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Let $X,Y$ be two normed spaces $T:X \rightarrow Y$ a compact operator.If $R(T)$ is complete then $dim(R(T))< \infty$

$R(T)$ denotes the range of $T$.

Here is my proof:

Assume that $dim(R(T))= \infty$

Consider the sets $B_n=\{x \in X:||x|| \leq n\}$

We have that $R(T)=\bigcup_{n=1}^{\infty}cl(T(B_n))$ because:

If $y \in R(T)$ then $y=Tx$ where $x \in X$ now we have that $||x|| \leq [||x||]+1=m \in \mathbb{N}$ thus $$x \in B_m \Rightarrow y \in T(B_m) \subseteq cl(T(B_m)) \subseteq \bigcup_{n=1}^{\infty}cl(T(B_n))$$

If $y \in \bigcup_{n=1}^{\infty}cl(T(B_n))$ then exists $N \in \mathbb{N}$ such that $y \in cl(T(B_N)) \subseteq R(T)$

So $R(T)=\bigcup_{n=1}^{\infty}cl(T(B_n))$

From our hypothesis $R(T)$ is complete thus from Baire's category theorem exists $n_0 \in \mathbb{N}$ such that $ cl(T(B_{n_0})$ has an empty interior.

But $cl(T(B_{n_0})$ is a compact set because $T$ is compact.

So we have a contradiction because a compact subset of an infinite dimensional space has empty interior.

Is my proof correct?

I'm not sure if i have used Baire's theorem properly or if Baire's theorem is needed at all.

I would appreciate some opinions.

Thank you in advance.

Marios Gretsas
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1 Answers1

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A shorter proof: Let $Z=X/\ker(T)$ with the quotient norm.

Then we have a map $S:Z \to R(T)$ given by $(x+\ker T) \mapsto Tx$. This map is clearly injective and surjective, so if $R(T)$ is complete then by the bounded inverse theorem, it is an isomorphism of Banach spaces.

You may check that $S$ is itself a compact operator (since it is induced from $T$), and the only time a compact operator between two Banach spaces can be surjective is in finite dimensions! Indeed, by the open mapping theorem, $S(B)$ must contain a ball in $R(T)$ for any ball $B \subset Z$, and it is well-known that a ball in any infinite-dimensional Banach space contains a sequence with no Cauchy subsequence.

shalin
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  • The statement "the only time a compact operator can be surjective is in finite dimensions" seems a little fishy to me. Consider a compact operator $T : X\to Y$ such that $\operatorname{im}(T)$ is infinite-dimensional. Now, consider $\tilde{T} : X\to \operatorname{im}(T)$ defined by $\tilde{T}(x) = T(x)$. $\operatorname{im}(T)$ has infinite dimensions, and $\tilde{T}$ is surjective. Also, $\tilde{T}$ is compact because if $T(S)$ is compact in $Y$ for some bounded set $S$, then $\tilde{T}(S) = T(S)$ is compact in the subspace topology on $\operatorname{im}(T)$. – Michael L. Aug 17 '17 at 21:21
  • @MichaelLee Is im$(T)$ a Banach space though? I meant that a compact operator can only be surjective onto another Banach space in finite dimensions. – shalin Aug 17 '17 at 21:24
  • True, if $\operatorname{im}(T)$ is not closed, then it is not a Banach space. – Michael L. Aug 17 '17 at 21:25
  • and a compact operator $T$ has closed image iff it has finite-dimensional image (see here). – Michael L. Aug 17 '17 at 21:31
  • @MichaelLee It's an equivalent statement. The answer given there is essentially the same argument as my answer here, and I think is the most natural way to proceed. – shalin Aug 17 '17 at 21:34
  • Sure, although some authors (such as Kreyszig and whatever author the poster got this problem from) don't restrict the definition of "compact operator" to Banach spaces only. I would agree with your paragraph the way it's written now. – Michael L. Aug 17 '17 at 21:40
  • @Shalop +1 from me..thanks for the second solution – Marios Gretsas Aug 18 '17 at 00:43