Coercivity and spectral gap: understanding the equivalence

Question

I am referring to this paper, p. 21.

First, there is the following definition of coercivity:

Let $L$ be an unbounded operator on a Hilbert space $\mathcal{H}$ with kernel $\mathcal{K}$ and let $\tilde{\mathcal{H}}$ be another Hilbert space continuously and densely embedded in $\mathcal{K}^\perp$, endowed with a scalar product $\langle\cdot,\cdot\rangle_{\tilde{\mathcal{H}}}$ and a Hilbertian norm $\Vert\cdot\Vert_{\tilde{\mathcal{H}}}$. The operator $L$ is said to be $\lambda$-coercive on $\tilde{\mathcal{H}}$ if $$ \forall h\in\mathcal{K}^\perp\cap D(L),\quad\Re\langle Lh,h\rangle_{\tilde{\mathcal{H}}}\geq\lambda\Vert h\Vert^2_{\tilde{\mathcal{H}}}, $$ where $\Re$ stands for the real part. The operator $L$ is said to be coercive on $\tilde{\mathcal{H}}$ if it is $\lambda$-coercive on $\tilde{\mathcal{H}}$ for some $\lambda>0$.

Afterwards, it is said that

[..] the most standard situation is when $\tilde{\mathcal{H}}=\mathcal{K}^\perp\simeq \mathcal{H}/\mathcal{K}$.

Moreover, in this particular case, it is said that

[..] it is equivalent to say that $L$ is coercive on $\mathcal{K}^\perp$, or that the symmetric part of $L$ admits a spectral gap.

Could you please explain to me why this equivalence holds?

As far as I understand, spectral gap means that there exists some $c>0$ such that $$ \sigma(L)\subset\{0\}\cup [c,\infty) $$

However, from this, I can neither conclude that $L$ is coercive on $\mathcal{K}^\perp$ (nor vice versa).

My "feeling" is that this has something to do with the relationship between the spectrum of $L$ and its numerical range (but I only know this for self-adjoint bounded operators and this seems to be another situation here).

Answer 1

I don't think there is anything deep here.

Note that

\begin{align} \Re \langle Lh,h \rangle &= \frac{1}{2} \left ( \langle Lh,h \rangle + \langle Lh,h \rangle^\ast \right )\\ &= \frac{1}{2} \left ( \langle Lh,h \rangle + \langle h,Lh \rangle \right ) \\ &= \frac{1}{2} \left ( \langle Lh,h \rangle + \langle L^\ast h,h \rangle \right ) \\ &= \langle Sh,h \rangle \end{align}

Where I just defined the symmetric part of $L$ to be

$$S = \frac{1}{2} \left ( L + L^\ast \right ). $$

The definition of spectral gap depends a bit on the context. It seems your definition is: the symmetric operator $A$ admits a spectral gap $\gamma \ge c$ if

$$ \sigma(A) \subset \{0\} \cup [c, \infty).$$

Note that we must require the operator to be symmetric or the spectrum need not be real.

The above definition is equivalent to the fact that

$$ \langle Ah,h \rangle \ge c \langle h,h \rangle \quad \forall h \notin \ker (A) $$

Now the equation

$$ \Re \langle Lh,h \rangle = \langle Sh,h \rangle \ge \lambda \langle h, h \rangle >0 $$

means that $h$ is not in the kernel of $S$ and so it means that $S$ has a spectral gap at least as big as $\lambda$.