What kind of “isomorphisms” is the mapping from $H_0^1$ to $H^{-1}$ defined by the elliptic operator?

The following is an excerpt from Evans’s Partial Differential Equations

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Here $L$ is the elliptic operator. The part about $\langle f,v\rangle$ is stated in this question.

Would anybody clarify what kind of “isomorphism” it should be in the last sentence? Of course it is a vector space isomorphism which means $L_\mu$ is a bijective linear maps.

My question concerns:

  • Is it a homeomorphism?
  • Is it an isometry?

[Added:] Here is Theorem 3 and its proof right before the quoted remark above.

enter image description here

Solutions Collecting From Web of "What kind of “isomorphisms” is the mapping from $H_0^1$ to $H^{-1}$ defined by the elliptic operator?"

It’s an isomorphism since $L_\mu$ and $L_\mu^{-1}$ are both bounded linear operators. This was established in the proof of Theorem 3 on the page before the remark in question.

As to whether it is an isometry, this is a bit more delicate. Recall the weak form of $Lu + \mu u = f$ is

$$B_\mu[u,v] = \langle f,v\rangle$$

where $B_\mu[u,v] = B[u,v] + \mu(u,v)$, and $B[u,v]$ is the bilinear form associated with $L$ (so $(Lu,v) = B[u,v]$ for smooth $u,v$). Recall that $\mu$ was chosen so that

$$\beta \|u\|_{H^1_0(U)}^2 \leq B_\mu[u,u] \ \text{ and } \ B_\mu[u,v] \leq C\|u\|_{H^1_0(U)} \|v\|_{H^1_0(U)}.$$

Suppose we have no first order terms in the PDE, so $b^i= 0$. Then $B_\mu$ is symmetric: $B_\mu[u,v] = B_\mu[v,u]$. This means that the inner product $u\times v \mapsto B_\mu[u,v]$ is an equivalent inner product on $H^1_0(U)$, which induces the norm $B_\mu[u,u]^\frac{1}{2}$. If we endow $H^1_0(U)$ with this inner product, and there are no first order terms in the PDE, then $L_\mu$ is an isometry because

$$\|f\|_{H^{-1}(U)} = \sup_{B_\mu[v,v]=1} \langle f,v\rangle = \sup_{B_\mu[v,v]=1} B_\mu[u,v] = B_\mu[u,u]^\frac{1}{2}.$$

Notice that we are using a new norm on $H^{-1}(U)$ induced by the new norm on $H^1_0(U)$.

So when there are no first order terms, you can choose equivalent norms on $H^1_0(U)$ and $H^{-1}(U)$ so that $L_\mu$ is an isometry. Maybe this feels like cheating to you. If you prefer to stick with the usual norms, then $L$ is an isometry when

$L = -\Delta u + u.$

In this case the bilinear form associated with $L$ is $B[u,v] = \int \nabla u \cdot \nabla v + uv \, dx$ and

$$B[u,u] = \|u\|_{H^1_0(U)}^2.$$

When there are first order terms in the PDE, e.g., $L = -\Delta + \partial_{x_i}$, it will not be an isometry. It should not be too hard to work out a counterexample in a special case (say 1D with $U=(-1,1)$).