Continuous Function Spaces

Continuous Functions on an Interval

$a < x < b \to C (a, b)$ .

$C (a, b)$ is a vector space. Elements are the maps $f : C \to C$ , i.e. $f \in C (a, b)$ .

The set of values ${f (x)}$ represent the vector $| f ⟩ \in C (a, b)$ as $f (x) = ⟨ x | f ⟩$ in some dual basis ${⟨ x |}$ defined by this relationship, $⟨ x | f ⟩ = f (x)$ . C.f. $⟨ e_{i} | a ⟩ = a^{i}$ (in this case we would write $f (x)$ as $f^{x}$ ).

We define the inner product as $⟨ f | g ⟩_{w} = \int_{a}^{b} d x w (x) f (x)^{*} g (x)$ . This is an inner product space called $L_{2}^{w} (a, b)$ . C.f. $⟨ a | b ⟩ = \sum_{i} w_{i} a^{i *} b^{i}$ .

Where does the weight function come from? Consider change of coordinates: $∭ d^{3} x = ∭ d r d θ d φ r^{2} \sin θ$ .

We must require $⟨ f | f ⟩ < \infty$ , i.e. for $ℓ_{2}$ we must have ``square-integrable’’ functions (functions whose square has finite area over the prescribed interval).

However, $ℓ_{2}$ is not Cauchy-complete, while all $f$ are square integrable, we have sequences of functions in $ℓ_{2}$ that converge to functions not in the space. $(1 + \tanh n x) / 2$ converges to the step function, which is not square integrable.

So to get a Cauchy-complete inner product space (a Hilbert space) must include funtions with finite step discontinuitites.

Note that

Theorem

$L_{2}^{w} (a, b)$ is separable.

Result

$L_{2}^{w} (a, b)$ is a Hilbert space.

Orthogonality of $| x ⟩$ basis

Note that $⟨ f | g ⟩ = \int_{a}^{b} d x w (x) f (x)^{*} g (x) = \int_{a}^{b} d x w (x) ⟨ f | x ⟩ ⟨ x | g ⟩ = ⟨ f | \int_{a}^{b} d x w (x) | x ⟩ ⟨ x | g ⟩$ . So, $\int_{a}^{b} d x w (x) | x ⟩ ⟨ x | = I$ .

$| x^{'} ⟩ = I | x^{'} ⟩ = \int_{a}^{b} d x w (x) | x ⟩ ⟨ x | x^{'} ⟩ = \int_{a}^{b} d x w (x) | x ⟩ \frac{1}{w (x)} δ (x - x^{'}) = | x^{'} ⟩$ . Sometimes, $⟨ x | x^{'} ⟩ = \frac{δ (x - x^{'})}{\sqrt{w (x) w (x^{'})}}$ .

Exemplum Gratia

$d^{3} x = d x d y d z = r^{2} \sin θ d r d θ d φ$

$δ (\vec{r} - {\vec{r}}^{'}) = δ (x - x^{'}) δ (y - y^{'}) δ (z - z^{'}) = \frac{δ (r - r^{'})}{r} \frac{δ (θ - θ^{'})}{r \sin θ} δ (φ - φ^{'})$ .

Generalized Fourier Coefficients

Bases for $L_{2}^{w} (a, b)$

Aside

There are also $C^{k} (a, b)$ functions which are functions that are $k$ times continuously differentiable. $C^{ω}$ are the set of analytic functions, they are $C^{\infty}$ and their taylor series converges to the function.

Motivation

What kinds of things might we find in the Algebraic dual space of $L_{2}^{w} (a, b)$ . Everybody’s favorite example, the delta function $⟨ δ_{c} |$ defined by $⟨ δ_{c} | f ⟩ = f (c)$ , $a < c < b$ . Is there a function in $L_{2}^{w} (a, b)$ that is the dual of $⟨ δ_{c} |$ ? According to Reitz Representation Theorem, we expect a function such that $δ_{c} (x) = ⟨ x | δ_{c} ⟩$ for which $⟨ δ_{c} | f ⟩ = (| δ_{c} ⟩, | f ⟩) = \int_{a}^{b} d x w (x) δ_{c}^{*} (x) f (x) = f (c)$ for all $| f ⟩$ . The problem is that there is no Lebesgue square integrable function that will do that.

Lets see why not, what characteristics would such a function have to have. Choose $w (x) = 1$ . Then, $\int_{a}^{b} d x w (x) δ_{c} (x)^{*} f (x) = \int_{a}^{b} d x δ_{c} (x)^{*} f (x) = f (c)$ . But the integrand is zero except the measure zero value at $x = c$ of $f (x) / d x$ . Choose $d_{n} (x - c) = {n : c - 1 / (2 n) < c < + 1 / (2 n), 0 : else}$ . Then, $\int_{a}^{b} d x d_{n} (x - c) = 1$ . Then, $\int_{a}^{b} d x f (x) d_{n} (x - c) \approx [\int_{a}^{b} d_{n} (x - c)] f (c)$ . The assumption with the approximate is that $f$ is well behaved in the small slice. Taking the limit, $\int_{a}^{b} lim_{n \to \infty} d x δ_{c} (x - c) f (x) = \int_{a}^{b} d x δ_{n} (x - c) f (c) = f (c)$ . The problem is that $lim_{n \to \infty} d_{n} (x - c)$ does not exist, $lim = {\infty : x = c, 0 : otherwise}$ .

Thus, we cannot get the delta function, but can get close. Even though the limit does not exist, $lim_{n \to \infty} \int_{a}^{b} d x d_{n} (x - c) f (x) = f (c)$ , the integral does exist. Here is a case where you can’t interchange the limit and the integral.

So let’s get mathy, $d_{n} (x - c) \in L_{2} (a, b)$ but the limit as $n$ goes to infinity is not in $L_{2} (a, b)$ . But the limit of integrals with them (all) does. Start out with good functions - $C^{\infty}$ (infinitely differentiable) for which it does and its derivatives does not increase faster than any power of $| x |$ at infinity - Schwartz Space $Φ$ . If ${α_{n}}$ is any sequence of functions for which the limit of the integrals exists, $\forall f \in Φ$ , $lim_{n \to \infty} \int d x α_{n} (x) f (x) \equiv \int d x χ (x)^{*} f (x) \equiv ⟨ χ | f ⟩$ defines a perfectly good linear functional, $⟨ χ |$ on $L_{2} (a, b)$ and write $lim_{n \to \infty} α_{n} (x) = χ (x)$ (i.e. $lim_{n \to \infty} d_{n} (x - c) = δ_{c} (x)$ ).

$Φ \subset L_{2} (a, b)$ and $⟨ χ | \in Φ^{*}$ . These $⟨ χ |$ are called distributions, tempered distributions, or generalized functions.

$Φ \subset V \subset Φ^{- X}$ - rigged Hilbert space.

Rapid Oscillation

Rapid oscillations, oscillations more rapid than the function is changing, causes the integral to cancel out. Riemann-Lebesgue lemma: for a smooth $f (x)$ $lim_{a \to \infty} \int d x \exp (i a x) f (x) = 0$ .

Quantum Mechanic Aside

In Quantum Mechanics, we mean $⟨ x | \equiv ⟨ δ_{x} |$ and $⟨ p | \equiv ⟨ {\tilde{δ}}_{p} |$ .