2
$\begingroup$

I am wondering about when an operator norm coincides with the maximum eigenvalue of an operator and there is one particular aspect that confuses me quite a lot.

Let's say we have a symmetric positive continuous linear operator $$ T : L^2(\Omega) \rightarrow L^2(\Omega) $$ with maximum eigenvalue $\lambda>0$ so that $T u = \lambda u$ for some eigenfunction $u\in L^2(\Omega)$. Then (if I understand it correctly), it should hold $$ \lambda = \sup_{v \in L^2(\Omega) \setminus \{ 0 \}} \frac{\| Tv \|_{L^2(\Omega)}}{\| v \|_{L^2(\Omega)}}. $$ Next, let's assume that the operator has a smoothing effect such that $\mbox{image}(T) \subset H^1_0(\Omega)$ and that it is also $H^1$-continuous (I am thinking of $T$ as the inverse of an elliptic differential operator). In this case we can interpret the operator as $$ T : H^1_0(\Omega) \rightarrow H^1_0(\Omega) $$ The spectrum should remain unchanged, so that I would think that $$ \lambda = \sup_{v \in H^1_0(\Omega) \setminus \{ 0 \}} \frac{\| Tv \|_{H^1_0(\Omega)}}{\| v \|_{H^1_0(\Omega)}}. $$ However, the statement $$ \sup_{v \in H^1_0(\Omega)} \frac{\| Tv \|_{H^1_0(\Omega)}}{\| v \|_{H^1_0(\Omega)}} = \lambda = \sup_{v \in L^2(\Omega)} \frac{\| Tv \|_{L^2(\Omega)}}{\| v \|_{L^2(\Omega)}}. $$ looks wrong to me. Is it? If so, where is the mistake in my arguments? I feel like I have a very basic misunderstanding here.

$\endgroup$
10
  • 2
    $\begingroup$ If the operator is not self-adjoint/symmetric, then the formula is for singular value, not eigenvalue. $\endgroup$ – Piyush Grover Sep 20 '20 at 20:19
  • $\begingroup$ Thanks! I missed the point with the singular values. Let's say the operator is also symmetric, is the statement true then? $\endgroup$ – Peppermint Sep 20 '20 at 20:28
  • $\begingroup$ A brief remark (though not essential for the question): your assumption that $T$ be continuous from $L^2$ to $H^1_0$ is redundant; this follows automatically from the closed graph theorem. $\endgroup$ – Jochen Glueck Sep 20 '20 at 20:43
  • $\begingroup$ If we have the identity $Tu = \lambda u$ for $u\not = 0$ (strongly in $H^1(\Omega)$), then $(T - \lambda I) u = 0$, which means that $(T - \lambda I)$ cannot be invertible. This should not change depending on if we interpret $T$ as an operator on $L^2(\Omega)$ or on $H^1_0(\Omega)$. What do I miss here? $\endgroup$ – Peppermint Sep 20 '20 at 20:44
  • 1
    $\begingroup$ Do we know that $T$ is self-adjoint as an operator on $H_0^1(\Omega)$? $\endgroup$ – Mateusz Kwaśnicki Sep 21 '20 at 7:11
5
$\begingroup$

The major point here is that, for an operator $S$ on a Banach space (or Hilbert space) $X$, the number $\sup_{x \in X \setminus\{0\}} \frac{\|Sx\|}{\|x\|}$ is not the spectral radius of $S$ but the operator norm. The operator norm is always $\ge$ the spectral radius, but we cannot expect equality in general.

On a Hilbert space, one sufficient condition for equality of the operator norm and the spectral radius is that the operator be self-adjoint or, more generally, normal.

But as pointed out in a comment by Mateusz Kwaśnicki, if $T$ is self-adjoint on $L^2$, this does not imply that $T$ is self-adjoint on $H^1_0$ since the inner product there is different from the inner product on $L^2$.

Here is a concrete counterexample:

Let $\Omega = (0,2\pi)$ (endowed with the non-normalised Lebesgue measure) and define $z,v \in H^1_0 := H^1_0(\Omega)$ by \begin{align*} z(x) & = \frac{|\sin(x)|}{\sqrt{\pi}}, \\ v(x) & = \sin(\frac{1}{2}x) \end{align*} for all $x \in (0,2\pi)$.

We define the operator $T$ on $L^2$ by $$ Tf = \langle f, z\rangle_{L^2} \cdot z $$ for all $f \in L^2$. Then $T$ is a self-adjoint rank-$1$ projection on $L^2$ whose norm and spectral radius are thus equal to $1$. Clearly, the range of $T$ is a subset of $H^1_0$.

The restriction of $T$ to $H^1_0$ is again a non-zero projection and thus still has spectral radius $1$. But the operator norm of $T$ on $H^1_0$ is strictly larger than $1$. Indeed, we have $$ \|T\|_{H^1_0 \to H^1_0} \ge \frac{\|Tv\|_{H^1_0}}{\|v\|_{H^1_0}} = \frac{\sqrt{512}}{\sqrt{45}\pi} > 1 $$ (we need to compute a few integrals to obtain the equality in the middle, but the computations are rather straightforward).

This proves that the operator norms of $T$ on $L^2$ and on $H^1_0$ are distinct, although the spectral radius on both spaces is $1$. In particular, $T$ cannot be self-adjoint (and not even normal) on $H^1_0$.

EDIT: An additional observation. While, in the example above, equality of the spectral radii on both spaces follows from the fact that $T$ acts as a projection on both spaces, I thought it might be worthwhile to point out that the equality of the spectral radii is actual a general fact:

Proposition. (Equality of spectral radii) Let $V,X$ be complex Banach spaces such that $V$ is continuously embedded in $X$. Let $T: X \to X$ be a bounded linear operator such that $TX \subseteq V$. Then the spectral radius of the operator $T: X \to X$ coincides with the spectral radius of the restriction $T|_V: V \to V$.

Proof. We use the spectral radius formula $$ (*) \qquad r(S) = \lim_{n \to \infty} \|S^n\|^{1/n} $$ which holds for the spectral radius $r(S)$ of each bounded linear operator $S$ on a complex Banach space.

For each $n \in \mathbb{N}$ the operator $(T|_V)^n = (T^n)|_V: V \to V$ factorizes as $$ V \hookrightarrow X \overset{T^{n-1}}{\longrightarrow} X \overset{T}{\longrightarrow} V, $$ so the spectral radius formula $(*)$ implies that $r(T|_V) \le r(T)$. But conversely, the operator $T^n: X \to X$ factorizes as $$ X \overset{T}{\longrightarrow} V \overset{(T|_V)^{n-1}}{\longrightarrow} V \hookrightarrow X, $$ so the spectral radius formula $(*)$ also implies that $r(T) \le r(T|_V)$. This proves the proposition.

Remark. What is quite nice about the proposition above is that it does not rely on eigenvalues, so no compactness assumption on the operator $T$ is needed.

$\endgroup$
1
  • $\begingroup$ Great answer! Thanks a lot! That really clarifies my misunderstanding. $\endgroup$ – Peppermint Sep 21 '20 at 20:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.