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Let $-\infty<a<b<+\infty$. Consider the norms $\|\cdot\|_{L^2}$, $\|\cdot\|_0$ and $\|\cdot\|_{H^1}$ defined on suitable spaces and given by

$$\|f\|_{L^2}=\left(\int_a^bf^2\right)^{\frac{1}{2}},\;\|f\|_0=\|f'\|_{L^2}\text{ and }\|f\|_{H^1}=\|f\|_{L^2}+\|f'\|_{L^2}.$$

In the book Sobolev Spaces (Robert Adams) we can see (page 184) that the norms $\|\cdot\|_0$ and $\|\cdot\|_{H^1}$ defined on $H_0^1(a,b)$ are equivalent. Hence, $H_0^1(a,b)$ equipped with the norm $\|\cdot\|_0$ is complete.

My question is: is the space $H^2(a,b)\cap H_0^1(a,b)$ equipped with the norm $\|f\|_0=\|f'\|_{L^2}$ complete?

Thanks.

Pedro
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2 Answers2

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Note that $$C_0^\infty(a,b) \subset H^2(a, b)\cap H^1_0(a, b) \subset H^1_0(a, b)$$

and $C^\infty_0(a, b)$ is dense in $H^1_0(a, b)$ in $H^1$ norm. Thus $H^2(a, b)\cap H^1_0(a, b)$ is complete in $H^1$ norm if and only if $H^2(a, b)\cap H^1_0(a, b) = H^1_0(a, b)$. But this is not true.

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The answer is No.
A similar question: Will $C[a,b]$ be complete in the norm $\|\cdot\|_{L^2}$?
(The answer will be the same No).

For your question, to construct a counterexample choose $a=0$, $b=\pi$, and take a Fourier series, say, wrt an orthogonal basis $\;\{\sin{nx}\}_{n-1}^{\infty}\;$ in $ H_0^1(0,\pi)$ $$ f(x)\overset{def}{=}\sum\limits_{n=1}^{\infty}c_n\sin{nx},\quad x\in [0,\pi],\tag{$\ast$}$$ with any coefficients $c_n$ satisfying the conditions $$\sum\limits_{n=1}^{\infty}n^2|c_n|^2<\infty,\quad \sum\limits_{n=1}^{\infty}n^4|c_n|^2=\infty.$$ Given such coefficients $c_n\,$, a function $f\in H_0^1(0,\pi)$ but $f\notin H^2(0,\pi) \cap H_0^1(0,\pi)$ due to convergence of series $(\ast)$ in the norm $\|\cdot\|_{H^1}$ and its divergence in the norm $\|\cdot\|_{H^2}\,$. Now consider a sequence of partial sums $$ f_m(x)\overset{def}{=}\sum\limits_{n=1}^{m}c_n\sin{nx},\quad x\in [0,\pi],\tag{$\ast\ast$}$$ which will be a Cauchy sequence in the norm $\|\cdot\|_{H^1}$ due to its convergence in the norm $\|\cdot\|_{H^1}$ to the element $f\in H_0^1(0,\pi)$. If the space $H^2(0,\pi)\cap H_0^1(0,\pi)$ were complete in the norm $\|\cdot\|_{H^1}\,$, the Cauchy sequence $ f_m\in H^2(0,\pi)\cap H_0^1(0,\pi)$ would converge to some element $g\in H^2(0,\pi)\cap H_0^1(0,\pi)$ which should coincide with $f\notin H^2(0,\pi)\cap H_0^1(0,\pi)$. Hence, the space $H^2(0,\pi)\cap H_0^1(0,\pi)$ cannot be complete in the norm $\|\cdot\|_{H^1}\,$.

mkl314
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  • Do you know any norm $|\cdot|$ such that $H^2\cap H_0^1$ equipped with $|\cdot|$ is complete? – Pedro Jan 27 '14 at 00:11
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    Pedro: It is the norm of $H^2,$ , namely $|f|{H^2}=|f|{L^2}+|f''|{L^2},$, or its equivalent $|f|{H^2}=|f|{L^2}+|f'|{L^2}+|f''|{L^2},$. Alternatively, one can take $|f|{H^2}=\bigl(|f|^2_{L^2}+ |f''|^2_{L^2}\bigr)^{1/2},$ when $H^2$ is to be employed as a hilbert space. – mkl314 Jan 27 '14 at 00:32
  • @mkl314, I posted here$^*$ a question related to your comment please, can you give me a Hint or a reference to prove such remark.

    $^*$: http://math.stackexchange.com/q/2006756/244803

     – Achaire Nov 10 '16 at 13:03