0
$\begingroup$

Let $\;I\;$ be a compact subset of $\;\mathbb R\;$ and consider sequence $\;\{ f_n \}\;$ such that: $\;f_n \to f\;$ weakly on $\;W^{1,2}(I,\mathbb R^m)\;$.

This is enough in order to claim that, the derivative $\;{f_n}' \to f'\;$ weakly in $\;L^2\;$ norm (according to an answer below). Why does this hold?

I'm really new to Sobolev Spaces, so it might be quite silly what I'm asking..I would appreciate any help.

Thanks in advance!!

$\endgroup$
1
$\begingroup$

No it is not.

Assume that $f_n\to f$ in $W^{1,2}$-weak. That means that for any fixed $g\in W^{1,2}$, it is $$\langle f_n, g\rangle_{W^{1,2}} \to \langle f, g\rangle_{W^{1,2}}$$ Now putting in the definition of the $W^{1,2}$ scalar-product this means $$ \langle f_n, g\rangle_{L^2} + \langle f'_n, g'\rangle_{L^2} \to \langle f,g\rangle_{L^2} + \langle f', g'\rangle_{L^2}$$

As this holds for for arbitrary $g$ you can show that this implies $$ \langle f_n, g\rangle_{L^2} \to \langle f,g\rangle_{L^2} $$ and $$ \langle f'_n, g'\rangle_{L^2} \to \langle f', g'\rangle_{L^2}$$ . The second one is precisely weak-$L^2$ convergence of $f_n'$.

But you won't be able to get norm-$L^2$ convergence of $f_n'$.

On a side note: weak-$W^{1,2}$ convergence of $f_n$ does imply norm-$L^2$ convergence of $f$. But that is non-trivial so I won't attempt a proof here.

$\endgroup$
  • $\begingroup$ I just edited my post. I wanted to say $\;f_n' \to f'\;$.. Why is this true? Could you explain this to me, because it's not very obvious...? Thanks! $\endgroup$ – kaithkolesidou May 14 '17 at 16:08
  • $\begingroup$ I made the answer somewhat more elaborate. hope it helps. $\endgroup$ – Simon May 14 '17 at 17:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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