How to prove that the mapping $f\longmapsto f'$ from $H^1(\mathbb{R})$ to $L^2(\mathbb{R})$ is closed?

Question

Let $D:H^1(\mathbb{R})\to L^2(\mathbb{R})$ be the operator given by $D(f)=Df$ where $Df\in L^2(\mathbb{R})$ is the weak derivative of $f$, that is, the function $Df$ satisfies $$\int_\mathbb{R}f(x)\varphi'(x)dx=-\int_\mathbb{R}Df(x)\varphi(x)dx$$ for all $\varphi\in C^\infty_0(\mathbb{R})$. I think this operator is closed, then I'm trying to prove it and need help.

Let $(h_n)$ be a sequence in $H^1(\mathbb{R})$. Suppose there exists $h,g\in L^2(\mathbb{R})$ such that $\|h_n-h\|_{L^2}\to 0$ and $\|Dh_n-g\|_{L^2}\to 0$. In order to prove that $D$ is closed we need to show that $h\in H^1(\mathbb{R})$ and $Dh=g$. In other words, we need conclude

$$\int_\mathbb{R}h(x)\varphi'(x)dx=-\int_\mathbb{R}g(x)\varphi(x)dx\tag{1}$$ for all $\varphi\in C^\infty_0(\mathbb{R})$.

Notice that

$$\int_\mathbb{R}|h(x)\varphi'(x)+g(x)\varphi(x)|dx\leq 0\Rightarrow\left|\int_\mathbb{R}h(x)\varphi'(x)+g(x)\varphi(x)dx\right|=0\Rightarrow(1).\tag{2}$$

Therefore it's enough to show that the inequality in $(2)$ holds. Since

$$\begin{align*} |h(x)\varphi'(x)+g(x)\varphi(x)|&\leq|h(x)-h_n(x)||\varphi'(x)| +|h_n(x)\varphi'(x)+Dh_n(x)\varphi(x)|+ \\ &+|g(x)-Dh_n(x)||\varphi(x)|, \end{align*}$$ we can write

$$\int_\mathbb{R}|h(x)\varphi'(x)+g(x)\varphi(x)|\leq\int_\mathbb{R}|h(x)-h_n(x)||\varphi'(x)|dx +\\+\int_\mathbb{R}|h_n(x)\varphi'(x)+Dh_n(x)\varphi(x)|dx +\int_\mathbb{R}|g(x)-Dh_n(x)||\varphi(x)|dx.\tag{3}$$

It seems that (by Hölder's inequality) we can conclude that the first and the last integral in the left-hand side of $(3)$ goes to zero, but I don't if this conclusion is true for the second one. Is it a good approach? How can we finish it?

Thanks.

Answer 1

If we look at the graph of $D$,

$$\Gamma(D) = \left\lbrace (f, Df) : f \in H^1(\mathbb{R})\right\rbrace \subset L^2(\mathbb{R}) \times L^2(\mathbb{R}),$$

we see that $\iota \colon H^1(\mathbb{R}) \to \Gamma(D);\; \iota(f) = (f, f')$ is a bijection that is isometric with respect to the canonical norm on $H^1(\mathbb{R})$ and the Hilbert-product norm on $L^2(\mathbb{R})\times L^2(\mathbb{R})$. Thus $\Gamma(D)$ is a complete subspace of $L^2(\mathbb{R})\times L^2(\mathbb{R})$, hence closed. Thus $D$ is a closed operator.

We could also continue the way you have started, if $h_n \xrightarrow{L^2} h$ and $Dh_n \xrightarrow{L^2} g$, then $(h_n)$ and $(Dh_n)$ are both Cauchy sequences in $L^2$, and that means $(h_n)$ is a Cauchy sequence in $H^1$. Since $H^1$ is complete, there is a $h^\ast \in H^1$ with $h_n \xrightarrow{H^1} h^\ast$. But that means nothing but

• $h_n \xrightarrow{L^2} h^\ast$, and
• $Dh_n \xrightarrow{L^2} Dh^\ast$.

Which implies $h = h^\ast$ and $g = Dh^\ast$, so $(h,g) \in \Gamma(D)$.