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Let $X,Y$ be Banach space and $T:X\to Y$ bounded . If $T(X)$ has finite dimensional, then $T$ is compact.

Assume $\{y_i\}_{i=1}^N$ be a basis of $T(X)$ , $\{x_n\}$ be a bounded sequence in $X$, we have $$T(x_n)=\sum_1^N a_i(x_n)y_i$$ I want to show $\sup_{1\le i\le N}\sup_n |a_i(x_n)|\le M \lt \infty$. Since $\{z\in C| \,\,|z|\le M\}$ is compact , we can use subsequence trick to extract a convergent subsequence $\{T{x_n}_k\}$ .

Since $T(X)$ is compact in $Y$ , in particular , $T(X)$ is bounded . If for a subsequence ${x_n}_k$ , we have $|a_i({x_n}_k)| \to \infty$ , it seems that this can lead to a contradiction . So let's consider the following question:

Let $Y$ be a Banach space , $A=span\{y_1,...,y_n\}$ be a $n$ dimensional subspace of $Y$ . $B$ is a bounded subspace of $A$. Then for each $y=\sum a_i(y) y_i \in B$ can we show that $$\sup_{1\le i \le n} \sup_{y \in B} |a_i(y)| \lt \infty $$ ?

If $Y$ is a Hilbert space , then we can orthogonormal $\{y_i\}$ and then the result follows . If we only assume $Y$ be a Banach space , can we have the same result?

gdj
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    All norms are equivalent on a finite-dimensional space. You can assume $A$ is a Hilbert space if you find it useful. – Captain Lama Jan 12 '20 at 17:46

2 Answers2

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Let $B$ be the closed unit ball of $X$. Then, $T(B)$ is a bounded subset of a finite-dimensional space and is therefore precompact. Therefore, $T$ is compact.

ncmathsadist
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You want to show that the Theorem of Heine-Borel holds in any finite-dimensional Banach space (ie a set is compact iff it is bounded and closed).

Now consider the linear isomorphism $$F: T(X)\rightarrow \mathbb{R}^n, \sum_{j=1}^n a_j y_j \mapsto (a_1, \dots, a_n).$$ We know that the Heine-Borel theorem is true for $\mathbb{R}^n$ and that both boundedness and compactness are preserved by linear isomorphism. Hence, we get the Heine-Borel theorem also for $T(X)$ and we can conclude that $T$ is compact.

Severin Schraven
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