# 7. Basic properties#

**Definition:** A **vector space** \(V\) over \({\mathbb R}\) is a set with
the operations \(+ : V \times V \rightarrow V\) and \(\cdot : {\mathbb R} \times V
\rightarrow V\) such that for all \(u,v \in V\) and \(\lambda, \mu \in {\mathbb R}\)
there holds

\(u + v = v + u\)

\((u+v)+w = u + (v+w)\)

\(\lambda \cdot (u+v) = \lambda \cdot u + \lambda \cdot v, \quad (\lambda + \mu) \cdot u = \lambda \cdot u + \mu \cdot u\)

Examples are \({\mathbb R}^n\), the continuous functions \(C^0([0,1])\), or the Lebesgue space \(L_2(\Omega)\).

**Definition:** A **normed** vector space \((V,\|\cdot\|)\) is a vector
space with the operation \(\| . \| : V \rightarrow {\mathbb R}\) being a norm, i.e.,
for \(u,v \in V\) and \(\lambda \in {\mathbb R}\) there holds

\(\| u + v \| \leq \| u \| + \| v \|\)

\(\| \lambda \, u \| = | \lambda | \, \| u \|\)

\(\| u \| = 0 \Leftrightarrow u = 0\)

Examples are \((C^0, \|\cdot\|_{\sup})\), or \((C^0, \|\cdot\|_{L_2})\).

**Definition:** In a **complete** normed vector space, Cauchy sequences
\((u_n) \in V^{\mathbb N}\) converge to an \(u \in V\). A complete normed vector space
is called **Banach space**.

Examples of Banach spaces are \((L_2, \|\cdot\|_{L_2})\), \((C^0, \|\cdot\|_{\sup})\), but not \((C^0, \|\cdot\|_{L_2})\).

**Definition:** The **closure** of a normed vector-space \((W, \| \cdot \|_V)\),
denoted as \(\overline{W}^{\| \cdot \|_V}\) is the smallest complete space containing \(W\).

Example: \(\overline{ C }^{\| \cdot \|_{L_2}} = L_2\).

**Definition:** A **functional** or a **linear form** \(l(\cdot)\) on \(V\) is a linear mapping
\(l(\cdot) : V \rightarrow {\mathbb R}\).
The canonical norm for linear forms is the **dual norm**

A linear form \(l(\cdot)\) is called bounded if the norm is finite. The vector space of all bounded linear forms on \(V\) is called the dual space \(V^\ast\).

An example for a bounded linear form is \(l(\cdot) : L_2 \rightarrow {\mathbb R} : v \rightarrow \int v \, dx\).

**Definition:** A **bilinear form** \(A(\cdot,\cdot)\) on \(V \times W\) is a mapping
\(A : V \times W \rightarrow {\mathbb R}\) which is linear in \(u\) and in \(v\).
It is called **symmetric** if \(V=W\) and \(A(u,v) = A(v,u)\) for all \(u,v \in V\).

Examples are the bilinear form \(A(u,v) = \int u v \, dx\) on \(L_2\), or \(A(u,v) := u^T A v\) on \({\mathbb R}^n\), where \(A\) is a (symmetric) matrix, or \(b(u,p) = \int \operatorname{div} u \, q \, dx\) with \(V = [C^1(\Omega)]^d\) and \(W=L_2(\Omega)\).

**Definition:** A symmetric bilinear form \(A(\cdot,\cdot)\) is
called an **inner product** if it satisfies

\(A(v,v) \geq 0 \; \forall \, v \in V\)

\(A(v,v) = 0 \Leftrightarrow v = 0\)

Often, it is denoted as \((\cdot,\cdot)_A\), \((\cdot,\cdot)_V\), or simply \((\cdot,\cdot)\).

An example on \({\mathbb R}^n\) is \(u^T A v\), where \(A\) is a symmetric and positive definite matrix.

**Definition:** An **inner product space** is a vector space \(V\) together
with an inner product \((\cdot, \cdot)_V\).

**Lemma:** **Cauchy Schwarz inequality.** If \(A(\cdot,\cdot)\) is a symmetric
bilinear form such that \(A(v,v) \geq 0\) for all \(v \in V\), then there holds

*Proof:* For any \(t \in {\mathbb R}\) there holds

If \(A(v,v) = 0\), then \(A(u,u)-2tA(u,v) \geq 0\) for all \(t \in {\mathbb R}\), which forces \(A(u,v) = 0\), and the inequality holds trivially. Else, if \(A(v,v) \neq 0\), set \(t = A(u,v) / A(v,v)\), and obtain

which is equivalent to the statement. \(\Box\)

**Lemma:** \(\| v \|_V := (v,v)_V^{1/2}\) defines a norm on the inner product
space \((V, (\cdot,\cdot)_V)\).

**Definition:** An inner product space \((V,(\cdot,\cdot)_V)\) which is
complete with respect to \(\|\cdot\|_V\) is called a **Hilbert space**.

