Is weak derivative a bounded operator from $H^k(\mathbb{R}^n)$ to $L^2(\mathbb{R}^n)$?

Question

I'm struggling to become comfortable with the concept of the weak derivative and Sobolev space. In my textbook, it is proved that the Sobolev space $H^k(\mathbb{R}^n)$ is a Hilbert space, and it is also proved that the weak derivative $D$ is a closed operator from $H^k(\mathbb{R}^n)$ to $L^2(\mathbb{R}^n)$. Does this mean that $D$ is a bounded operator? I guess so because a closed operator defined everywhere in a Hilbert space is bounded. Is that also mean $D$ is bounded as an operator from $L^2(\mathbb{R}^n)$ to $L^2(\mathbb{R}^n)$ with domain $H^k(\mathbb{R}^n)$?

Answer 1

You need to be careful on which inner product you put on $H^k(\mathbb{R}^n)$.

• If you equip $H^k(\mathbb{R}^n)$ with its own inner product, then the differentiation operator $D$ is bounded (and in particular, closed).

• If you equip $H^k(\mathbb{R}^n)$ with the $L^2$ inner product, then the differentiation operator $D$ is an unbounded operator from $H^k(\mathbb{R}^n)$ to $L^2(\mathbb{R}^n)$, or in other words, a densely defined unbounded operator on $L^2(\mathbb{R}^n)$ with domain $H^k(\mathbb{R}^n)$. The operator is closed, but $H^k(\mathbb{R}^n)$ is not a Banach space with the $L^2$ inner product and therefore we cannot use the closed graph theorem to conclude that it is bounded.

The issue is that the $H^k$ inner product yields a topology which is too strong - sure, $D$ is continuous, but there are too few converging sequences in its domain for it to be useful. Therefore, most of the times it is more useful to consider $D$ as a densely defined unbounded operator on $L^2(\mathbb{R}^n)$.