Invertibility of compact operator on infinite dimensional Hilbert space

Question

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.

Answer 1

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 $Y$.

Answer 2

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.

Answer 3

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.