**Definition:** A **closed subspace** \(S\) of an Hilbert space \(V\)
is a subset which is a vector space, and which is complete with respect to
\(\|\cdot \|_V\).

A finite dimensional subspace is always a closed subspace.

**Lemma:**
Let \(T\) be a continuous linear operator from the Hilbert space \(V\)
to the Hilbert space \(W\). The kernel of \(T\), \(\operatorname{ker} T := \{ v \in V : T v = 0 \}\) is a closed subspace of \(V\).

*Proof:* First we observe that \(\operatorname{ker} T\) is a vector space.
Now, let \((u_n) \in \operatorname{ker} T^{\mathbb N}\) converge to \(u \in V\). Since
\(T\) is continuous, \(T u_n \rightarrow T u\), and thus \(T u = 0\) and \(u \in \operatorname{ker} T\).
\(\Box\)

**Lemma:** Let \(S\) be a subspace (not necessarily closed) of \(V\). Then

is a closed subspace.

The proof is similar to the Lemma on the kernel.

**Definition:** Let \(V\) and \(W\) be vector spaces. A **linear operator** \(T : V \rightarrow W\) is a linear mapping from \(V\) to \(W\). The operator is called **bounded** if its operator-norm

is finite.

An example is the differential operator on the according space \(\frac{d}{dx} : (C^1(0,1), \| \cdot \|_{sup} + \| \frac{d}{dx} \cdot \|_{sup}) \rightarrow ( C(0,1), \| \cdot \|_{sup})\).

**Lemma:** A bounded linear operator is continuous.

*Proof:* Let \(v_n \rightarrow v\), i.e. \(\| v_n - v \|_V \rightarrow 0\).

Then \(\| T v_n - T v \| \leq \| T \|_{V \rightarrow W } \| v_n - v \|_V\) converges to 0,
i.e. \(T v_n \rightarrow T v\). Thus \(T\) is continuous.

**Definition:** A **dense subspace** \(S\) of \(V\) is such that every element of \(V\)
can be approximated by elements of \(S\), i.e.

**Extension principle:** Let \(S\) be a dense subspace of the normed space \(V\),
and let \(W\) be a complete space. Let \(T : S \rightarrow W\) be a bounded linear operator
with respect to the norm \(\| T \|_{V \rightarrow W}\). Then, the operator can be uniquely extended onto \(V\).

*Proof:* Let \(u \in V\), and let \(v_n\) be a sequence in \(S\) such that \(v_n \rightarrow u\). Thus, \(v_n\) is Cauchy. \(T v_n\) is a well defined sequence in \(W\). Since \(T\) is bounded, \(T v_n\) is also Cauchy. Since \(W\) is complete, there exists a limit \(w\) such that \(T v_n \rightarrow w\). The limit is independent of the sequence, and thus \(T u\) can be defined as the limit \(w\).

**Definition:** A bounded linear operator \(T : V \rightarrow W\) is called **compact** if for every bounded sequence \((u_n) \in V^{\mathbb N}\), the sequence \((T u_n)\) contains a convergent sub-sequence.

**Lemma:** Let \(V, W\) be Hilbert spaces. An operator is
compact if and only if there exists a complete orthogonal system
\((u_n)\) for \((\operatorname{ker} T)^\bot\) and values \(\lambda_n \rightarrow 0\) such that

This is the eigensystem of the operator \(K : V \rightarrow V^\ast : u \mapsto (T u, T \cdot)_W\).

*Proof:* (sketch) There exists an maximizing element of \(\frac{ (Tv, Tv)_W } { (v,v)_V }\).
Scale it to \(\| v \|_V = 1 \) and call it \(u_1\), and \(\lambda_1 = \frac{ (Tu_1, Tu_1)_W } { (u_1,u_1)_V }\). Repeat the procedure on the \(V\)-complement of \(u_1\) to generate \(u_2\), and so on.