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I'm trying to show that the space $W^{1,2}$ is an Hilbert space, i found this answered question: Showing Sobolev space $W^{1,2}$ is a Hilbert space

which offer a proof, but I'm having trouble with one of the basic steps: in order to claim that if $f_n$ is Cauchy in $W^{1,2}$ than $f_n \rightarrow f$ in $L^2$ , I need to show that $f_n$ is Cauchy in $L^2$ , correct? How can I show that?

Thanks Shahar

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user3350919
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    Well, $\lVert \bullet\rVert_{1,2}\ge \lVert \bullet\rVert_2$. –  Jul 27 '16 at 09:26
  • yes but to prove that $f_n$ is Cauchy in $L^2$ its not enough since I need $<f_n,f_m>{L^2}<\epsilon$ , this is not the norm but a inner product, and since $<f_n,f_m>{W^{1,2}}=<f_n,f_m>{L^2}+<f'_n,f'_m>{L^2}$ I need to bound $<f'n,f'_m>{L^2}$ so it will be true. no? – user3350919 Jul 27 '16 at 09:59
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    No. In an inner-product space, just like in normed spaces, you need $\lVert f_n-f_m\rVert<\varepsilon$, not $\langle f_n,f_m\rangle<\varepsilon$ (which is hardly ever true: when $f_n\to f$, you have that $\langle f_n,f_m\rangle\to \lVert f\rVert^2$). –  Jul 27 '16 at 10:10
  • Wow, you are right! I was so incredibly confused! Thanks! – user3350919 Jul 27 '16 at 10:27

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Since $\lVert f\rVert_{1,2}\ge \lVert f\rVert_2$, $$\limsup_{\min(n,m)\to \infty}\lVert f_n-f_m\rVert_{2}\le\limsup_{\min(n,m)\to \infty}\lVert f_n-f_m\rVert_{1,2}=0$$