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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 $$\|y\|_X \leq \liminf \|y_n\|_X \ .$$

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Your point 2 is not the same as above: it follows immediately from the continuity of the norm. –  András Bátkai May 11 '13 at 21:25
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Why is the second inequality true? If the sequence converges only in $L^2$, why is the left side necessarily finite? –  Deane Yang May 16 '13 at 12:29
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1 Answer

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.

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