Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
add comment

1 Answer 1

up vote 5 down vote accepted

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.