spectral radius in normed space is subadditive.

Question

I am reading the book "characterizations of c*-algebras" by Doran and Belfi. On the page 299, the book reads "It can also be shown that if $xy=yx$, then $\nu(x+y)\le\nu(x)+\nu(y)$." The definition of $\nu$ is as follows: in normed algebra, $$\nu(x)=\limsup_{n\to\infty}\|x^n\|^{\frac1n}$$

Note that the algebra is not assumed to be complete and is also not assumed to be complex scalar(we can have real scalars)

This is my trial: I just tried to mimic the proof of the Minkowski inequality, but failed.

Thank you

Answer 1

Fix $\varepsilon > 0$, and take $m$ sufficiently large so that $\|x^n\|^{1/n} < \nu(x) + \varepsilon$ and $\|y^n\|^{1/n} < \nu(y) + \varepsilon$ for all $n \geq m$. For convenience, write $A = \nu(x) + \varepsilon$ and $B = \nu(y) + \varepsilon$, so this means $\|x^n\| < A^n$ and $\|y^n\| < B^n$ for $n \geq m$. Then for $n \geq 2m$, we have \begin{align*} \|(x + y)^n\| &= \left\|\sum_{k=0}^n \binom{n}{k} x^k y^{n-k}\right\| \\ &\leq \sum_{k=0}^n \binom{n}{k} \|x^k\|\|y^{n-k}\| \\ &= \sum_{k=0}^{m-1} \binom{n}{k} \|x^k\| \|y^{n-k}\| + \sum_{k=m}^{n-m} \binom{n}{k} \|x^k\| \|y^{n-k}\| + \sum_{k=n-m+1}^n \binom{n}{k} \|x^k\| \|y^{n-k}\| \\ &\leq \sum_{k=0}^{m-1} \binom{n}{k} \|x^k\| B^{n-k} + \sum_{k=m}^{n-m} \binom{n}{k} A^k B^{n-k} + \sum_{k=0}^{m-1} \binom{n}{k} \|y\|^k A^{n-k} \\ &\leq p(n) B^n + (A + B)^n + q(n) A^n \end{align*} where $p(n)$ and $q(n)$ are the polynomials in $n$ of degree $m-1$ given by $p(n) = \sum_{k=0}^{m-1} \binom{n}{k} \|x^k\| B^{-k}$ and $q(n) = \sum_{k=0}^{m-1} \binom{n}{k} \|y^k\| A^{-k}$. It follows that \begin{align*} \limsup_{n \to \infty} \|(x + y)^n\|^{1/n} &\leq \limsup_{n \to \infty} \left(p(n) B^n + (A + B)^n + q(n) A^n \right)^{1/n} \\ &= (A+B) \cdot \limsup_{n \to \infty} \left(1 + p(n) \left(\frac{B}{A+B}\right)^n + q(n) \left(\frac{A}{A+B}\right)^n \right)^{1/n} \\ &= A+B \end{align*} So $\nu(x + y) \leq \nu(x) + \nu(y) + 2 \varepsilon$. Since this holds for each $\varepsilon > 0$, $\nu(x + y) \leq \nu(x) + \nu(y)$.

Answer 2

Begin by fixing a $k\in\mathbb{N}$ and then suppose that $n$ is an even multiple of $k$. Since $x$ and $y$ commute we have that $||(x+y)^n||^{1/n}=||\sum\limits_{i=0}^n {n\choose i}x^iy^{n-i}||^{1/n}\leq (\sum\limits_{i=0}^n {n\choose i}||x^i||\text{ }||y^{n-i}||)^{1/n}=(\sum\limits_{0\leq i\leq n, \text{ }i\equiv 0 \bmod{k}}{n\choose i}||x^i||\text{ }||y^{n-i}||+\sum\limits_{0\leq i\leq n, \text{ }i\equiv 1 \bmod{k}} {n\choose i}||x^i||\text{ }||y^{n-i}||+\ldots+\sum\limits_{0\leq i\leq n, \text{ }i\equiv k-1 \bmod{k}} {n\choose i}||x^i||\text{ }||y^{n-i}||)^{1/n}.$

We now aim at estimating the above in terms of binomial expansions of $||x^k||$ and $||y^k||$, but to do that we have to use the submultiplicativity of the norm and pull out some factors; the above is no greater than:

$(\sum\limits_{0\leq i\leq n, \text{ }i\equiv 0 \bmod{k}}{n\choose i}||x^i||\text{ }||y^{n-i}||+||x||\text{ }||y^{k-1}||\sum\limits_{0\leq i\leq n, \text{ }i\equiv 1 \bmod{k}} {n\choose i}||x^{i-1}||\text{ }||y^{n-(k-1)-i}||+\ldots+||x^{k-1}||\text{ }||y||\sum\limits_{0\leq i\leq n, \text{ }i\equiv k-1 \bmod{k}} {n\choose i}||x^{i-(k-1)}||\text{ }||y^{n-1-i}||)^{1/n}.$

If we now make subsitution $j=i-1$ in the second sum, $j=i-2$ in the third sum etc. the above becomes:

