Lecture 10 - Solvability of Linear Operator Equations

#math

Conditions of the solvability of linear operator equations

We are interested in solving $T u = f$ for a given $T$ and $f$ , where $T$ is a bounded linear operator on a Hilbert space $H$ .

Proposition

If $T \in B (H)$ (that is $T$ is bounded operator from $H$ to $H$ ), then $K e r (T^{*}) = R (T)^{⊥}$
In particular $\overset{―}{R (T)} = K e r (T^{*})^{⊥}$

Proof

We know that the $R (T) \subset K e r (T^{*})^{⊥}$ and $K e r (T^{*}) \subset R (T)^{⊥}$ . We want to prove that $R (T)^{⊥} \subset K e r (T^{*})$ .
Let $v \in R (T)^{⊥}$ , then for any arbitrary $u \in H$ we have:

\begin{aligned} ⟨ u, T^{*} v ⟩ & = ⟨ T u, v ⟩ \\ = 0 & [Since T u \in R (T)] \end{aligned}

thus $T^{*} v = 0$ , i.e, $v \in K e r (T^{*})$ .
Note that if $E \subset X$ then $E^{⊥}$ is closed, one can show that $\overset{―}{E} = {E^{⊥}}^{⊥}$ thus $\overset{―}{R (T)} = K e r (T^{*})^{⊥}$

Corollary

If $T \in B (H)$ and $T$ has a closed $R (T)$ , then:

R (T) = K e r (T^{*})^{⊥}

That is to say $T u = f$ has a solution iff $f \in K e r (T^{*})^{⊥}$ .

Definition

If $T$ is any linear operator, we define the $Rank (T) = d i m (R (T))$ , and we say that $T$ is a finite rank operator whenever the $Rank (T)$ is finite.

Recall that any finite dim sub-space is closed.

Corollary

If $T \in B (H)$ is a finite rank operator then $R (T) = K e r (T^{*})^{⊥}$ .

Example

Let $H = L^{2} ([0, 1])$ and

T u (x) = \int_{0}^{1} x y u (y) d y

Then $R (T) = span {e}$ where $e (x) = x$ , so $Rank (T) = 1$ . Here $T$ is self-adjoint.

K e r (T^{*}) = K e r (T) = {e}^{⊥}

so the conclusion of the of corollary is trivial.

In general in $L^{2} (Ω)$ , $Ω \subset R^{N}$ with

T u (x) = \int_{Ω} K (x, y) u (y) d y

with $K (x, y) = \sum_{k = 1}^{M} φ_{k} (x) ψ_{k} (y)$ for some $φ_{k}, ψ_{k} \in L^{2} (Ω)$ . We assume that $φ_{k}, ψ_{k}$ are linearly independent. In this case:

T u (x) = \sum_{k = 1}^{M} φ_{k} (x) \int_{Ω} ψ_{k} (y) u (y) d y

So $R (T) = span {φ_{k}}$ , so that $rank (T) = M$ . The condition $f ⊥ K e r (T^{*})$ amounts to requiring infinitely many consistency conditions since $K e r (T^{*})$ has infinite dimension.
Since $K e r (T^{*})^{⊥}$ is always a closed subspace, $R (T) = K e r (T^{*})^{⊥}$ can only hold if $T$ has a closed range. The range is generally not always closed.

Fredholm operators and the Fredholm alternative

Definition

$T \in B (H)$ is of Fredholm type (a.k.a a Fredholm operator) if

$K e r (T), K e r (T^{*})$ are both finite dimensional.
$R (T) = \overset{―}{R (T)}$
For such an operator we define the index of $T$ , as

i n d (T) = d i m K e r (T) - d i m (K e r (T^{*}))

Fredholm operators of index 0 will be the most important. If one can show that an operator $T$ belongs to this class, then we obtain immediately the conclusion:

Uniqueness ⟺ Existence

In other words $T u = f$ has at most one solution for any $f \in H ⟺$ to the property that $T u = f$ has at least one solution. This is clear since if we have uniqueness then $d i m (K e r (T)) = 0 = d i m (K e r (T^{*}))$ and since $R (T) = K e r (T^{*})^{⊥}$ , thus $R (T)$ is the entire space.

Theorem

Let $T \in B (T)$ be a Fredholm operator of index 0. Then, either

$K e r (T) = K e r (T^{*}) = {0}$ and $T u = f$ has a unique solution for every $f \in H$
$d i m K e r (T) = d i m K e r (T^{*}) = M > 0$ then the equation $T u = f$ has a solution iff $f$ satisfies M compatibility conditions. $f ⊥ K e r (T^{*})^{⊥}$ and the general solution of $T u = f$ can be written as

{u = u^{*} + v, v \in K e r (T)}