0
$\begingroup$

Recall that Fatou's Lemma says that for every sequence $f_n$ of non-negative measurable functions $$\int \liminf_{n\to \infty} f_n \ d\mu\leq \liminf_{n\to \infty} \int f_n\ d\mu \ .$$ If I am not mistaken, it implies, in particular, that for any sequence of smooth functions $f_n : \Omega\subset \mathbb{R}^n \to \mathbb{C}$, which converges to some $f$ in $L^2(\Omega)$, that $$\|f\|_{H^k(\Omega)} \leq \liminf \|f_n\|_{H^k(\Omega)} \ . \tag{$*$}$$ Here $H^k(\Omega)$, $k \in \mathbb{N}_0$, denotes the $L^2$-Sobolev space of functions, all of whose derivatives up to order $k$ belong to $L^2(\Omega)$.

We use $\|f\|_{H^k(\Omega)} = \infty$ if $f \not \in H^k$.

Questions:

1.) Does $(*)$ extend to fractional Sobolev spaces, e.g. $H^{1/2}(\Omega)$?

2.) Is there an abstract principle that whenever a Banach space $X$ is continuously embedded into a Banach space $Y$ and a sequence $y_n \in X$ converges in $Y$ to an element $y$, then $$\|y\|_X \leq \liminf \|y_n\|_X \ .$$

$\endgroup$
2
  • 2
    $\begingroup$ Your point 2 is not the same as above: it follows immediately from the continuity of the norm. $\endgroup$ May 11, 2013 at 21:25
  • 1
    $\begingroup$ Why is the second inequality true? If the sequence converges only in $L^2$, why is the left side necessarily finite? $\endgroup$
    – Deane Yang
    May 16, 2013 at 12:29

1 Answer 1

-1
$\begingroup$

If $X$ is a Banach space and $x_n$ converges WEAKLY to $x$, then $$ \|x\|\leq \liminf \|x_n\|. $$ Moreover, if $X$ is a reflexive Banach space and $x_n$ is a bounded sequence then

a) there exists $x\in X$ such that $x_n$ converges weakly to $x$

and

b) the previous inequality holds.

In particular, the previous result is valid for every $H^s$. Concerning your second point: usually you have a bounded sequence in $H^k(\Omega),$ which for bounded $\Omega$ is compact in some $L^p$. From the boundedness in $H^k$ you obtain (*) while from the compactness in $L^p$ you get $$ \|x\|_{L^p}=\lim \|x_n\|_{L^p}. $$

I hope this helps you.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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