Why an Infinite-Rank Compact Operator Has Nonclosed Range

A compact operator can have infinite rank, but its range cannot then be closed—provided the domain and codomain are Banach spaces. In particular, this applies to bounded compact operators between Hilbert spaces.

The key reason is that compactness would make the unit ball of a closed range relatively compact. But the unit ball of an infinite-dimensional normed space is never compact. The quotient by the kernel lets us make this argument even when the operator is not one-to-one.

The theorem and its assumptions

Let $X$ and $Y$ be Banach spaces, and let $T:X\to Y$ be a compact bounded linear operator. If $T(X)$ is closed in $Y$, then $T(X)$ is finite-dimensional. Equivalently, if $T$ has infinite rank, its range is not closed.

Here, compact means that $T$ maps every bounded sequence to a sequence with a convergent subsequence. The completeness assumptions matter: they allow use of the bounded inverse theorem and ensure that a closed subspace of the codomain is complete. Hilbert spaces satisfy these assumptions.

Proof using the quotient by the kernel

The kernel $\ker T$ is closed because $T$ is bounded. Hence the quotient space $X/\ker T$, with its quotient norm, is a Banach space. The operator induces a map

$$\widetilde T:X/\ker T\to T(X),\qquad \widetilde T(x+\ker T)=Tx.$$

This induced map is linear and bijective: quotienting removes precisely the obstruction to injectivity, and its image is the range of $T$.

Now suppose the range $T(X)$ is closed in $Y$. It is then a Banach space with the norm inherited from $Y$. The bounded inverse theorem says that the inverse $\widetilde T^{-1}:T(X)\to X/\ker T$ is bounded. Thus there is a constant $C$ such that

$$\|x+\ker T\|_{X/\ker T}\le C\|Tx\|_Y.$$

This estimate means that every element in the unit ball of $T(X)$ has a preimage coset of bounded quotient norm. Each such coset has a representative in a fixed bounded ball of $X$ (allowing a small extra amount when choosing a representative close to the quotient norm). Since $T$ is compact, the image of that bounded ball is relatively compact. Consequently, the unit ball of $T(X)$ is relatively compact.

The unit ball is also closed in $T(X)$. Because $T(X)$ is complete, it follows that this unit ball is compact. Riesz’s lemma tells us that the closed unit ball of an infinite-dimensional normed space is not compact. Therefore $T(X)$ must be finite-dimensional.

A concrete example on ℓ²

Define $T:\ell^2\to\ell^2$ by

$$T(x_1,x_2,\ldots)=(x_1,x_2/2,x_3/3,\ldots).$$

This operator has infinite rank: each standard basis vector $e_n$ is mapped to the nonzero vector $e_n/n$, and these images are linearly independent.

It is compact. For each $N$, truncate the output after coordinate $N$ to obtain a finite-rank operator $T_N$. The operator norm of the difference is

$$\|T-T_N\|=\frac{1}{N+1},$$

which tends to zero. A norm limit of finite-rank operators is compact, so $T$ is compact.

To see directly that its range is not closed, take $y=(1,1/2,1/3,\ldots)$. This belongs to $\ell^2$ because the sum of $1/n^2$ converges. Its truncations $y^{(N)}=(1,1/2,\ldots,1/N,0,0,\ldots)$ are all in the range: each is the image of $(1,1,\ldots,1,0,0,\ldots)$ with $N$ initial ones. The truncations converge to $y$ in $\ell^2$. But $y$ itself is not in the range, because a preimage would have every coordinate equal to $1$, and that sequence is not in $\ell^2$. Thus the range contains a convergent sequence whose limit it does not contain.

Common mistakes

• Assuming compactness means finite rank. Compact operators can have infinite rank, as the diagonal example shows. The theorem says their range is then nonclosed, not that the operator cannot exist.
• Forgetting the completeness assumptions. The quotient-space proof uses the bounded inverse theorem. State that the spaces are Banach (or Hilbert) and that the operator is bounded.
• Treating the range as closed merely because it is a subspace. A linear subspace of a Banach space need not be closed. Closedness is an additional hypothesis being tested.
• Confusing closed range with a bounded inverse on the original domain. The operator need not be injective. The correct bijection is from $X/\ker T$ to $T(X)$.

A useful related fact is that a compact operator with closed range has finite rank, even if finite rank was not assumed in advance. In particular, the identity operator on an infinite-dimensional Banach space is not compact: its range is closed and infinite-dimensional, which would contradict that fact.