How to show that $A_m^s \to A^s$ if $A_m \to A\ $?

Question

Let $\{A_m\}_{m \geq 1}$ be a sequence of $n \times n$ complex self-adjoint matrices converging to an $n \times n$ matrix $A$ in norm. Then for any $s \in \mathbb N$ show that $A_m^s$ converges to $A^s$ in norm.

I have tried to write $A_m^s - A^s = (A_m - A)\ p(A_m,A)$ for some polynomial $p$ of two variables but I failed. Could anybody provide me some suggestion in this regard?

Thanks for investing your valuable time in reading my question.

Answer 1

Convergence in the norm is equivalent to convergence of all the entries of the matrices. So $(A_m)_{ij} \to A_{ij}$ for la $i,j$. Now $(A_m^{s})_{ij}$ is a finite sum of numbers of the type $(A_m)_{ik_1}(A_m)_{k_1k_1}\cdots (A_m)_{k_{s-1},j}$ so it converges to the $(i,j)-$ element of $A^{s}$.

Answer 2

You can very well use your idea. The key would be: Use that in finite dimensions all norms are topologically equivalent, so you can choose your norm. Take any submultiplicative matrix norm. Then use the submultiplicative property and the regular norm properties to bound $||p(A_m,A)||$ by some function of $\max\{||A||,||A_i||\}$. Then you get your result.

Edit: That is of course only if you find such Polynomial $P_m$ that are uniformly bounded, which I assumed have been found from the way the question was asked.

Alternatively have a look at this more general principle:

If $V,W,Z$ are metric spaces and $f_n,f:V\to W$ and $g_n,g:W\to Z$ equicontinuous. If $f_n\to f $ and $g_n\to g$ unif, then $f_n\circ g_n \to f\circ g$ unif.

Indeed: If $\epsilon>0$ there is a $\delta>0$ so that for $w,v\in W$ with $|w-v|<\delta$ we have $|g(w)-g(v)|<\epsilon$ and $|g_n(w)-g_n(v)|<\epsilon$. For n large enough we’ll have $||f-f_n||<\delta$ and $||g-g_n||<\epsilon$. So we have: $$ || g(f(x))-g_n(f_n(x)) || \leq || g(f(x)) - g_n(f(x))|| + || g_n(f(x)) - g_n(f_n(x)) < 2\epsilon $$