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Let $E$ and $F$ be normed vector spaces and let $\lambda_{n}$ be a sequence in $L(E,F)$, the set of continuous linear maps from $E\rightarrow F$. Assume $F$ is complete.

Let $v\in E$ and suppose that $\lambda_{n}(v)$ converges to a point in $F$. Write $\lambda(v)=\lim_{n\rightarrow \infty} \lambda_{n}(v)$.

How can I use this result to generalize to the statement that $\lambda_{n}(w)\rightarrow\lambda(w)$ $\forall w\in E$ i.e. how do I go from a single vector to an arbitrary vector?

Regarding a linear map as a matrix I can see that if $\lambda_{n}(v)\rightarrow\lambda(v)$ then the matrix $\lambda_{n}\rightarrow\lambda$ and so it can be applied to any vector but would anyone be willing to provide a more rigorous answer to the above question?

Another way to rephrase this question: Let $x,y\in E$ and suppose $\lambda_{n}(x)\rightarrow \lambda(x)$ and $\lambda_{n}(y)\rightarrow \lambda^{*}(y)$. Prove that $\lambda(z)=\lambda^{*}(z)$ $\forall z\in E$.

Mael
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    This is wrong in general, already for finite dimensional spaces. Set e.g. $v=0$. Then $\lambda_n(v)=0\rightarrow 0=\lambda(v)$ is trivially true. Also if you ignore that trivial case, it's still not true. – J.R. Jan 22 '14 at 17:26

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In general this is wrong, already for matrices. For example, set $v=0$. Then $\lambda_n(v)=0\rightarrow 0=\lambda(v),$ no matter what $\lambda_n,\lambda$ are exactly.

But also if we ask that $v\not=0$, the statement is wrong. Consider for example

$$\lambda_n = \pmatrix{1 & n\\ 0 & n}$$

Then $\lambda_n(e_1)=(1,0)^T$ is constant, i.e. converges, but for example $\lambda_n(e_2)=(n,n)^T$ clearly diverges.

J.R.
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  • That's a nice counter example. Thank you. Suppose I added the condition that $\lambda_{n}$ is a Cauchy sequence in $L(E,F)$, how would that change things? – Mael Jan 22 '14 at 18:16
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    Since $F$ is complete, $L(E,F)$ is a Banach space. Therefore $\lambda_n$ would converge (in the norm of $L(E,F)$). – J.R. Jan 22 '14 at 18:22