Lecture 09 - Linear Operator Equations

#math

Plaucherel Identity

If $f, g \in L^{1} (R^{n})$ , then $f * g \in L^{1} (R^{n})$ and $\hat{f * g} = \hat{f} \cdot \hat{g} \cdot (2 π)^{\frac{n}{2}}$
For any $f i n L^{2} (R^{n}), \exists \hat{f} \in L^{2} (R^{n})$ such that

\hat{f} (x) = \frac{1}{(2 π)^{\frac{n}{2}}} = \int_{R^{n}} f (x) e^{- i x y} d y

whenever $f \in L^{1} (R^{n}) \cap L^{2} (R^{n})$ and

| | f | |_{L^{2}} = | | \hat{f} | |_{L^{2}}

Linear Operator Equations

Given a linear operator $T : D (T) \subset X \to Y$ .
We want to solve:

T u = f

where $T$ and $f$ is given, i.e. we are looking for $u$ .

If $T$ is one to one (injective), that is if $K e r (T) = {0_{X}}$ then we may define the inverse operator $T^{- 1}$ :

T^{- 1} : R (T) \to D (T)

$T^{- 1}$ is linear whenever it exists. $T^{- 1}$ needs not be bounded even if $T$ is bounded, or it may be bounded even if $T$ is not.

Definition

$H^{1} ([0, 1]) = {f : [0, 1] \to R : f \in L^{2} ([0, 1]) & f^{^{'}} \in L^{2} ([0, 1])}$

$f^{'}$ is in a weak sense.
The norm ${| | f | |_{H^{1}}}^{2} = {| | f | |_{L^{2}}}^{2} + {| | f^{'} | |_{L^{2}}}^{2}$

Problem

Show that the following linear operators are bounded and find their norms:

$A : X = C ([0, 1]) \to C ([0, 1])$ with $A x (t) = \int_{0}^{t} x (s) d s$
$A : X \to X$ with $A x (t) = t^{2} x (0)$ .
$A : C ([- 1, 1]) \to C ([0, 1])$ with $A x (t) = x (t)$
$A : X \to X$ with $A x (t) = x (t^{2})$
$A : C^{1} ([0, 1]) \to X$ with $A x (t) = x (t)$
$A : C^{1} ([a, b]) \to C [(a, b)]$ with $A x (t) = \frac{d}{d t} x (t)$
$A : Y = L^{2} ([0, 1]) \to L^{2} ([0, 1])$ with $A x (t) = t \int_{0}^{1} x (s) d s$
$A_{λ} : Y \to Y$ with $A_{λ} x (t) = {\begin{cases} x (t) & t \leq λ \\ 0 & t > λ \end{cases}$
$A : Y \to Y$ with $A x (t) = \int_{0}^{t} x (τ) d τ$
$A : H^{1} ([0, 1]) \to H^{1} ([0, 1])$ with $A x (t) = x (t)$
$A : H^{1} ([0, 1]) \to H^{1} ([0, 1])$ with $A x (t) = t x (t)$
$A : H^{1} ([0, 1]) \to L^{2} ([0, 1])$ with $A x (t) = t x (t)$

For which $f$ does $\exists$ a solution $u$ of $T u = f$ ? In other words what is the range of $T$ and iIf a solution exists, is it unique? If not, how can we describe the set of all solutions?
Let us focus on Hilbert spaces. Let $X = Y = C^{n}$ , so $T u = A u$ for some matrix $A = a_{k j}$ then $R (T)$ is the column space of $A$ , that is the set of all linear combinations of the columns of $A$ .

R (T) = N (T^{*})^{⊥}

where $T^{*}$ is the matrix operator with matrix $A^{*}$ (the conjugate transpose aka adjoint of A).
A solution of $T u = f$ exists iff $f ⊥ v$ for every $v \in K e r (T^{*})$ . If ${v_{1}, . . ., v_{p}} = B_{K e r (T^{*})}$ , then it is equivalent to requiring $⟨ f, v_{k} ⟩ = 0$ for $k = 1, . ., p$ . So $p$ consistency conditions on $f$ , which are necessary and sufficient for the existence of a solution of:

T u = f

So $T^{*}$ plays a key role in solving $T u = f$ . $T^{*}$ satisfies the property that $⟨ T u, v ⟩ = ⟨ u, T^{*} v ⟩$ .