# On Fredholm operator on Hilbert spaces

Let $u: H \to H'$ be a continuous linear operator and $H,H'$ be Hilbert spaces. Let $u^\ast$ denotes its adjoint.

By definition, an operator $u$ is called Fredholm if and only if $\ker u$ has finite dimension and $\mathrm{im}(u)$ has finite codimension.

My book states that

"...$u$ is Fredholm if and only if $u(H)$ is closed and $\ker u$ and $\ker u^\ast$ are both finite dimensional"

I am not sure but I beleive that closedness of $u(H)$ is redundant, that is, the following is true:

$u$ is Fredholm if and only if $\ker u$ and $\ker u^\ast$ are finite-dimensional.

My proof:

$\implies$: If $u$ is Fredholm then by definition $\ker u$ is finite dimensional. Also, $(u(H))^\bot =\ker u^\ast$. If $V$ is a vector space and $U$ a subspace then the dimension of all complements of $U$ are equal, in particular, equal to the orthogonal complement. Hence $\ker u^\ast$ is finite dimensional.

$\Longleftarrow$: Let $\ker u$ and $\ker u^\ast$ be finite dimensional. By the same argument as before, $u(H)$ has finite codimension. Since $\ker u$ is finite dimensional it follows that $u$ is Fredholm.

Where is the mistake in my proof? What am I missing here?

• I'm confused by your proof. If the definition of "$u$ is Fredholm" is "$u(H)$ is closed, $\ker u$ is finite dimensional, and $\ker u^*$ is finite dimensional", then the $\implies$ direction of your theorem is trivial. And in the reverse direction I don't see that you have even tried to show that having $\ker u$ and $\ker u^*$ finite dimensional implies that $u(H)$ is closed (and indeed this implication is not true). – Nate Eldredge Jan 17 '15 at 3:40
• Whether $u(H)$ has finite codimension depends on how you define "codimension". Either way, it doesn't imply that $u(H)$ is closed. – Nate Eldredge Jan 17 '15 at 3:48
• As a counterexample, there are plenty of operators that are injective and have dense range (so that the adjoint is injective) but do not have closed range. One example is the operator $u$ on $L^2([0,1])$ defined by $(uf)(x) = xf(x)$. – Nate Eldredge Jan 17 '15 at 3:51
• @NateEldredge No the definition is different, I added it to my question. – user167889 Jan 17 '15 at 4:12

The problem is that "$u(H)^\perp$ has finite dimension" does not imply "$u(H)$ has finite codimension".

"Codimension" is meant in the algebraic sense. For a (not necessarily closed) linear subspace $E \subset H$, "$E$ has finite codimension" means (among other equivalent definitions) "there exists a finite dimensional subspace $F \subset H$ such that for every $x \in H$, there exist unique $y \in E$, $z \in F$ such that $x=y+z$". In other words, $H = E \oplus F$ as a direct sum of vector spaces.

Let's consider the example I gave in my comment: $H = L^2((0,1))$ (it's simpler if we leave off the endpoints) and $u$ defined by $(uf)(x) = xf(x)$. Note that $u$ is injective since if $uf = 0$ we have $x f(x) = 0$ and hence $f(x) = 0$ for almost every $x \in (0,1)$. So $\ker u = 0$ which is finite dimensional.

It's easy to check that $u$ is self-adjoint, so we have $\ker u^* = \ker u = 0$ as well.

Also, we can see directly that $u(H)^\perp = 0$. Suppose $g \in u(H)^\perp$; then in particular $\langle ug, g \rangle = 0$. This says that $\int_0^1 x g(x)^2\,dx = 0$, so that $x g(x)^2 = 0$ and hence $g(x) = 0$ almost everywhere.

But $u(H)$ has infinite codimension. For $0 < p < 1/2$, let $f_p(x) = x^{-p}$. Note that $f_p \notin u(H)$ since if we had $u g_p = f_p$, we would have to have $g_p(x) = x^{-p-1}$ almost everywhere; but those functions are not in $L^2((0,1))$. Moreover, the uncountable set of functions $\{f_p\}$ is linearly independent. So if we let $F$ be the linear span of all the $\{f_p\}$, then $F$ has infinite (even uncountable) dimension and $u(H) \cap F = 0$; the codimension of $u(H)$ is infinite (even uncountable).

So for this operator $u$, we have that $\ker u$ and $\ker u^*$ are both finite dimensional (actually zero dimensional) but $u$ is not Fredholm.

