b_a|f(x)|^2dx\leq M\int^b_a|f^\prime(x)|^2dx$?

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## 1 Answer

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 $M=(b-a)^2/2$.

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