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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 $A$.

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    $\begingroup$ 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? $\endgroup$
    – t.b.
    Commented Sep 2, 2012 at 3:22

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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 $x=e_n$.

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