1

In "A First Course in Sobolev Spaces" by Leoni, the proof for Theorem 12.83 when $p=N=1$ is left as an exercise. However, I have no idea how to prove it; can anyone provide some guidance to the proof for this part of the theorem?

The theorem goes as follows (with $p=N=1)$:

Let $q\in[1,\infty]$, $\theta\in[0,1]$ and $r=\frac{q}{\theta}$.

Then there exists a constant $c > 0$ (depending only on $q$ and $\theta$) such that $$ ||u||_{L^r(\mathbb{R})} \le c\, ||u||_{L^q(\mathbb{R})}^\theta ||\nabla u||_{L^1(\mathbb{R})}^{1-\theta} $$ for every $u\in L^q(\mathbb{R})\cap\dot{W}^{1,1}(\mathbb{R})$.

Tham
  • 774
  • 3
  • 9

1 Answers1

1

Suppose ${\|\nabla u\|}_{L^1(\mathbb R)} = \int|u'(x)| \, dx = L$. Since $u \in L^q(\mathbb R)$, there must be a sequence $x_n \to -\infty$ such that $u(x_n) \to 0$. Therefore, $u(x) = \lim_{n\to \infty} \int_{-x_n}^x u'(y) \, dy$. It follows that $|u(x)| \le L$. Hence $$ \int_{\mathbb R} |u(x)|^r \, dx \le L^{r-q} \int_{\mathbb R} |u(x)|^q \, dx ,$$ and taking $r$th roots, the result follows.

Stephen Montgomery-Smith
  • 26,965
  • Thank you for the simple answer. Just wanted to ask, why must there exist a sequence $x_n\to-\infty$ with $u(x_n)\to0$? I read in this article that tails of $L^q$ functions might not necessarily decay to $0$. – Tham Feb 20 '22 at 05:22
  • If there doesn't exist such a sequence, then there exists some $A>0$, $B\in\mathbb R$ such that $|u(x)|>A$ for $x<B$. But then the $L^q$ norm must be infinite. – Stephen Montgomery-Smith Feb 20 '22 at 05:26
  • Got it, thank you so much! I sincerely appreciate it! – Tham Feb 20 '22 at 05:35