Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A sequence $\{A_n\}_{n=1}^\infty$ of linear operators $A_n\in \mathcal{B}(X)$ converges uniformly (or in the norm) to an operator $A\in \mathcal{B}(X)$ if $\|A_n-A\|\rightarrow 0$ as $n\rightarrow \infty$, and the sequence $A_n\in \mathcal{B}(X)$ is said to converges strongly to $A\in \mathcal{B}(X)$ if $\|A_nx-Ax\|\rightarrow 0$ as $n\rightarrow \infty$ for every $x\in X$.

What is wrong with the following argument. If $A_n$ converges strongly to $A$, then $\|A_n-A\|=\sup\limits_{x\ne 0}\frac{\|(A-A_n)x\|}{\|x\|}=\sup\limits_{x\ne 0}\frac{\|Ax-A_nx\|}{\|x\|}\rightarrow 0$, showing that $A_n$ converges uniformly to $A$.

share|improve this question
2  
How do you justify writing $\to 0$ in your last line? (You tacitly swap a limit and $\sup$). Consider what happens when $S$ is the left shift operator on $\ell^p(\mathbb{N})$ and $A_n = S^n$, $A = 0$. You didn't say what $X$ is supposed to be, by the way: A Banach space, a Hilbert space, something else? –  t.b. Sep 2 '12 at 3:22

1 Answer 1

up vote 1 down vote accepted

In a case like this, it is useful to see what happens in an appropriate example. Let $H$ be a separable Hilbert space. Fix an orthonormal basis $\{e_n\}$ and let $\{E_n\}$ be the family of projections where $E_n$ is the orthogonal projection onto $\mathbb{C}e_n$.

Then $\|E_n\|=1$ for all $n$, but $E_nx\to0$ for all $x\in H$. Indeed, $E_nx=x_ne_n$, where $x_n$ is the $n^{\rm th}$ coefficient of the representation of $x$ in the basis $\{e_n\}$.

The $\sup$ in your last line is $$ \sup_{x\ne0}\frac{\|E_nx\|}{\|x\|}=\sup_{x\ne0}\frac{x_n}{\left(\sum_n|x_n|^2\right)^{1/2}}=1. $$ That the $\sup$ equals $1$ is easily seen by evaluating at $x=e_n$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.