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.

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    $\begingroup$ 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
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    $\begingroup$ 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

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.

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

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