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I am trying to solve an exercise from the book Analysis for Applied Mathematics by Ward Cheney. Exercise 2.3.25 states: "Prove that if $X$ is an infinite-dimensional Hilbert space, then a compact operator on $X$ cannot be invertible."

I interpret the question to mean that the operator maps $X$ to a general normed linear space $Y$. I have been able to show that if the inverse exists, then it cannot be bounded/continuous. This is also covered in several questions on this site. This also takes care of the case where the operator maps to a Banach space because the interior/open mapping theorem then implies that if the inverse exists then it must be bounded/continuous.

Is the assertion in the exercise true if the operator maps $X$ to $Y$, where $Y$ is a normed linear space? After failing to prove this myself I have tried searching both this site, as well as google, but still have not found a proof. As I am learning functional analysis for the first time, I realize I might have overlooked something I did not quite understand. I have seen some mentions of spectral theory when searching, but this is not covered before the next chapter in the book.

Thanks in advance to anyone that took the time to read this post.

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  • $\begingroup$ You shall modify slightly the title of your question as it is not exactly what you ask at the end of your question... $\endgroup$
    – Nicolas
    Jun 13 at 12:46
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Let $(x_n)$ be a sequence in the closed unit ball of $X$. Then $T(x_n)$ has a convergent subsequence in $Y$, say $T(x_{n_k})$ by compactness of $T$. If $T$ has continuous inverse then $(x_{n_k})=(T^{-1} (Tx_{n_k}))$ would also be convergent. But this shows that close unit ball of $X$ is compact and this implies that $X$ is finite dimensional.

EDIT: It is customary in FA to say that a bounded operator is invertible if it has a bounded inverse. If you just want $T$ to be a bijection onto some normed linear space there are many examples. Let $T:\ell^{2} \to \ell^{2}$ be defined ny $T(x_n)=(\frac 1 n x_n)$. Then $T$ is a compact one-to-one operator . If $Y$ is the range of $T$ then $T$ is a bijection from $X$ onto $Y$.

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  • $\begingroup$ Thank you for your reply! If I understand correctly, this only shows that $T$ cannot have a continuous inverse, which as I wrote above I have been able to show myself in a similar fashion. This in turn implies that $T$ cannot have an inverse when$Y$ is a Banach space. What about the case where $T^{-1}$ is not required to be continuous? $\endgroup$
    – tfjaervik
    Jun 13 at 12:30
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    $\begingroup$ @tfjaervik I have edited my answer. $\endgroup$ Jun 13 at 12:45
  • $\begingroup$ Thanks a lot! I was just trying to check whether the previous example you gave satisfied the conditions in the question (I am a little slow being new to functional analysis). Really appreciate it! $\endgroup$
    – tfjaervik
    Jun 13 at 12:51
  • $\begingroup$ My first example was defined in a Banach space which was not a HIlbert space. So I changed the example. @tfjaervik $\endgroup$ Jun 13 at 12:52
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Alternatively, recall the following two facts:

(1) The identity operator $\operatorname{id}_X: X \to X$ is compact if and only if $X$ has finite dimension.

(2) If $S$ is compact and $T$ is any bounded operator, then $ST$ is compact.

Using this, we can easily prove your claim:

Assume that $S$ is compact and invertible. Then $\operatorname{id}= SS^{-1}$ is compact by $(2)$, and by $(1)$ it follows that $X$ has finite dimension.

Note: This does not rely on any properties of Hilbert spaces, i.e. the claim is true in a Banach space.

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  • $\begingroup$ Thanks! I had been able to show it in a similar fashion, but I was not aware that inverses of bounded operators usually means bounded inverses. Hence, I had been struggling to prove something that was not true, as the example in the accepted answer shows. $\endgroup$
    – tfjaervik
    Jun 13 at 12:58
  • $\begingroup$ @tfjaervik Usually, invertibility is meant to happen in some ring/algebra, here the algebra of bounded operators. But, in fact it doesn't matter in this context, because by the bounded inverse theorem, the set-theoretic inverse of a continuous linear operator between Banach spaces is automatic continuous again. See en.wikipedia.org/wiki/…. $\endgroup$ Jun 13 at 13:00
  • $\begingroup$ Thanks for the input! Indeed I mentioned the last part of your comment in my original post. I was struggling to prove the assertion for general normed linear spaces $Y$ ($T: X\rightarrow Y$) $\endgroup$
    – tfjaervik
    Jun 13 at 13:03
  • $\begingroup$ @tfjaervik In that case, you are right. You need to require that $X \to Y$ has continuous inverse. $\endgroup$ Jun 13 at 13:05
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I add an alternative to the (excellent) answer of Kavi Rama Murthy. This concerns infinite dimensional Hilbert spaces and assume as well that invertibility implies continuity of the inverse as it is generally required.

Let $K$ be a compact operator and let $(K_n)_{n\geq0}$ be a sequence of finite rank operators converging in norm topology to $K$. If $K$ is invertible, then $$K_nK^{-1}=(K_n-K+K)K^{-1}=(K_n-K)K^{-1}+\mathrm{Id}$$ is invertible for $n$ large enough by a Neumann series argument, and thus the finite rank operator $K_n$ is invertible, which is absurd.

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  • $\begingroup$ @KaviRamaMurthy The OP's background is that of infinite dimensional Hilbert spaces. $\endgroup$
    – Nicolas
    Jun 13 at 12:42
  • $\begingroup$ @KaviRamaMurthy Actually the end of the OP asks for something else indeed, so I will clarify what I am doing in my answer. $\endgroup$
    – Nicolas
    Jun 13 at 12:47

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