Inner Product Proof on Square Integrable Functions

[SunnyBoyNY]

Homework Statement

Consider the linear space S, which consists of square integrable continuous
functions in [0,1]. These are continuous functions x : [0,1] -> R such that the integral is less than infinity.

Homework Equations

Show that the operation

∫x(t)y(y)dt at [0,1] is an inner product in this space.

The Attempt at a Solution

Apparently there are three axioms:

1] <x,x> is >= 0 (<x,x> = 0 only if x is a trivial function). I think that know how to prove this one. Simply, if the integrand is positive at any point then the function must be positive in the neighborhood of that point. Therefore, the integral cannot be <= 0. It can be zero only if the function is trivial such that x(t) = 0.

2] Linearity - how would I prove the linearity argument for general continuous functions? Could I replace such function with its Taylor series and show the linearity on the resulting polynomials?

3] Symmetry - no clue so far.

Could you guys give me some pointers?

Thanks a lot,

SunnyBoy

 

[Kreizhn]

Should that be $\int x(t) y(t) dt$?

I think you may be over complicating things. First off, linearity and symmetry are actually the two easier parts. What does linearity of the inner product mean? Write it out and you'll see it follows immediately. Same thing with symmetry.

Edit: Forgot to address the first part. This is actually the harder part. You've shown that if the inner-product is non-negative then x cannot be trivial. This is the contrapositive of the statement $\langle x ,x \rangle = 0 \Rightarrow x \equiv 0$. However, this you still need to show the first part: You want to show that $\langle x, x \rangle \geq 0$. Again, write this out and you'll see why it's true.

 

[SunnyBoyNY]

Hello Kreizhn,

you are right- the integrands are both functions of one variable (time).

I have done what you suggested - the algebra is not really that difficult. However, how do I incorporate the "square-integrable" statement into the problem? Or is that just a generic statement which limits our function selection so that the inner product can be evaluated?

Linearity: $<ax,y> = a<x,y>$
And:
$<x+y,z> = <x,z> + <y,z>$

So:

$$<af(t),g(t)> = \int_{0}^{1}af(t)g(t)dt = a\int_{0}^{1}f(t)g(t)dt = a<f(t),g(t)>$$
And:
$$<f(t)+g(t),h(t)> = \int_{0}^{1}(f(t)+g(t))h(t)dt = \int_{0}^{1}f(t)h(t)+g(t)h(t)dt = <f(t),h(t)>+<g(t),h(t)>$$

Symmetry:

$$<f(t),g(t)> = \int_{0}^{1}f(t)g(t)dt = \int_{0}^{1}g(t)h(t)dt = <g(t),f(t)>$$

First axiom:

The integrand must be non-negative at all times since it is a square of a function. Therefore, if the function is continuous then no point can be negative. Therefore, the integral value must be nonnegative.

$$<f(t),f(t)> = \int_{0}^{1}f(t)f(t)dt= \int_{0}^{1}f^2(t)dt >= 0$$

Thanks!

SunnyBoy

 

[Kreizhn]

However, how do I incorporate the "square-integrable" statement into the problem? Or is that just a generic statement which limits our function selection so that the inner product can be evaluated?

...

$$<f(t),f(t)> = \int_{0}^{1}f(t)f(t)dt= \int_{0}^{1}f^2(t)dt >= 0$$

I'm not sure if you didn't just answer this question yourself, but in case it's not clear, let me just explain it quickly. Look at your last line of math. How do you know $\langle f(t), f(t) \rangle$ is finite? Well, it is, and the reason it is finite is precisely because f is square integrable on [0,1].

 

[SunnyBoyNY]

Seems pretty straightforward then. Thank you very much, Kreizhn!

Should I edit my second post so that folks who are solving the same problem will have the same pointer as I had? I am not sure what the policy is but i have gotten the impression that pple are encouraged to do the thinking themselves (no direct answers provided).

SunnyBoy

 

[Kreizhn]

I think it's fine. It's more important that I don't give you the answer. You are allowed to demonstrate that you have successfully solved the problem.