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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 $A+K$?
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1 Answer 1

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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 $(0,\frac{1}{2}, 0,\frac{1}{4},0,\frac{1}{6},\ldots)$.

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    $\begingroup$ 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 $\ker(A+K)$. $\endgroup$ Mar 24, 2020 at 16:12
  • $\begingroup$ 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. $\endgroup$ Mar 24, 2020 at 16:57

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