# Perturbing by a compact operator preserves closed range

Let $$A,K$$ be bounded operators on a Hilbert space $$H$$ such that $$K$$ is compact. How can we prove the following?

1. If $$A$$ has closed range, finite-dimensional kernel and infinite-dimensional cokernel, then so does $$A+K$$.
2. Same as above, but switching the words "kernel" and "cokernel".
3. If $$A$$ has infinite-dimensional kernel and cokernel, but closed range, can we make any conclusions regarding $$A+K$$?

## 1 Answer

For 1: Let $$\{x_n\}$$ be a sequence of elements in the unit ball of the $$\ker(A+K)$$; we will prove some subsequence converges, and this will imply that the unit ball in $$\ker(A+K)$$ is compact, proving that this kernel is finite-dimensional. By compactness, after passing to a subsequence, since $$||x_n||\le 1$$ we see $$Kx_n=-Ax_n$$ converges in $$H$$. Since $$A$$ has closed range, we have $$Kx_n \to -Ax$$ for some $$x\in H$$, so $$A(x_n-x)\to 0$$. Let $$y_n:=x_n-x$$; it suffices to prove some subsequence of the bounded sequence $$\{y_n\}$$ converges to an element of $$\ker(A)$$.

Since $$\ker(A)$$ is finite dimensional, we have $$z_n\in\ker(A)$$ with $$||y_n-z_n||=d(y_n,\ker(A))$$. Furthermore $$A(y_n-z_n)=Ay_n\to 0$$. Since $$\{z_n\}$$ is a bounded sequence in a finite dimensional space, after passing to some subsequence we find $$z_n\to z\in\ker(A)$$. We claim $$y_n\to z$$; it suffices to prove $$y_n-z_n\to 0$$. Letting $$w_n:=y_n-z_n$$ we have $$\{w_n\}$$ is bounded and $$Aw_n\to 0$$, plus $$||w_n||=d(w_n,\ker(A))$$, so $$w_n\in(\ker A)^\perp$$. Notice $$A:(\ker A)^\perp \to Ran(A)$$ is a bijection of Hilbert spaces and so by the Inverse Mapping Theorem we have $$||w||\le C||Aw||$$ for some $$C>0$$ for all $$w\in(\ker A)^\perp$$. So $$Aw_n\to 0$$ implies $$w_n\to 0$$, as desired.

Next we prove $$Ran(A+K)$$ is closed. Let $$f_n\in Ran(A+K)$$ and $$f_n\to f$$ and assume au contraire $$f\notin Ran(A+K)$$. Write $$f_n=Au_n+Ku_n$$, then by finite-dimensionality of the kernel we can find $$v_n\in\ker(A+K)$$ with $$||u_n-v_n||=d(u_n,\ker(A+K))$$. Then $$f_n=A(u_n-v_n)+K(u_n-v_n)$$. If $$w_n=\frac{u_n-v_n}{||u_n-v_n||}\in(\ker(A+K))^\perp$$ then $$\frac{f_n}{||u_n-v_n||} = Aw_n+Kw_n$$. This implies $$\{||u_n-v_n||\}$$ is bounded, otherwise for some subsequence we would have $$Aw_n+Kw_n\to 0$$. But $$||w_n||=1$$ so by compactness, some subsequence has $$Kw_n\to y$$ and so $$Aw_n\to -y$$ implying $$-y=Az$$ for some $$z\in(\ker A)^{\perp}$$ (since $$A$$ has closed range). Thus $$A(w_n-z)\to 0$$. But the previous paragraph also proves that if $$\pi: H\to\ker A$$ is the projection, then $$\sqrt{||w||^2-||\pi w||^2} \le C||Aw||$$ for any $$w\in H$$, thus $$||w_n-z||^2 - ||\pi(w_n-z)||^2\to 0$$, i.e. $$d(w_n-z, \ker A)\to 0$$. Since $$\{||w_n-z||\}$$ is bounded and $$\ker A$$ is finite-dimensional, after passing to a subsequence we can find $$z'\in\ker A$$ with $$w_n-z\to z'$$, i.e. $$w_n\to z+z'$$, which also implies $$(A+K)w_n\to (A+K)(z+z')$$ so $$z+z'\in\ker(A+K)$$. But $$d(w_n, \ker(A+K))=1$$ while $$d(z+z',\ker(A+K))=0$$ so $$w_n\to z+z'$$ is absurd.

