functional_analysis.md

Functional analysis

Given a linear operator $K:\mathcal{U}\rightarrow\mathcal{V}$ we denote

• the domain of $K$: $\mathcal{D}(K) = \mathcal{U}$,
• the kernel or null-space of $K$: $\mathcal{N}(K) = {u \in \mathcal{D}(K) , | , Ku = 0}$,
• the range of $K$: $\mathcal{R}(K) = {Ku, | , u \in \mathcal{D}(K)}$,
• the operator norm of $K$: $|K| = \sup_{u\in\mathcal{U} \backslash {0}} \frac{|Ku|{\mathcal{V}}}{|u|{\mathcal{U}}}$.

We say that $K$ is continuous if for all $\epsilon > 0$ there exists a $\delta > 0$ such that

$$ |Ku - Kv|{\mathcal{V}} \leq \epsilon, \quad \text{with} \quad |u - v|{\mathcal{U}} \leq \delta. $$ It can be verified that for linear operators, continuity is equivalent to boundedness ($|K| < \infty$).

If in addition, $\mathcal{U}$ and $\mathcal{V}$ are Hilbert spaces we have the following

• the adjoint operator $\adjoint{K}$ is (uniquely) defined by $\langle Ku, v \rangle_{\mathcal{V}} = \langle u, \adjoint{K}v \rangle_{\mathcal{U}}$
• orthogonal complement of a subspace $\mathcal{X} \subset \mathcal{U}$: $\mathcal{X}^\perp = {u\in\mathcal{U},|, \langle u, v \rangle_{\mathcal{U}} = 0 , \forall v \in \mathcal{X}}$. If $\mathcal{X}$ is closed we have $\mathcal{U} = \mathcal{X} + \mathcal{X}^\perp$.
• orthogonal projection: let $\mathcal{X} \subset \mathcal{U}$ be a closed non-empty subspace, then the orthogonal projection onto $\mathcal{X}$ is denoted by $P_{\mathcal{X}}$ and obeys the following properties:
  • $\adjoint{P_{\mathcal{X}}} = P_{\mathcal{X}}$
  • $|P_{\mathcal{X}}| = 1$
  • $I - P_{\mathcal{X}} = P_{\mathcal{X}^{\perp}}$
• $\mathcal{R}(K)^\perp = \mathcal{N}(\adjoint{K})$, $\mathcal{R}(\adjoint{K})^\perp = \mathcal{N}(K)$, $\mathcal{N}(\adjoint{K})^\perp = \overline{\mathcal{R}(K)}$, $\mathcal{N}(K)^\perp = \overline{\mathcal{R}(\adjoint{K})}$
• orthogonal decomposition: $\mathcal{U} = \mathcal{N}(K) \oplus \overline{\mathcal{R}(\adjoint{K})}$ and $\mathcal{V} = \mathcal{N}(\adjoint{K}) \oplus \overline{\mathcal{R}(K)}$

An important class of operators are the \emph{compact operators}. An operator $K \in BL(\mathcal{U},\mathcal{V})$ in the space of bounded linear maps from Hilbert space $\mathcal{U}$ to $\mathcal{V}$ is finite dimensional if its range $\mathcal{R}(K)$ is a finite dimensional subspace of $\mathcal{V}$. $\mathcal{F}$ denotes the space of all finite dimensional operators.

An operator $K$ is compact (completely continuous) if

• the image of a bounded sequence ${u_j}$ contains a convergent subsequence ${Ku_j}$, or
• the closure of the image $\overline{K(B)}$ of any bounded $B$ set is compact, or
• if it is in the closure $\bar{\mathcal{F}}$ of $\mathcal{F}$, i.e. $K = \lim_{n \rightarrow \infty} K_n$ with $K_n$ finite dimensional operators.

Hence $\mathcal{C} := \bar{\mathcal{F}}$ is the space of all compact (completely continuous) operators. Compact operators have many nice properties, but also a particularly nasty one; if $\mathcal{R}(K)$ is infinite-dimensional then the pseudo-inverse of $K$ is not continuous.

We will broaden the setting a little and let $\mathcal{U}$ be a Banach space with some topology that is not necessarily induced by the underlying norm. We need this generality to be able to formally tackle some of the more fancy regularisation techniques. As before, we will not focus on the proofs too deeply and focus on the concepts. We will need the following concepts, however.

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With every Banach space $\mathcal{U}$ we can associate the dual space consisting of linear, continuous functionals on $\mathcal{U}$. For a given $v \in \mathcal{U}^*$ we denote the application of $v$ on $u$ as the dual product $\langle v,u\rangle$. As such, we can think of this as a way to generalise the concept of an inner product. However, the dual product is not generally symmetric.

The dual product also allows us to define the adjoint of a linear operator $K:\mathcal{U} \rightarrow \mathcal{F}$ as

$$\langle g, Ku\rangle = \langle K^*g, u\rangle \quad \forall u \in \mathcal{U}, g \in \mathcal{F}^*.$$

The main technical difficulties will arise when showing convergence of sequences in the usual way. For this, we need to introduce the notion of weak convergence.

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A sequence $\{u_k\}_{k\in\mathbb{N}} \subset \mathcal{U}$ is said to converge weakly to $u \in \mathcal{U}$ iff for all $v \in \mathcal{U}^*$ we have

$$\langle v, u_k\rangle \rightarrow \langle v,u\rangle.$$

We denote weak convergence by $u_k\rightharpoonup u$.