$(\sum\limits_{0\leq i\leq n, \text{ }i\equiv 0 \bmod{k}}{n\choose i}||x^i||\text{ }||y^{n-i}||+||x||\text{ }||y^{k-1}||\sum\limits_{0\leq j< n, \text{ }j\equiv 0 \bmod{k}}{n\choose {j+1}}||x^j||\text{ }||y^{n-k-j}||+\ldots+||x^{k-1}||\text{ }||y||\sum\limits_{0\leq j< n, \text{ }j\equiv 0 \bmod{k}}{n\choose {j+k-1}}||x^j||\text{ }||y^{n-k-j}||)^{1/n}.$

The nice thing now is that we are only summing over indices that are multiples of $k$. Hence the above is not greater than:

$(\sum\limits_{0\leq i\leq n, \text{ }i\equiv 0 \bmod{k}}{n\choose i}||x^k||^{\frac{i}{k}}\text{ }||y^k||^{\frac{n}{k}-\frac{i}{k}}+||x||\text{ }||y^{k-1}||\sum\limits_{0\leq j< n, \text{ }j\equiv 0 \bmod{k}}{n\choose {j+1}}||x^k||^{\frac{j}{k}}\text{ }||y^k||^{\frac{n}{k}-1-\frac{j}{k}}+\ldots+||x^{k-1}||\text{ }||y||\sum\limits_{0\leq j< n, \text{ }j\equiv 0 \bmod{k}}{n\choose {j+k-1}}||x^k||^{\frac{j}{k}}\text{ }||y^k||^{\frac{n}{k}-1-\frac{j}{k}})^{1/n}.$

Now the sums above are very close to binomial expansions of $||x^k||$ and $||y^k||$. In fact, if in the first sum the coefficients $n\choose i$ were replaced by $n/k \choose i/k$ the sum would be a genuine binomial expansion. Similarly, if the coefficients in the other sums read $n/k-1 \choose j/k$ they would also be binomial expansions! The natural thing to do then is to in, for instance, the second sum rewrite $n \choose j+1$ as $\frac{n \choose j+1}{n/k-1 \choose j/k} {n \choose j+1}{n/k-1 \choose j/k}$ and they key observation is that this is no greater than $\frac{n!}{((n/2)!)^2}{n/k-1 \choose j/k}$ (those of you who are really paying attention will notice that here we use that $n$ is even). This sort of estimation will work for all coefficients so we will use it for all, but for ease we will write $C_n$ instead of $\frac{n!}{((n/2)!)^2}$. The above then becomes less than or equal to

$(C_n\sum\limits_{0\leq i\leq n, \text{ }i\equiv 0 \bmod{k}}{n/k \choose i/k}||x^k||^{\frac{i}{k}}\text{ }||y^k||^{\frac{n}{k}-\frac{i}{k}}+C_n||x||\text{ }||y^{k-1}||\sum\limits_{0\leq j< n, \text{ }j\equiv 0 \bmod{k}}{n/k-1 \choose j/k}||x^k||^{\frac{j}{k}}\text{ }||y^k||^{\frac{n}{k}-1-\frac{j}{k}}+\ldots+||x^{k-1}||\text{ }||y||\sum\limits_{0\leq j< n, \text{ }j\equiv 0 \bmod{k}}{n/k-1 \choose j/k}||x^k||^{\frac{j}{k}}\text{ }||y^k||^{\frac{n}{k}-1-\frac{j}{k}})^{1/n}=(C_n(1+||x||\text{ }||y^{k-1}||+||x^2||\text{ }||y^{k-2}||+\ldots ||x^{k-1}||\text{ }||y||))^{1/n}((||x^k||+||y^k||)^{n/k}+(||x^k||+||y^k||)^{n/k-1}+\ldots +(||x^k||+||y^k||)^{n/k-1})^{1/n}.$

Recall that there are $k$ sums in the last factor. Stirling's approximation shows that $C_n^{1/n}=(\frac{n!}{((n/2)!)^2})^{1/n} \rightarrow 1$ as $n \rightarrow \infty$. Hence $\limsup\limits_{n \rightarrow \infty}(C_n(1+||x||\text{ }||y^{k-1}||+||x^2||\text{ }||y^{k-2}||+\ldots ||x^{k-1}||\text{ }||y||))^{1/n}=1$. Moreover, $\limsup\limits_{n \rightarrow \infty}((||x^k||+||y^k||)^{n/k}+(||x^k||+||y^k||)^{n/k-1}+\ldots +(||x^k||+||y^k||)^{n/k-1})^{1/n}$ is easily seen to be $(||x^k||+||y^k||)^{1/k}$. Finally, this is less than or equal to $||x^k||^{1/k}+||y^k||^{1/k}$. This argument shows that for arbitrary $k\in \mathbb{N}$ that are even multiples of $n$, we have: $\limsup\limits_{n \rightarrow \infty} ||(x+y)^n||^{1/n}\leq ||x^k||^{1/k}+||y^k||^{1/k}$. Hence $\limsup\limits_{n \rightarrow \infty} ||(x+y)^n||^{1/n}\leq \limsup\limits_{k \rightarrow \infty}||x^k||^{1/k}+||y^k||^{1/k}$ as desired.