# Operator norm and spectrum

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.

• If the operator is not self-adjoint/symmetric, then the formula is for singular value, not eigenvalue. – Piyush Grover Sep 20 '20 at 20:19
• Thanks! I missed the point with the singular values. Let's say the operator is also symmetric, is the statement true then? – Peppermint Sep 20 '20 at 20:28
• 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. – Jochen Glueck Sep 20 '20 at 20:43
• 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? – Peppermint Sep 20 '20 at 20:44
• Do we know that $T$ is self-adjoint as an operator on $H_0^1(\Omega)$? – Mateusz Kwaśnicki Sep 21 '20 at 7:11

## 1 Answer

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.

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