# Continuous Function Spaces

## Continuous Functions on an Interval

\(a< x < b\to\mathcal{C}(a,b)\).

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

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

We define the inner product as \(\langle f|g\rangle_w = \int_a^b dx w(x)f(x)^*g(x)\). This is an inner product space called \(L_2^w(a,b)\). C.f. \(\langle a|b\rangle = \sum_i w_ia^{i*}b^{i}\).

Where does the weight function come from? Consider change of coordinates: \(\iiint d^3 x = \iiint dr d\theta d\varphi r^2\sin\theta\).

We must require \(\langle f|f\rangle < \infty\), i.e. for \(\ell_2\) we must have ``square-integrable’’ functions (functions whose square has finite area over the prescribed interval).

However, \(\ell_2\) is not Cauchy-complete, while all \(f\) are square integrable, we have sequences of functions in \(\ell_2\) that converge to functions not in the space. \((1+\tanh nx)/2\) 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 \(|x\rangle\) basis

Note that \(\langle f|g\rangle = \int_a^b dx w(x)f(x)^*g(x) = \int_a^b dxw(x)\langle f|x\rangle\langle x|g\rangle = \langle f|\int_a^bdxw(x)|x\rangle\langle x|g\rangle\). So, \(\int_a^b dxw(x)|x\rangle\langle x| = \mathbb{I}\).

\(|x'\rangle = \mathbb{I}|x'\rangle = \int_a^b dx w(x)|x\rangle\langle x|x'\rangle = \int_a^b dx w(x)|x\rangle \frac{1}{w(x)}\delta(x-x') = |x'\rangle\). Sometimes, \(\langle x|x'\rangle = \frac{\delta(x-x')}{\sqrt{w(x)w(x')}}\).

### Exemplum Gratia

\(d^3x=dxdydz=r^2\sin\theta drd\theta d\varphi\)

\(\delta(\vec{r}-\vec{r}')=\delta(x-x')\delta(y-y')\delta(z-z') = \frac{\delta(r-r')}{r}\frac{\delta(\theta-\theta')}{r\sin\theta}\delta(\varphi-\varphi')\).

## Aside

There are also \(\mathcal{C}^k(a,b)\) functions which are functions that are \(k\) times continuously differentiable. \(\mathcal{C}^\omega\) are the set of analytic functions, they are \(\mathcal{C}^\infty\) 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 \(\langle \delta_c|\) defined by \(\langle \delta_c|f\rangle = f(c)\), \(a< c< b\). Is there a function in \(L_2^w(a,b)\) that is the dual of \(\langle \delta_c|\)? According to Reitz Representation Theorem, we expect a function such that \(\delta_c(x) = \langle x|\delta_c\rangle\) for which \(\langle \delta_c|f\rangle = (|\delta_c\rangle,|f\rangle) = \int_a^b dxw(x)\delta_c^*(x)f(x) = f(c)\) for all \(|f\rangle\). 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, \(\int_a^bdxw(x)\delta_c(x)^*f(x) = \int_a^b dx\delta_c(x)^*f(x) = f(c)\). But the integrand is zero except the measure zero value at \(x=c\) of \(f(x)/dx\). Choose \(d_n(x-c) = \{n: c-1/(2n)< c < + 1/(2n), 0: \text{else}\}\). Then, \(\int_a^b dx d_n(x-c)=1\). Then, \(\int_a^b dxf(x)d_n(x-c)\approx \left[\int_a^b d_n(x-c)\right]f(c)\). The assumption with the approximate is that \(f\) is well behaved in the small slice. Taking the limit, \(\int_a^b \lim_{n\to\infty} dx\delta_c(x-c)f(x)=\int_a^b dx\delta_n(x-c)f(c) = f(c)\). The problem is that \(\lim_{n\to\infty}d_n(x-c)\) does not exist, \(\lim = \{\infty: x=c,0: \text{otherwise}\}\).

Thus, we cannot get the delta function, but can get close. Even though the limit does not exist, \(\lim_{n\to\infty}\int_a^b dxd_n(x-c)f(x) = f(c)\), 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* - \(\mathcal{C}^\infty\) (infinitely differentiable) for which it does and its derivatives does not increase faster than any power of \(|x|\) at infinity - *Schwartz Space* \(\Phi\).
If \(\{\alpha_n\}\) is any sequence of functions for which the limit of the integrals exists, \(\forall f\in\Phi\), \(\lim_{n\to\infty}\int dx \alpha_n(x)f(x) \equiv \int dx\chi(x)^*f(x) \equiv \langle\chi|f\rangle\) defines a perfectly good linear functional, \(\langle\chi|\) on \(L_2(a,b)\) and write \(\lim_{n\to\infty}\alpha_n(x)=\chi(x)\) (i.e. \(\lim_{n\to\infty}d_n(x-c)=\delta_c(x)\)).

\(\Phi\subset L_2(a,b)\) and \(\langle\chi|\in\Phi^*\). These \(\langle \chi|\) are called distributions, tempered distributions, or generalized functions.

\(\Phi\subset\mathcal{V}\subset\Phi^{-X}\) - 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 \(f(x)\) \(\lim_{a\to\infty}\int dx\exp(iax)f(x)=0\).

### Quantum Mechanic Aside

In Quantum Mechanics, we mean \(\langle x|\equiv\langle \delta_x|\) and \(\langle p|\equiv \langle\tilde{\delta}_p|\).