Common orthogonal basis for $L^2$ and $H^1$

Question

How can we obtain a common orthogonal basis for the space $L^2(U)$ and $H^1(U)$ for some bounded open subset of $\mathbb{R}^n$? That this can be done is mentioned in Evans's Partial Differential Equations on page 380 of first edition. Should not the basis of $L^2$ be necessarily larger?

Answer 1

The trick is here that the norms (and scalar products) of $L^2(U)$ and $H^1(U)$ are not the same, which means that the closure of the orthogonal basis (or complete orthogonal set) will be different, depending on the process of completion.

The general idea is that the inclusion of the derivative in the Sobolev norm will not only make sure that for a Cauchy sequence, the limit is not only an $L^2$-function, but also that the derivatives behave similarly.

A similar effect can be observed in sequence spaces, although I can only offer a Banach space setting where different norms give rise to different closures of the same set. Consider the set of sequences (roughly speaking the unit vectors): $$e_1= (1,0,\dots), e_2=(0,1,\dots), \dots$$

Then we know that $\ell^1, \ell^2$ or generally $\ell^p (p \in [1,\infty))$ are different Banach spaces, but the unit vectors above are dense, i.e. their closure are the respective spaces (since the every $p-$summable sequence can be approximated by truncated series: the convergence of partial sums ensures this).

We see here: Different norms generate different closures.