I have this homework question and I'm in need of some assistance:

"Let there be a function $f:\left[a,b\right] \rightarrow\mathbb{R}$ continuously derivatable (every derivative is continuous), and $f(a)=0$. Prove the following:


Hint: Use the fact that $|f|$ is a continuous function, you can also use the following theorems (which were proved in earlier exercises):

1) $$|\int_{a}^{b}f\left(t\right)dt|\leq\int_{a}^{b}|f\left(t\right)|dt $$

2) $$\exists c\epsilon\left[a,b\right]\,\, s.t\,\,\int_{a}^{b}f\left(t\right)dt=f(c)(b-a)$$

I tried to play around with the hints and also with other theorems that I know, but I haven't made too much progress, my main problem is with the absolute values which seem to work against me each time I try something, its gotten to the point where I'm starting to confuse myself over nothing.

Any help is appreciated, Thanks!

  • 2
    $\begingroup$ Use that $|f(t)|=\left|\int_a^t f'(x)dx\right|\leq (b-a)\int_a^b |f'(x)|dx $ $\endgroup$ – dinosaur Apr 29 '14 at 16:31

This is Poincaré-Friedrich's inequality

By mean value theorem,since $f(a)=0$ such that $$|f(t)|=\left|\int_a^t f'(x)dx\right|\leq \int_a^b |f'(x)|dx$$

by integrating against t variable we get, $$\int_a^b|f(t)|dt\leq (b-a)\int_a^b |f'(x)|dx$$ this shows that, $$\|f\|_{L^1(a,b)}\leq (b-a)\|f'\|_{L^1(a,b)}$$

More generally we can drop the condition that $f(a)=0. $ Indeed, By mean value theorem, there exists $c\in(a,b)$ such that $$f(c) = \frac{1}{b-a}\int_a^b f(x)dx$$ By fondamental theorem of calculus for all $t$, we have $$ f(t)=f(c)+\int_c^t f'(x)dx= \frac{1}{b-a}\int_a^b f(x)dx+\int_c^t f'(x)dx\leq \int_a^b |f'(x)|dx$$ taking the supremum we have

$$\|f\|_{L^\infty(a,b)}\leq (b-a)^{-1}\|f\|_{L^1(a,b)}+\|f'\|_{L^1(a,b)}= ((b-a)^{-1}+1)\|f\|_{W^{1,1}(a,b)}$$. that is $$\|f\|_{L^\infty(a,b)}\leq C\|f\|_{W^{1,1}(a,b)}$$.

  • $\begingroup$ There is something wrong with the last equation on the yellow box. Essentially it says that if we drop the condition that $f(a)=0$,then $f(t)\le\int_a^b|f'(x)|dx$, which in general is false. For example if $f(t)=1$, so $f'(t)=0$, this will give $1\le0$. $\endgroup$ – Anders Beta Feb 15 '18 at 14:57
  • $\begingroup$ @AndersBeta that inequality is supposed to there thank I reomved it $\endgroup$ – Guy Fsone Feb 15 '18 at 15:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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