I have the Sobolev space $W^{1,2}$ consisting of all continuous functions $f \in L^2(\mathbb{R})$ such that there exists an $f'$ with $f(b) - f(a) = \int_a ^b f'(t) dt$. $W^{1,2}$ has inner product $<f,g> = \int fg + f'g'$. I'm trying to show this is a Hilbert space but I'm having some difficulties.

  • I can show that as $t \to+/- \infty$ we must have $f(t) \to 0$ because $f$ is continuous. From this and using the relation $f(x)^2 = \int_{-\infty} ^x f(t)f'(t)$ I can obtain the estimate $||f||_{\infty} \le K||f||_{1,2}$
  • If we now take a sequence $f_n$ in $W^{1,2}$ such that $f_n \to f$ in $W^{1,2}$ norm. I want to show that in fact $f$ is in $W^{1,2}$. We know from the above estimate that $f_n \to f$ uniformly and hence $f$ is continuous and in $L^2$ so it remains to show that $f$ has the weak derivative property.

I'm struggling to show this last part. I know that for each $f_n$ there is an $f_n'$ such that $f_n(b) - f_n(a) = \int_a ^b f_n'(t) dt$ but how can we show that these $f_n'$ converge to a derivative for $f$?

Thanks for any help here!

  • $\begingroup$ I suppose that $W=W^{1,2}$ and that you mean $f_n\to f$ in $W^{1,2}$? $\endgroup$ – Tomás May 15 '15 at 13:56
  • $\begingroup$ Yes thanks - I've added the $1,2$. I have $f_n \to f$ in $L^2$ and then I'm trying to show that in fact $f \in W^{1,2}$ $\endgroup$ – Wooster May 15 '15 at 14:22
  • $\begingroup$ If the convergence is only in $L^2$ then, it is no true. If you want to prove that $W^{1,2}$ is closed, you have to assume that the convergence is in $W^{1,2}$ (by definiton). $\endgroup$ – Tomás May 15 '15 at 14:24
  • $\begingroup$ Oh sorry yes - I meant that $f \in L^2$ but the convergence is happening in $W^{1,2}$ norm. I'll make that clearer in the question $\endgroup$ – Wooster May 15 '15 at 14:24
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    $\begingroup$ The method I've usually seen here is to say that if a sequence is Cauchy in $W^{1,2}$, then it is Cauchy in $L^2$, so it converges in $L^2$ (from some previous proof that $L^2$ is complete). Then you show that the $L^2$ limit is in fact $W^{1,2}$ using the dominated convergence theorem. This is a pretty standard approach: as a rule of thumb it is easier to show a sequence is convergent when you already have its limit. This might not be convenient given the definitions you appear to have (which seem to be related to absolute continuity rather than distribution theory). $\endgroup$ – Ian May 15 '15 at 14:27

Suppose that $f_n$ is a Cauchy sequence in $W^{1,2}$ and write $$f_n(b)-f_n(a)=\int_a^b f_n'(t)dt.\tag{1}$$

By one hand, there is $f\in L^2$ such that $f_n\to f$. On the other hand, there is $g\in L^2$, such that $f'_n\to g$.

From the estimate $\|f_n\|_\infty\le K\|f_n\|_{1,2}$, we must conclude that $f_n(x)\to f(x)$ for all $x\in\mathbb{R}$ and by using Lebesgue theorem, we also have that $$\int_a^b f'_n(t)dt \to \int_a^b g(t)dt.$$

It follows from $(1)$ that $$f(b)-f(a)=\int_a^b g(t)dt.$$

Thus $f\in W^{1,2}$ and $f'=g$.

  • $\begingroup$ Thank you for your answer - I'm just wondering how you know there is a $g$ such that $f_n' \to g$ in $L^2$? $\endgroup$ – Wooster May 15 '15 at 20:42
  • $\begingroup$ Because $f_n$ is a Cauchy sequence in $W^{1,2}$. $\endgroup$ – Tomás May 16 '15 at 3:17
  • $\begingroup$ Oh yes of course! Thanks $\endgroup$ – Wooster May 16 '15 at 8:15
  • $\begingroup$ When you say Lebesgue theorem is this dominated convergence? $\endgroup$ – Wooster May 16 '15 at 8:17
  • $\begingroup$ Yes, that's right. $\endgroup$ – Tomás May 16 '15 at 18:10

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