Another way to think about the issue is that the orthogonal complement $E^\perp$ is not in general an algebraic complement of $E$; it's an algebraic complement of the closure of $E$. (You may be confused since in finite dimensions, $E^\perp$ is an algebraic complement of $E$; but in finite dimensions all linear subspaces are closed.) In particular, having $E^\perp = 0$ does not imply $E = H$; it only implies that $E$ is dense in $H$, but dense subspaces can still have large codimension.

As a final remark, it is not true in general that a linear subspace $E$ with finite codimension is closed (consider the kernel of a discontinuous linear functional, which has codimension 1). However, it is true in the special case that $E$ is the image of a continuous operator on a complete space.

• In the second paragraph, why is $f(x) =0$ only for almost every $x \in (0,1)$? – user167889 Jan 18 '15 at 2:35
• @student: Maybe the phrasing could have been clearer: from $uf=0$ we can only conclude that $x f(x) = 0$ for almost every $x$, and for those $x$ we have $f(x)=0$. – Nate Eldredge Jan 18 '15 at 2:40
• I am sorry if my questions are stupid, I am really trying to understand this. Regarding the very first sentence in your answer: it seems to contradict the sentence from Encyclopedia of Maths that I quote here. The encyclopedia claims that all the complements have the same dimension... or at least it is how I understand it. (I am now about to study the second half of your answer, maybe it will already resolve my confusion....) – user167889 Jan 18 '15 at 2:41
• @student: When I say "$uf=0$", I mean "equal as elements of $L^2([0,1])$". Remember that formally, the elements of $L^2([0,1])$ are equivalence classes of functions mod almost everywhere equality. To be very pedantic, I should have defined $u([f])$ to be $[g]$ where $g$ is the function defined by $g(x)=xf(x)$, and verified that this is well defined. Then saying "$uf=0$" really means $u([f]) = [0]$, that is, the equivalence class $u([f])$ is the equivalence class of 0. – Nate Eldredge Jan 18 '15 at 2:48
• Regarding the Encyclopedia of Mathematics article: it's poorly written, since in a general vector space there is no notion of an orthogonal complement. In any vector space $X$, a complement of a vector subspace $E$ is, as in my second paragraph, a subspace $F$ such that for any $x \in X$ there exist unique $y \in E, z \in F$ such that $x=y+z$. In a Hilbert space, if $E$ is a closed subspace, then the orthogonal complement is a complement in this sense, but as my example shows, that is not true if $E$ is not closed. – Nate Eldredge Jan 18 '15 at 2:52

Although your proof is not right, your statement that closedness of u(H) is redundant in the definition of Fredholm operator is true, in that

$$H/uH$$ is finite-dimensional implies $$uH$$ is closed.

Proof. Suppose that $$[e_1],\cdots,[e_n]$$ is a basis of $$H/uH$$, and $$H_0=\operatorname{span}\{e_1,\cdots,e_n\}\subset H.$$ Then $$H=uH\oplus H_0$$.($$e_1,\cdots,e_n\notin uH$$, so $$H_0\cap uH=\{0\};$$ for any $$h\in H$$, suppose $$[h]=k_1[e_1]+\cdots+ k_n[e_n]$$, then $$h-k_1e_1-\cdots-k_ne_n\in uH$$ because $$[h-k_1e_1-\cdots-k_ne_n]=[h]-[h]=0$$, thus $$H=uH+H_0$$.)

Let $$\Phi: H/\ker u\times H_0\to H(=uH\oplus H_0), (h+\ker u,k)\mapsto uh+k$$, then $$\Phi$$ is a continuous linear bijection:

(1)$$\Phi$$ is well-defined. If $$(h+\ker u, k)=(h'+\ker u, k')$$, then $$h-h'\in \ker u$$ and $$k=k'$$, so $$u(h)+k=u(h')+k'$$.

(2)$$\Phi$$ is injective. If $$h\in H$$ and $$k\in H_0$$ satisfy $$uh+k=0$$, then $$uh=0$$ and $$k=0$$ for $$H=uH\oplus H_0$$, i.e., $$(h+\ker u,k)=0$$

(3)$$\Phi$$ is surjective. Of course.

Since that $$H_0$$ is finite-dimensional, $$H/\ker u\oplus H_0$$ is a Banach space, so we can apply the Open Mapping Theorem to $$\Phi$$, $$\Phi$$ has a continuous linear invert $$\Phi^{-1}$$, and thus is bounded below. Hence $$uH=\Phi(H/\ker u\times \{0\})$$ is closed.