# Convergence of operator norm

I have a linear bounded operator $A:L_2(0,1) \rightarrow L_2(0,1)$ satisfying $\|A^n\|^{1/n} \rightarrow 0$. Thus, for some sufficiently large $N$, $\|A^N\| < 1$ and then from Gelfand's formula, I can show that the spectral radius of $A$ satisfies $\rho(A)<1$.

My question is, knowing these properties of the operator $A$, what can we say about the spectral radius of the operator $A+A^\star$, where $A^\star$ is the adjoint of $A$?

Since I can find the spectral radius as the limit of a sequence of operator norms, I want to construct $\|(A+A^\star)^n\|$ and I know that $\|A^n\|=\|(A^\star)^n\|$, but I don't know how to relate $\|(A+A^\star)^n\|$ and $\|A^n\|$.

If any one's interested, the operator $A$ is the volterra operator given by $(Aw)(x)=\int_0^x K(x,\xi)w(\xi)d \xi$ for a continuous $K$. One can show that $\rho(A)=0$ and $\rho(A^\star)=0$, but I have no idea about $\rho(A+A^\star)$.

Any help would be much appreciated.

From Gelfand's formula, which holds for every element of a unital complex Banach algebra more generally, $$\lim_{n\rightarrow +\infty}\|A^n\|^\frac{1}{n}=0\quad\iff\quad \rho(A)=0.$$ This characterizes the set of quasinilpotent operators, which extends the set of nilpotent operators.

There is nothing you can say about $\rho(A+A^*)$ without further assumptions. Indeed, for every $t\geq 0$, $$A=\pmatrix{0&t\\0&0}\quad\Rightarrow\quad A+A^*=\pmatrix{0&t\\t&0}\quad\Rightarrow\quad \sigma(A+A^*)=\{\pm t\}\quad\Rightarrow\quad \rho(A+A^*)=t.$$ So $\rho(A+A^*)$ could be any nonnegative number and yet $\rho(A)=0$.

Note: the question is more interesting if you add the condition that the self-adjoint operator $A+A^*$ be positive (i.e. has nonnegative spectrum). In finite dimension, we have $$\rho(A)=0\quad\mbox{and}\quad A+A^*\geq 0\quad \Rightarrow\quad A=0.$$ Indeed, the condition $\rho(A)=0$ implies $\mbox{Tr} A=0$, whence $\mbox{Tr}(A+A^*)=0$. Now if $A+A^*\geq 0$ has null trace, it follows that every eigenvalue is zero. Since $A+A^*$ is diagonalizable, this entails $A+A^*=0$. So $A=-A^*$ is normal, whence diagonalizable with null spectrum. Therefore $A=0$.

This argument seems to extend without difficulty to the trace-class quasinilpotent operators when $H$ is infinite-dimensional separable. I don't know at this point what happens in full generality. Help anybody?

Edit: I was given the answer on MO by Mike Jury. In infinite dimension, you can have $A$ quasinilpotent, nonzero, and yet $A+A^*\geq 0$. It suffices to consider the Volterra operator...

• Hi Julien, thanks for providing a example and yes, I am aware that these operators are quasinilpotent. There is an additional property that the operator $A+A^\star$ is a non-negative operator. However, I don't think that has any bearing on my problem. – Amit May 2 '13 at 14:31
• I was wondering, are there any results on the spectra of the product of quasinilpotent operators? – Amit May 2 '13 at 14:55
• @Amit If they commute, the product is still quasinilpotent. If they don't, the spectral radius of the product could be anything. – Julien May 2 '13 at 14:57
• Ok, I think I have sufficient information now to at least attempt a stab at obtaining my desired results. Thanks a lot Julien. – Amit May 2 '13 at 15:00
• @Amit You're welcome. I've added some thoughts about the case when you assume $A+A^*\geq 0$. Let me know what you think. – Julien May 2 '13 at 17:58