# Uniform limit of finite-rank operators with the same rank.

Let $\{T_n\in\mathcal{B}(X)\,|\,\text{rank}(T_n)=R\,\}^{\infty}_{n=1}$ is a sequence of linear bounded finite-rank operators on a Banach space with the same rank $R$. Let it converge uniformly to an operator $T\in\mathcal{B}(X)$, that is $\Vert T_n - T\Vert_{\mathcal{B}(X)}\longrightarrow 0\;(n \rightarrow\infty)$, that is $\forall\epsilon>0\;\exists N\in\mathbb N\;\forall n\ge N\;\forall x\in X\;|\;\Vert x\Vert_{X}\le 1$ holds $\Vert T_n x - T x\Vert_X\le\epsilon$.

Prove that $\text{rank}(T)\le R$.

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Michael, you will usually get more effective answers here if you explain what you tried. As is, my suggestion is: try arguing by contradiction. – user53153 Dec 26 '12 at 20:05
@PavelM Thanks! I'll try. – mgozhev Dec 26 '12 at 20:20

I'll assume we are working over a real Banach space; the complex case is similar.

Suppose $x_1, \dots, x_{R+1} \in X$. We will show that $Tx_1, \dots, Tx_{R+1}$ are linearly dependent.

For each $n$, the rank of $T_n$ is $R$, and so $T_n x_1, \dots, T_n x_{R+1}$ are linearly dependent. Thus there are coefficients $a_{n,1}, \dots, a_{n,R+1} \in \mathbb{R}$, not all zero, such that $\sum_{i=1}^{R+1} a_{n,i} T_n x_i = 0$. By rescaling, we may assume that $\sum_{i=1}^{R+1} |a_{n,i}|^2 = 1$, i.e. $(a_{n,1}, \dots, a_{n,R+1})$ lies in the unit sphere $S^R$ of $\mathbb{R}^{R+1}$. Now $S^R$ is compact, so we can extract a subsequence $(a_{n_k, 1}, \dots, a_{n_k, R+1})$ converging to some $(a_1, \dots, a_{R+1}) \in S$. In particular, for each $i$ we have $a_{n_k, i} \to a_i$, and the $a_i$ are not all zero. But since $T_{n_k} \to T$ uniformly, we have $$\sum_{i=1}^{R+1} a_i T x_i = \lim_{k \to \infty} \sum_{i=1}^{R+1} a_{n_k, i} T_{n_k} x_i = 0. \quad (\ast)$$

Note it was sufficient that $\operatorname{rank}(T_n) \le R$ for all but finitely many $n$, so we have in fact shown that rank is lower semicontinuous with respect to the uniform topology on $\mathcal{B}(X)$.

The same argument goes through if $T_n \to T$ in the strong operator topology (SOT), or even the weak operator topology (WOT, intepreting the limit in (*) as a weak limit in $X$). We could also replace the sequences with nets to get the strongest statement: rank is WOT-lower semicontinuous on $\mathcal{B}(X)$.

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@julien: The third sentence is "We will show that $Tx_1, \dots, Tx_{r+1}$ are linearly dependent". Does that help? This will imply that the rank of $T$ is at most $R$. – Nate Eldredge Apr 18 '13 at 1:11
Oh boy...I read independent wherever you wrote dependent. That's what I said, I'm dumb. Thank you for your patience. – 1015 Apr 18 '13 at 1:14
And now that I can see, this is a nice answer, +1. – 1015 Apr 18 '13 at 1:16