So $$\{||u_n-v_n||\}$$ is indeed bounded, and compactness gives that after passing to a subsequence, $$K(u_n-v_n)\to z$$. Since $$f_n=(A+K)(u_n-v_n)\to f$$, we see $$A(u_n-v_n)\to f-z=Ay$$ for some $$y$$ since $$A$$ has closed range. So $$A(u_n-v_n-y)\to 0$$ giving by the same arguments as above, $$d(u_n-v_n-y, \ker(A))\to 0$$. Since $$\{||u_n-v_n-y||\}$$ is bounded and $$\ker(A)$$ is finite-dimensional, after passing to a subsequence we get $$u_n-v_n-y\to y'\in\ker(A)$$. So $$u_n-v_n\to y+y'$$ which implies $$(A+K)(u_n-v_n)\to (A+K)y+y'$$ so $$f=(A+K)(y+y')$$, proving $$f\in Ran(A+K)$$, as desired.

Finally, to prove $$Coker(A+K)$$ is infinite-dimensional, assume otherwise. Then $$A+K$$ is a Fredholm operator and so $$(A+K)-K=A$$ is also Fredholm , contradicting the fact that $$Coker(A)$$ is infinite-dimensional.

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For 2: apply 1 to $$A^*$$ and $$K^*$$ (which is also compact), using that $$Ran(A)$$ is closed iff $$Ran(A^*)$$ is closed (by the Closed Range Theorem), and that $$Coker(A)=H/Ran(A) \cong Ran(A)^\perp = \ker(A^*)$$ and $$\ker(A)\cong Coker(A^*)$$ and analogously for $$A+K$$.

For 3: No conclusion can be made. Use $$H=\ell_2$$, $$A(x_1,x_2,\ldots) = (x_1,0,x_3,0,x_5,\ldots)$$ which has closed range but infinite-dimensional kernel and cokernel, and $$K(x_1,x_2,\ldots) = (0,\frac{x_2}{2^2},0,\frac{x_4}{4^2},\ldots)$$. Then

$$(A+K)(x_1,x_2,\ldots) = \left(x_1, \frac{x_2}{2^2}, x_3, \frac{x_4}{4^2},x_5, \frac{x_6}{6^2},\ldots\right)$$ doesn't have closed range. Indeed, every sequence with $$0$$s everywhere in except finitely many places (i.e. "finite sequence") is in $$Ran(A+K)$$, and the closure of the set of finite sequences is $$\ell_2$$ itself, but $$A+K$$ isn't surjective because it doesn't hit $$(0,\frac{1}{2}, 0,\frac{1}{4},0,\frac{1}{6},\ldots)$$.

• In the third paragraph, the existence of $v_n$ has nothing to do with finite-dimensionality. As $A$ and $K$ are bounded, $\ker(A+K)$ is a closed subspace, so in particular it is a closed convex set. And in any Hilbert space you can realiza the distance to a closed convex set. But here all you need to do is take $v_n=Pu_n$, where $P$ is the orthogonal projection onto $\ker(A+K)$. Mar 24, 2020 at 16:12
• You're right! I was thinking that since the kernel is finite dimensional, any ball $B$ in it is compact, and so the function $B\to [0.\infty)$ taking $v\to ||u_n-v||$ must realize its minimum. So here we don't even need Hilbert-ness, as long as we take $B$ to have radius $2||u_n||$ or something. But I do end up using a lot of projection arguments later on, so it didn't make much sense to exclude it there. Mar 24, 2020 at 16:57