# An inner product must exist on the set of all functions in Hilbert space

1. Aug 25, 2010

### bjnartowt

1. The problem statement, all variables and given/known data
Show that $\int {{f^*}(x)g(x) \cdot dx}$ is an inner product on the set of square-integrable complex functions.

2. Relevant equations
Schwarz inequality:
$$\left| {\int {{f^*}(x)g(x) \cdot dx} } \right| \le \sqrt {\int {{{\left| {f(x)} \right|}^2} \cdot dx} \int {{{\left| {g(x)} \right|}^2} \cdot dx} }$$

3. The attempt at a solution

Rephrase what we want to prove: see if:
$\int {{f^*}(x)g(x) \cdot dx} < \infty$

...is true.

Since we are considering the set of square-integrable-functions: “f” and “g” are in the set of square-integrable functions: they are elements of Hilbert space:
$$\left\{ {f(x),g(x)} \right\} \in {L^2}$$

This means the following integrals exist:
$$\int_{ - \infty }^{ + \infty } {{f^*}(x)f(x) \cdot dx} = \int_{ - \infty }^{ + \infty } {{{\left| {f(x)} \right|}^2} \cdot dx} < \infty {\rm{ }}\int_{ - \infty }^{ + \infty } {{g^*}(x)g(x) \cdot dx} = \int_{ - \infty }^{ + \infty } {{{\left| {g(x)} \right|}^2} \cdot dx} < \infty$$

In turn: this gaurantees:
$$\sqrt {\int {{{\left| {f(x)} \right|}^2} \cdot dx} \int {{{\left| {g(x)} \right|}^2} \cdot dx} } < \infty$$

Schwarz inequality says:
$$\left| {\int {{f^*}(x)g(x) \cdot dx} } \right| \le \sqrt {\int {{{\left| {f(x)} \right|}^2} \cdot dx} \int {{{\left| {g(x)} \right|}^2} \cdot dx} }$$

Together: the previous two equations require:
$$\left| {\int {{f^*}(x)g(x) \cdot dx} } \right| < \infty$$

Another inequality is that:
$$\left| {\int {{f^*}(x)g(x) \cdot dx} } \right| < \int {\left| {{f^*}(x)g(x)} \right| \cdot dx}$$

…which actually ruins what we’re trying to prove. Gah. Well, it doesn't counter-prove it, but I've obviously used the wrong inequality. Somehow it must be that:
$$\int {\left| {{f^*}(x)g(x)} \right| \cdot dx} \le \sqrt {\int {{{\left| {f(x)} \right|}^2} \cdot dx} \int {{{\left| {g(x)} \right|}^2} \cdot dx} }$$

but i'm not sure how to get that. Triangle inequality? Perhaps I am staring too hard, but it doesn't seem so?

2. Aug 25, 2010

### diazona

Are you trying to prove that $\int f^*(x)g(x) \mathrm{d}x < \infty$, or that $\int \lvert f^*(x)g(x)\rvert \mathrm{d}x < \infty$?

3. Aug 25, 2010

### bjnartowt

Am trying to prove $\int {{f^*}(x)g(x) \cdot dx} < \infty$.

I think we're valid up to the step:
$$\left| {\int {{f^*}(x)g(x) \cdot dx} } \right| < \infty$$

If you think of something, let me know. I'm off quantum, and practicing electro-magnetism problems for Physics GRE...

4. Aug 25, 2010

### diazona

OK, that's what I thought... now can you think of circumstances in which $\lvert\int f^*(x)g(x)\mathrm{d}x\rvert$ is convergent but $\int f^*(x)g(x)\mathrm{d}x$ is not?

5. Aug 25, 2010

### bjnartowt

Well....gosh, now that I think about it, I can't. It seems sensible, so dare I say:

$\lvert\int f^*(x)g(x)\mathrm{d}x\rvert>\int f^*(x)g(x)\mathrm{d}x$

Oh yeah, the absolute-value is that of just a number, so that's as true as |-6| > -6, and > only becomes >= if the number is 0.

Duh. Thanks so much. ...err.. I think. I think it's obvious. I'm way too cautious after a summer's research of "oops"-es coming from an inattentiveness to subtleties.

Now let's have some fun by just haphazardly going way off topic : )

Do null kets demand that case, where <generic bra|0> = 0, where |0> is the null ket? Like in vacuum-states?

6. Aug 25, 2010

### diazona

That's it, pretty much If you want something more like a proper justification, start by expressing the value of the integral in complex polar form, re.

By the way, it has to be |A| ≥ A, in case A is a positive number.

Now about those "null kets": the inner product doesn't necessarily have to be zero. For instance, the normalized ground state of a harmonic oscillator satisfies $\langle 0 \vert 0 \rangle = 1$. Something similar applies for a vacuum state in QFT; it has a nonzero norm. But then again, I don't think it really makes sense to call it a "null ket" since there's nothing "null" about it. Remember that the quantum number that appears in the ket is really just a label for the state, and the label is kind of arbitrary. The ground state of a SHO could just as well have been labeled $\lvert 1\rangle$ (of course you'd have to change the formulas for the energy levels etc.), or $\lvert \psi_0\rangle$, or $\lvert G\rangle$.

Anyway, the point is that just because you write a particular ket as $\lvert 0 \rangle$, doesn't mean the associated function is actually equal to zero.