Are both sides of a distributional equation in the same space?

Question

I'm trying to understand a paper, but will try to hold this question more general. (So if you need more information i can state the "real" problem.)

Say I have $$u \in L^2\\h\in W^{-1,2}$$ and the distributional equation $$u=h.$$

Does this imply $h\in L^2$?

From my understanding the distributional equation is to be taken like $$<u,\phi>=<h,\phi>\quad \text{for every }\phi \in C^\infty\text{ or }\phi \in W^{1,2}$$

In my PDE lecture we had the following corollary of Hahn-Banach:

Corollary (“The dual separates the points"): For $x_1,x_2\in X$ ($X$ is a normed space) with $x_2\neq x_1$ there is always some $x^\star\in X^\star$ with $$<x^\star,x_2>\neq<x^\star,x_1>$$

But since $u,h\in W^{-1,2}$ this implies for my problem that $u=h$, because otherwise this would contradict the Corollary. But in what sense? strong in $W^{-1,2}$ or since $u$ is in the better space and $L^2\subset W^{-1,2}$ in $L^2$?

(Please let $h$ be in $L^2$ otherwise the proof doesn't make sense to me. :D)

Answer 1

As you anticipate, yes, $u=h$, to put it bluntly.

Still, as you may suspect, there are some potential technicalities, which mostly don't matter, although can be used as trick questions, and, more rarely, can have genuine mathematical consequences.

So, yes, obviously, if we have a set containment $X\subset Y$, and $x\in X$ and $y\in Y$, then we can talk about equality-or-not of $x$ and $y$. And, if $x=y$, then, yes, $y\in X$.

The chief potential hazard arises when we declare that $L^2$ functions are equivalence classes of pointwise-valued functions, introducing some potential pitfalls to the notion of "equality". So Sobolev imbedding does not quite say that $H^s(\mathbb R)\subset C^0(\mathbb R)$ for $s>1/2$, but only that there is a continuous representative in that equivalence class.

In the example at hand, there is a different potential hazard, namely about limits. With $u=h$ and $u_n\to u$ in $L^2$ and $h_n\to h$ in $W^{-1,2}$ do not imply that $h_n\to h$ in $L^2$. Once stated, this may be fairly obvious. The possibly slightly hidden dangers are that various operations (integrals, or infinite sums) involve limits, and the notion of limit depends on the topology.

A tangible example of topological discrepancies appears in Fourier series. E.g., (by Baire category, for example) we know that the Fourier series of most continuous functions do not converge to them everywhere-pointwise. (The Fejer result asserts that other finite sums of exponentials, but not the finite partial sums of the Fourier series, do reliably converge to them...). The Fourier series of a $C^1$ function $f$ does converge uniformly pointwise to $f$. Good. However, that Fourier series will not in general have derivatives converging to $f'$, even though $f'$ is continuous. Slightly counter-intuitive, perhaps.

But, in the present example, unless someone is trying to prank you, yes, $u=h$.

Answer 2

So in Evans I found the following Theorem:

Theorem 1 (characterization of $H^{-1}$) Assume $f\in H^{-1}(U)$ then there exist functions $f^0, f^1,\dots , f^n$ in $L^2(U)$ such that $$ <f,v>=\int f^0 v+\sum_{i=1}^n f^i v_{x_i}dx\qquad v\in H^1_0(U) $$

So now the proof seems easy.

$$\int u\phi dx =<u,\phi>=<h,\phi>=\int h^0 \phi dx +\sum_i \int h^i \phi_{x_i} dx$$ and since $\phi$ is arbitrary (in $H^1_0$) follows $$ h^i=0 $$ and therefore $$<h,\phi>=\int h^0 \phi dx$$ and so $h\in L^2$.

And I guess the same holds similar for general $h \in W^{-m,q}, u\in W^{-k,q}$ and the distributional equation $$h=u$$ by the Definition $W^{-m,q}=\lbrace \sum_{|\alpha|\leq m } \partial^\alpha v_\alpha\in \mathcal{D}': (v_\alpha)_{|\alpha|\leq m}\in (L^q)^{p(m)}\rbrace$ ( where $p(m) = \#\lbrace \alpha:|\alpha|\leq m\rbrace $)

So that the function is always in the better space. :)