Proof of a functional inequality

Question

I am reading the lecture notes "Lecture Notes on the Course “Entropy Methods and Related Functional Inequalities”" by Daniel Matthes.

https://www.asc.tuwien.ac.at/~matthes/lecpavia.pdf

On p.54, the inequality $$ \sup_{\Omega} u\le 1+\int_{\Omega} |u_x|dx $$ is used. It seems this inequality holds for rather general $u$. My question is how to prove it?

Any reference, suggestion, idea, or comment is welcome. Thank you!

Answer 1

The proof is easy in the context of these notes, since $\Omega \subset\mathbb R$, and is similar to the main step in proving the usual Sobolev inequalities,

$$ |u(x)| =\left| u(x_0) + \int_{x_0}^x u_x (y) dy\right| \le |u(x_0)| + \int_\Omega |u_x(y)|dy$$ Now, it is assumed that $\int_\Omega |u| = 1$. Since $|u|$ is presumably continuous, by an integral Mean Value Theorem, this means that there exists $x_0$ such that $|u(x_0)|=1$. Pick that one, then take a supremum over $x$. Toss away the absolute values on $u$, because its also assumed that $u>0$, and we have the result.

From the proof, some smoothness is required, and the following additional assumptions are at least sufficient:

• $u>0$
• dimension is 1
• existence of $x_0$ such that $u(x_0)=1$

Dropping the first assumption means you should have absolute values. (And if you don't, then you might run into trouble by considering $-u$.)

Dropping the last assumption means you should have $u(x_0)$ instead.

Dropping the middle one, you can use a directional derivative instead.