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!
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 $f'=g$.