# 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 $$\langle p|\equiv \langle\tilde{\delta}_p|$$.

Created: 2023-06-25 Sun 02:24

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