Relevant concepts from functional analysis#

Vector space

A definition can be found on wiki.

Metric space

A metric space is a set \(X\) which is equipped with a distance function (or metric)

\[d(x,y): X \times X \to \RR.\]
Complete metric space

A metric space is called complete if every Cauchy sequence converges to some \(x \in X\).

Normed vector space

A vector space \((V, \|\cdot\|_V)\) consist of a vector space \(V\) which is equipped with a norm

\[\| \cdot \|_V : V \to \RR\]

Note that every norm induces a natural metric \(d(x, y) := \|x-y\|_V\). Typically we do not use the verbose notation \((V, \|\cdot\|_V)\), instead we simply speak of a normed vector space \(V\), and we omit the subscript \(_V\) in the norm symbol when the norm is clear from the context.

Banach space

A normed vector space which is complete with respect to the induced metric.

Inner product space

An inner product space \(\bigl(V, (\cdot, \cdot)\bigr)_V\) is a real (or complex) vector space \(V\) equipped with a inner product

\[(\cdot, \cdot)_V : V \times V \to \RR \quad (\text{or } \CC )\]

Every inner product induces a natural norm \(\| \cdot \| := \sqrt{(\cdot, \cdot)}\), and thereby a metric. Again, we typically do not use the verbose notation \(\bigl(V, (\cdot, \cdot)\bigr)\), instead we simply speak of a inner product space \(V\), and we often omit the subscript \(_V\) in \((\cdot, \cdot)_V\) symbol when the inner product is clear from the context.

Inner products satisfy the Cauchy-Schwarz inequality:

\[(u,v)_V \leqslant \|u\|_V \|v\|_V.\]
Bounded linear operator

A linear operator \(L: V \to W\) between two normed vector spaces \((V, \|\cdot\|_V)\) and \((W, \|\cdot\|_W)\) is call bounded if there is a constant \(C \in \RR^+_0\) such that

\[\| L v \|_W \leqslant \|v\|_V.\]

The operator norm \(\|T\|_{V\to W}\) of \(T\) is then the smallest such constant given by

\[\begin{split}\|L\| &= \inf \{C \in \RR^+_0 : \|L v \|_W \leqslant \|v\|_V \, \forall v \in V\} \\ & = \sup_{v \in V \setminus \{0\}} \dfrac{\|L v \|_W}{\|v\|_V} \\ & = \sup_{v \in V, \|v\|_V = 1} \|L v \|_W.\end{split}\]

It can be shown that the the following statements are equivalent for linear operators:

  • \(L: V \to W\) is bounded

  • \(L: V \to W\) is continuous

Exercise 1

Before you look up the proof, try to prove the previous claim yourself.

A linear operator \(l : V \to \RR \; (\text{or } \CC )\) is often called a linear functional or a linear form on \(V\).

Dual space

The dual space \(V^*\) for a normed vector space \((V, \|\cdot\|)\) consist of all continuous linear functionals defined on \(V\).

Note that for inner product spaces \(V\), every \(u \in V\) give rise to a continuous linear functional \(l_u\) defined by

\[l_u(v) := (v, u)_V \quad \forall v \in V.\]

For Hilbert space \(H\), that is in essence all the continuous linear functionals you can construct on \(H\) thanks to the following theorem.

Riesz representation theorem

Theorem 1 (Riesz representation theorem)

Let \(H\) be a Hilbert space with a inner product \((\cdot, \cdot)\). Then for every continuous functional \(l:H \to \RR\), there is a unique vector \(u_l \in H\) such that

\[l(v) = (u_l ,v) \quad \forall v \in H,\]

and we have that

\[\| l_u \|_{V^*} = \| u \|_{V}.\]

Proof. For a proof, we refer to Section 5.2 in [Brezis, 2011].

Later, when we have introduced the concept of weak formulation of partial differential equations, we will make heavily use of the Lax-Milgram theorem.

Theorem 2 (Lax-Milgram)

Given a Hilbert space \((V,\| \cdot\|)\), a bilinear form \(a(\cdot, \cdot): V \times V \to \RR\) (or \(\CC\)), and a linear form \(l(\cdot): V \to \RR\) (or \(\CC\)). Then the problem: Find \(u \in V\) such that

(1)#\[a(u, v) = l(v) \quad \forall v\in V\]

possesses solution a solution \(u \in V\) if the following assumptions are satisfied.

  1. The linear form \(l\) is bounded, i.e. there exists a constant \(C_l \geqslant 0\) such that

    (2)#\[| l(v) | \leqslant C_l \| v\| \quad \forall v \in V.\]

  2. The bilinear form \(a\) is bounded, i.e. there exists a constant \(C_a \geqslant 0\) such that

    (3)#\[| a(v, w) | \leqslant C_a \| v\| \|w\| \quad \forall v,w \in V.\]

  3. The bilinear form \(a\) is coercive, i.e. there is a constant \(\alpha > 0\) such that

    (4)#\[a(v, v) \geqslant \|v\|^2 \quad \forall v \in V.\]

Moreover, the solution \(u\) satisfies the stability (or a priori) estimate

(5)#\[\|u\| \leqslant \dfrac{C_l}{\alpha}\]

and is (therefore!) uniquely defined.

Proof. For a complete proof in particular the existence of a solution \(u\), we refer to the nice presentation in [Evans, 2010]. Here, we only show to derive (5) and uniqueness of \(u\).

Assume \(u\) solves (1). Then set \(v=u\) and successively employ the coercivity of \(a\) and boundedness of \(l\) to see that

\[\alpha \|u\|^2 \leqslant a(u, u) = l(u) \leqslant C_l \|u\|.\]

Dividing the previous chain of inequalities by \(\alpha\) and \(\|u\|\) if \(\| u \| \neq 0\) yields (5). For \(\|u\| = 0\) the stability estimate is trivially satisfied.

If \(u_1\) and \(u_2\) both satisfy problem (1), then thanks to linearity of \(a\) in the first slot, the difference \(u_1-u_2\) satisfies problem~(1) but with \(f = 0\) instead. In that case \(C_l = 0\) and thus \(0\leqslant\|u_1 - u_2 \| \leqslant \tfrac{0}{\alpha} = 0\), and thus \(u_1 = u_2\).

Remark 1

The Lax-Milgram theorem ensures that problem (1) is well-posed, i.e.,

  • Existence of a solution

  • Uniquessness of the solution

  • Continuous dependency on the data (or Stability) of the solution. In the particular case of Lax-Milgram theorem, stability is guaranteed through (5) which implies that “small changes” in \(a\) and \(l\) will only lead to small changes in the solution \(u\).