Prove the following: $\int_{a}^{b}|f\left(t\right)|dt\leq\left(b-a\right)\int_{a}^{b}|f'\left(t\right)|dt$

Question

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:

$$\int_{a}^{b}|f\left(t\right)|dt\leq\left(b-a\right)\int_{a}^{b}|f'\left(t\right)|dt$$

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!

Answer 1

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 fundamental theorem of calculus for all $t$, we have $$ f(t)=f(c)+\int_c^t f'(x)dx= \frac{1}{b-a}\int_a^b f(x)dx+\int_c^t f'(x)dx$$ taking the supremum implies

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