I would like to understand which of the statements about the Sobolev space $H^1(\mathbb{R})$ remain true if one introduces a density/weight function in the definition.


The Sobolev space $H^1(\mathbb{R})$ are those square integrable functions whose first weak derivatives exist almost everywhere and are square integrable or briefly $f^2\in L^1$ and $(f')^2\in L^1.$ This space $H^1(\mathbb{R})$ has the following properties

  1. It is a reproducing kernel Hilbert space with inner product $$\langle f,g\rangle=\int_{\mathbb{R}} f(x)g(x)\,dx + \int_{\mathbb{R}} f'(x)g'(x)\,dx.$$
  2. The functions in $H^1(\mathbb{R})$ are continuous.

Let $w:\mathbb{R}\rightarrow\mathbb{R}$ be a weight function or density, which is a strictly positive function with $\int w(x)\,dx=1.$ Now define the weighted $L^1$ space as $L^1(w):=\left\{ f:\mathbb{R}\rightarrow\mathbb{R} \mid fw\in L^1\right\}$ with norm $\lVert f\rVert_w=\int_{\mathbb{R}} |f(x)| w(x)\,dx$ and the weighted Sobolev space $H^1_w(\mathbb{R})$ by replacing $L^1$ with $L^1(w)$ in the definition above. This means the inner product of $H^1_w(\mathbb{R})$ would be $$\langle f,g\rangle_w=\int_{\mathbb{R}} f(x)g(x)w(x)\,dx + \int_{\mathbb{R}} f'(x)g'(x)w(x)\,dx.$$


Does $H^1_w(\mathbb{R})$ still have the two properties? I.e. is it still a reproducing kernel Hilbert space consisting of continuous functions?

EDIT: The literature for weighted Sobolev spaces seems to focus on weights which are of "Muckenhoupt class" (see this related question) or "doubling measures". But finite measures are never doubling measures.

EDIT2 I would like to use the weights in applications to control the asymptotic behaviour of the functions in the RKHS. E.g. I would like to have spaces containing constant functions, polynomials (up to a certain degree) or exponential functions. This means I am quite relaxed about the properties of $w$. $w$ may be assumed to be continuous or even differentiable, if this helps. Typical examples for $w$ would be functions such as $\exp(-x^2)$, $\frac{1}{\cosh x}$ or $\frac{1}{(1+x^2)^k}$.

  • $\begingroup$ Could you elaborate on the norms of $L^1(w)$ abd $H_w^1(\mathbb{R})$? $\endgroup$
    – Meowdog
    Apr 2 at 13:31
  • $\begingroup$ Sure, I included the definition of the inner products and norms in the question. Hope it is clearer now. $\endgroup$
    – g g
    Apr 2 at 16:17
  • $\begingroup$ Yes, thank you a lot $\endgroup$
    – Meowdog
    Apr 2 at 16:20
  • $\begingroup$ Have you tried using the proof for the normal Sobolev space and adapting it for the weighted Sobolev space? What happens? $\endgroup$
    – supinf
    Apr 12 at 15:26
  • 1
    $\begingroup$ @supinf: See my additional edit. $w$ can be assumed to be continuous, no problem. $\endgroup$
    – g g
    Apr 16 at 10:03

Suppose $w$ is continuous and strictly positive.

Let $n\in\Bbb N$ be given. We will show that functions in $H_w^1(\Bbb R)$ continuous on $(-n,n)$ (which implies that they are continuous).

Let $c_n>0$ be the minimum of $w$ over $[-n,n]$ and let $f\in H_w^1(\Bbb R)$. Then we have $$ \|f\|_{H^1((-n,n))}^2 = \int_{-n}^n |f(t)|^2 + |f(t)'|^2 \,\mathrm dt \leq \int_{-n}^n |f(t)|^2 + |f(t)'|^2 c_n^{-1} w(t) \,\mathrm dt = c_n^{-1} \|f\|_{H_w^1((-n,n))}^2 \leq c_n^{-1} \|f\|_{H_w^1(\Bbb R)}^2. $$ Thus $\|f\|_{H^1((-n,n))}$ is finite and therefore $f\in H^1((-n,n))$ (if we restrict $f$ to $(-n,n)$). Since functions in $H^1((-n,n))$ are continuous, there exists a continuous representative of $f$ on $(-n,n)$.

Can the continuous representative depend on $n$?: Suppose $n<m$ and $f_n$, $f_m$ are continuous representatives on $[-n,n]$, $[-m,m]$. Then we have $f_n = f_m$ a.e. on $[-n,n]$, and since both functions are continuous on $[-n,n]$, this means that $f_n=f_m$ everywhere on $[-n,n]$. Thus, one can always extend the continuous representative to a larger interval, while still being continuous. In the end, we can define $f(x) = f_n(x)$, where $n$ is such that $x\in[-n,n]$, and similar to the above arguments, the function $f$ has to be continuous.

We can also obtain a second thing from the inequality: If for a given $x\in Bbb R$$x\in \Bbb R$ we choose $n$ such that $x\in (-n,n)$, we can use the RKHS-Property of the unweighted $H^1((-n,n))$ to obtain $$ |f(x)| \leq C\|f\|_{H^1((-n,n))} \leq Cc_n^{-1} \|f\|_{H_w^1(\Bbb R)}. $$ This inequality shows that the point evaluation functionals $\delta_x$ are continuous, which implies that $H_w^1$ is a RKHS (see Theorem 2.28 in the linked document) (note that the constants $C,c_n$ can depend on $x$, but this is ok).

  • $\begingroup$ Thanks for your input! This is very interesting indeed. Two questions though: (1) Why is it clear that continuous functions are dense in the Hilbert space norm? (2) I do not understand how to go from the statement on compact $[-n, n]$ to a statement on $\mathbb{R}$? For example: how can one exclude that the continuous representatives depend on $n$? $\endgroup$
    – g g
    Apr 17 at 8:12
  • $\begingroup$ @gg I adressed your second point in the edit. I will try to find something for density. I even think that smooth functions with compact support should be dense, but I agree that this deserves some explanation. $\endgroup$
    – supinf
    Apr 17 at 12:29
  • $\begingroup$ One more point: Right at the beginning you use $f(y) - f(x)=\int_x^y f'(t)dt$. But for this one needs absolute continuity of $f$. So continuity alone does not seem enough. This point is also not addressed in the lecture. $\endgroup$
    – g g
    Apr 18 at 11:01
  • $\begingroup$ Furthermore, if one can establish (1) then statement (2) follows as you demonstrate. Would (2) not already be enough to ensure continuity for all $x\in [-n,n]$? The fact that continuous functions are dense would not be required. $\endgroup$
    – g g
    Apr 18 at 11:08
  • $\begingroup$ I guess the argument would run like this: $wf'^2$ exists almost everywhere by assumption. Since $w$ is continuous and positive $f'$ exists a.e. on every closed interval, which makes $f$ absolutely continuous there. Now follows (1) and (2) as above ... $\endgroup$
    – g g
    Apr 18 at 11:16

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