How similar are $l_2$ and $L_2[a,b]$?

Question

The former space is the space of infinite sequence Having a norm similar to Euclidean space. The latter is square integrable functions. Is there isometry or isomorphism between them?

Answer 1

They are isometrically isomorphic. And the proof is very simple: all you need is that both have a countable orthonormal basis. So take orthonormal bases $\{e_n\}$ and $\{f_n\}$ of $\ell^2$ and $L^2$ respectively.

Now define, on the span of $\{e_n\}$, $$ W\sum_{n=1}^m a_n e_n=\sum_{n=1}^m a_n f_n. $$ We have \begin{align} \left\|W\sum_{n=1}^m a_n e_n\right\|^2 &=\left\|\sum_{n=1}^m a_n f_n\right\|^2=\sum_{n=1}^m|a_n|^2=\left\|\sum_{n=1}^m a_n e_n\right\|^2. \end{align} So $W$ is linear an isometric, and thus it extends to a linear operator $W:\ell^2\to L^2$, still isometric.

It only remains to see that $W$ is onto: $$ \sum_n a_n f_n=W\sum_n a_n e_n. $$

Note that the above has nothing to do with the particulars of $\ell^2$ and $L^2$: any two Hilbert spaces of the same dimension are isometrically isomorphic.

Answer 2

I'll explain David Ullrich's comment for the sake of completeness.

For convenience, I'll work with the case $a = 0$ and $b = 1$. Otherwise, one should modify the basis functions $\{e_{n}\}_{n \in \mathbb{Z}}$ (introduced below) accordingly.

Next, if $f \in L^{2}([0,1])$ and $n \in \mathbb{N}$, define the $n$th Fourier coefficient $\hat{f}(n)$ by $$\hat{f}(n) = \int_{0}^{1} f(x) e^{- i 2 \pi n x} \, dx.$$
I will prove that the function $\mathcal{F} : f \mapsto \hat{f}$ is an isometric isomorphism from $L^{2}([0,1])$ to $\ell^{2}(\mathbb{Z})$.

Let $e_{n} : [0,1] \to \mathbb{C}$ be given by $$e_{n}(x) = e^{i 2 \pi n x}.$$
First, I will show $\{e_{n}\}_{n \in \mathbb{Z}}$ is a complete orthonormal system in $L^{2}([0,1])$ (i.e. an orthonormal set with dense span). Orthonormality is a calculus exercise: $$\int_{0}^{1} e_{n}(x) \overline{e_{m}(x)} \, dx = \int_{0}^{1} e^{i 2 \pi (n - m) x} \, dx = \delta_{nm}.$$ Density is a bit more work.

Let $\mathcal{A} = \text{span} \{e_{n} \, \mid \, n \in \mathbb{Z}\}$. Then $\mathcal{A} \subseteq C([0,1])$ is a self-adjoint algebra that separates points and vanishes nowhere, i.e. if $x \neq y$, then there is a $f \in \mathcal{A}$ such that $f(x) \neq f(y)$, and if $x \in [0,1]$, then there is a $f \in \mathcal{A}$ such that $f(x) \neq 0$. I leave it to the reader to verify these claims. (Hint: Use trigonometric functions.) By the Stone-Weierstrass Theorem, $\mathcal{A}$ is dense in $C([0,1])$ (with respect to the supremum norm). Note that density of $\mathcal{A}$ can also be proved using the Fejer kernel (or other "approximations of the identity"), which is certainly worth looking into.

Since the inclusion $i : C([0,1]) \to L^{2}([0,1])$ is continuous with dense image, it follows that $\mathcal{A}$ is dense in $L^{2}([0,1])$. This proves that $\{e_{n}\}_{n \in \mathbb{Z}}$ is a complete orthonormal system in $L^{2}([0,1])$.

We conclude that if $f, g \in L^{2}([0,1])$, then \begin{align*} \int_{0}^{1} f(x) \overline{g(x)} \, dx &= \sum_{n \in \mathbb{Z}} \langle f, e_{n} \rangle_{L^{2}([0,1])} \langle e_{n}, g \rangle_{L^{2}([0,1])} \\ &= \sum_{n \in \mathbb{Z}} \hat{f}(n) \overline{\hat{g}(n)} \end{align*} where the last equality follows from the definition of the Fourier coefficients. This proves that $\mathcal{F} : f \mapsto \hat{f}$ is an isometric isomorphism.

The $L^{2}$-limit $f(x) = \sum_{n \in \mathbb{Z}} \hat{f}(n) e^{i 2 \pi n x}$ is one way to introduce the study of Fourier series and harmonic anlaysis. The RHS is called the Fourier series of $f$.

Answer 3

They are isomorphic as inner-product spaces.

First let us be clear about what $\ell^2$ is. It is the space of sequences $(c_n)$ for which $\displaystyle\sum_n |c_n| < \infty.$

If $c$ is the sequence of Fourier coefficients of a square-integrable function $f$ on $[a,b],$ then the correspondence $c \longleftrightarrow f$ is an isomorphism – a linear isometry. See Riesz–Fischer theorem.

Answer 4

You already know that $L^2[a,b]$ has a countable orthonormal basis $(f_n)_{n=1}^\infty$.

Define a linear map $A : L^{2}[a,b] \to \ell^2$ as

$$Af = \Big(\langle f, f_n\rangle\Big)_{n=1}^\infty = \Big(\langle f, f_1\rangle, \langle f, f_2\rangle, \langle f, f_3\rangle, \ldots\Big)$$

for all $f \in L^2[a,b]$.

$A$ is an isometry because of Parseval's identity:

$$\|f\|^2_2 = \sum_{n=1}^\infty \left|\langle f, f_n\rangle\right|^2 = \left\|\Big(\langle f, f_n\rangle\Big)_{n=1}^\infty\right\|^2_2 = \|Af\|^2_2$$

$\operatorname{Im} A$ is clearly dense in $\ell^2$ because it contains the canonical basis $(e_n)_{n=1}^\infty$ for $\ell^2$. Namely $Af_n = e_n$ for all $n \in \mathbb{N}$.

Also, $\operatorname{Im} A$ is complete because $L^2[a,b]$ is complete and linear isometries preserve completeness. In particular, $\operatorname{Im} A$ is a closed subspace of $\ell^2$ so:

$$\ell^2 = \overline{\operatorname{Im} A} = \operatorname{Im} A \subseteq \ell^2 \implies \operatorname{Im} A = \ell^2$$

Therefore, $A$ is an isometric isomorphism between $L^2[a,b]$ and $\ell^2$.

In general, any separable inner product space is isometrically isomorphic to a dense subspace of $\ell^2$, via the same isomorphism. Furthermore, separable Hilbert spaces are isometrically isomorphic to $\ell^2$.

Answer 5

Hint $L^2([a,b])$ is separable

Use this to construct a bijection between ONB.