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The following is the problem:

Let $A:X\rightarrow Y$ be a linear mapping between finite dimensional inner product spaces. Let $\{x_n\}$ be a bounded sequence in $X$. ( I am taking this to mean that $\|x_i\|_X \leq M$ for some scaler $M$ $\forall i$. And $\| \quad \|_X$ is the induced norm in $X$. If this is the wrong interpretation, please let me know) Show that the sequence $\{Ax_n\} \in Y$ has a convergent subsequence.

My idea is the following:
We can show that any linear mapping between finite dimensional normed spaces is a bounded operator, therefore it can be shown that the sequence $\{Ax_n\}\in Y$ is a bounded sequence in $Y$ as well. Now we can use t$f:Y\rightarrow \mathbb{R}^m$ an isomorphism, where $m=dim(Y)$ , to get the sequence $\{f(Ax_n)\} \in \mathbb{R}^m$, which must also be bounded by the same reasoning, and then by The Bolzano-Weirstrass Theorem, there exists a convergent subsequence $\{f(Ax_{n_j})\}$ that converges to a number $L$ in $\mathbb{R}^m$, and the hope now is to take the inverse image under $f$ of this subsequence to get a converging subsequence of $\{Ax_n\}$ in $Y$ that converges to $f^{-1}(L)$. However, I am not sure how to show this.
I am also not sure if this is the most efficient way to to this. Any help is appreciated. Thank you.

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  • $\begingroup$ Show that any bounded sequence in $X$ has a Cauchy sub-sequence and that $A$ maps a Cauchy sequence to a Cauchy sequence,... You are correct about the norm being defined by the inner product. . $\endgroup$ – DanielWainfleet Mar 7 '17 at 1:13
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If you start with an n-dimensional inner product space $X$ over $\mathbb{R}$ (or $\mathbb{C}$) then you can find an orthonormal basis $\{ e_1,e_2,\cdots,e_n \}$ for $X$, and define an isometric linear isomorphism $$ U : \mathbb{R}^n \mbox{(or $\mathbb{C}^n$)} \rightarrow X $$ by $$ U(\alpha_1,\alpha_2,\cdots,\alpha_n) = \alpha_1 e_1 + \alpha_2 e_2 +\cdots + \alpha_n e_n. $$ In this way, your problem is reduced to looking at a problem on (complex) finite-dimensional Euclidean space.

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Since $f$ is an isomorphism, it has a bounded inverse. Since $\{f(Ax_{n_j})\}$ is convergent to $L$, we know $\{Ax_{n_j}\}=\{f^{-1}f(Ax_{n_j})\}$ is convergent to $f^{-1}(L)$, since $$\|Ax_{n_j}-f^{-1}(L)\|=\|f^{-1}\left(f(Ax_{n_j})-L\right)\|\leq\|f^{-1}\|\|f(Ax_{n_j})-L\|,$$ and the right-hand side can be made arbitrarily small.


To address your last question, this is the most efficient way to show this that I'm aware of.

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