Lecture 5

In this lecture we discuss another notion of convergence than pointwise convergence.

Banach and Hilbert spaces¶

Banach spaces¶

Normed vector spaces¶

Let $V$ be a complex vector space. A norm on $V$ is given by a non-negative real valued function $\|\cdot\|$ on $V$ such that

• $\|x\|\geq 0$ and $\|x\| = 0$ if and only if $x= 0$ (i.e. is the zero element);
• $\|x+y\|\leq \|x\|+ \|y\|$;
• $\|\lambda x\| = |\lambda| \|x\|$.

Completeness¶

$V$ is called complete if for every sequence $x_n$ such that

there exists some element $x\in V$ such that $\lim_{n\to \infty}\|x - x_n\| = 0$.

A sequence $x_n$ in $V$ is called Cauchy if it satisfies (1). By definition, it means for every $\epsilon > 0$, there exists $N$ (depending on ε) such that $\|x_n - x_m\| < \epsilon$ for every $m,n> N$. Thus

Lemma. Let $x_n$ be a Cauchy sequence, then $\|x_n\|$ is uniformly bounded, i.e. there exists $C>0$ such that $\|x_n\|\leq C$ for every $n$.

Proof. Choose any $\epsilon > 0$, then for some $N$ we have $\|x_n - x_m\|\leq \epsilon$ whenever $m,n \geq N$. In particular, $\|x_n - x_N\|\leq \epsilon$ for all $n> N$, so $\|x_n\|\leq \|x_N\| + \epsilon$. Take $C = \max_{1\leq m \leq N}\|x_m\|+\epsilon$.

Pre-Hilbert spaces¶

Let $V$ be a vector space over $\mathbb{C}$. An inner product on $V$ is a map $(,):V\times V\to \mathbb{C}$ satisfying

• Conjugate symmetry: $(x,y) = \overline{(y,x)}$.
• $(\alpha x,y) =\alpha (x,y)$.
  • By the conjugate symmetry, $(x,\alpha y) = \bar{\alpha}(x,y)$.
• $(x,x)\geq 0$.

If $V$ is finite-dimensional, an inner product corresponds to a Hermitian matrix via a basis. Let $A=(e_i,e_j)_{ij}$, then $(x,y) = \overline{\mathbf{y}^t}A x$. If $V$ is a real vector space, an inner product is bilinear symmetric. In the finite dimensional case, it corresponds to a positive definite matrix.

The Cauchy-Schwartz inequality¶

Let $(,)$ be an inner product, then $|(x,y)|^2\leq (x,x)(y,y)$.

Proof. If $y=0$, it trivially holds; If $y\neq 0$, let $\frac{(x,y)}{\|y\|}\frac{y}{\|y\|} = \frac{(x,y)y}{(y,y)}$ be the projection of $x$ to $y$ direction. Then let $x^{\perp}:=x - \frac{(x,y)y}{(y,y)}$. Now $(x^{\perp},x^{\perp})\geq 0$ gives the desired inequality.

Corollary. Let $\|x\|:=\sqrt{(x,x)}$, then $\|x\|$ is a semi-norm.

Proof. $(\|x\|+\|y\|)^2 = \|x\|^2 + \|y\|^2 + 2\|x\|\|y\|\geq \|x\|^2+\|y\|^2 +2|(x,y)| \geq \|x\|^2+\|y\|^2 +2Re(x,y) = (x+y,x+y) = \|x+y\|^2$.

The inner product space $V$ is called a Hilbert space provided that the induced semi-norm is a norm and is complete. We’ll call an inner product space a pre-Hilbert space.

Example. The space of continuous functions on $\mathbb{T}$, denoted by $\mathscr{C}^0(\mathbb{T})$, equipped with the inner product

is a pre-Hilbert space. The norm induced by the inner product is the $L^2$-norm. Its completion is the space of square Lebesgue integrable functions (see form example this note), denoted by $L^2(\mathbb{T})$, which is a Hilbert space.

Orthogonality¶

$x,y\in V$ is called orthogonal provided that $(x,y) = 0$, we denote this by $x\perp y$. Then we have the analogy of Pythagoren theorem

Lemma (Best approximation)¶

If $f - f_0$ is orthogonal to $V$. Then for every $g \in V$, $\|f - f_0\|\leq \|f - g\|$.

Proof. Since $f_0 - g\in V$, $f-f_0 \perp f_0 - g$. Then by Pythagorean theorem (3)

$\|f - g\|= \|f - f_0 + f_0 - g\| = \|f - f_0 \|+ \|f_0 - g\| \geq \|f - f_0\|+0$.

Mean-Square convergence of Fourier series¶

Now we apply the functional analysis result to Fourier series. Let $V_N$ be the complex vector space spanned by trignomic polynomials of degree $N$. Then for integrable $f$, $f - S_N f$ is orthogonal to $V_N$. Apply the best approximation lemma we can derive

Theorem¶

For $f\in L^2(\mathbb{T})$, $\|S_N f - f\|_{L^2}\to 0.$

Proof. Let $f$ be a continuous function on $\mathbb{T}$. Let $\sigma_N$ be the Cesaro sum of $f$. Then $\sigma_N\in V_N$. By Fejer’s theorem, $\sigma_N\to f$ uniformly, this implies $\|f - \sigma_N\|_{L^2}\to 0$. By best approximation lemma, $\|f - S_N\|_{L^2} \leq\|f-\sigma_N\|_{L^2}\to 0$.

To conclude the $L^2$ case. We need the fact that continuous functions are dense in $L^2$. This means for $f\in L^2(\mathbb{T})$ there exists some $h$ in $\mathscr{C}^0(\mathbb{T})$ such that $\|f - g\|_{L^2}<\epsilon$. The mean square convergence implies that trignomic polynomials are $L^2$- dense in $\mathscr{C}^0(\mathbb{T})$. Hence trignomic polynomials are dense in the $L^2$ space. It follows that for any $f\in L^2(\mathbb{T})$, there exists a trignomic polynomial $g$ such that $\|f-g\|_{L^2}\leq \epsilon$. Then $g\in V_N$ for some $N$. But $S_N f$ is the orthogonal projection of $f$ on $V_N$ so $\|f - S_N f\|\leq \|f - g\|< \epsilon$. This shows $\|f - S_N f \|\to 0$.