Proving closure of square integrable functions.

[EquationOfMotion]

I'm trying to prove that the set of all square integrable functions f(x) for which ∫_a^b |f(x)|^2 dx is finite is a vector space. Everything but the proof of closure is trivial.

To prove closure, obviously we should expand out |f(x)+g(x)|^2, which turns our integral into one of |f(x)|^2 (finite), |g(x)|^2 (finite), and the two cross terms f*(x)g(x) and f(x)g*(x).

This question has been posted before, and it's resolved by using the Cauchy Schwarz inequality. Since <f|g> ≤ √(<f|f><g|g>) the two cross terms will be finite and we are done.

But why are we allowed to use the Cauchy Schwarz inequality in the first place? Isn't the inequality only true given that we are in an inner product (and thus vector) space? So since I'm trying to prove that what I have is a vector space, it seems using the inequality is illogical.

The way I tried to work around this problem is by arguing that f(x) and g(x) belong to the vector space of all functions, and my inner product is defined as ∫_a^b f*(x)g(x) dx. But then the inner product could be infinite! Which doesn't make much sense.

 

[fresh_42]

Consider $0 \leq \langle f+tg,f+tg \rangle = \langle f,f \rangle+2t \operatorname{Re}\langle f,g \rangle + t^2\langle g,g\rangle$ as a real quadratic polynomial in $t$. What can you say about its zeros in this case?

 

[EquationOfMotion]

Consider $0 \leq \langle f+tg,f+tg \rangle = \langle f,f \rangle+2t \operatorname{Re}\langle f,g \rangle + t^2\langle g,g\rangle$ as a real quadratic polynomial in $t$. What can you say about its zeros in this case?

I'm confused. Isn't the inequality on the left only true do to positive semi-definiteness, which is a feature of inner products of vectors in a vector space?

 

[fresh_42]

Well, you will have to have something. If you do not want to use it, then please define $\langle \; , \;\rangle$ as well as the set your functions are from. Everything else is guesswork.

 

[EquationOfMotion]

Well, you will have to have something. If you do not want to use it, then please define $\langle \; , \;\rangle$ as well as the set your functions are from. Everything else is guesswork.

The inequality you gave above essentially offers a way to prove the Cauchy Schwarz inequality given positive semi-definiteness right? Since you look at the discriminant and conclude it must be real.

I'm confused as to why you're allowed to use the inequality in the first place. I am trying to prove that something is a vector space. Given that you need a vector space to define an inner product (unless you don't?) the Cauchy Schwarz inequality doesn't seem to be a reasonable theorem to apply. Of course, we can argue that we could pull f(x) and g(x) from the vector space of all functions, but then the inner product as defined by ∫_a^b f*(x)g(x) dx isn't necessarily finite. But does the inner product have to be finite?

 

[fresh_42]

The inequality you gave above essentially offers a way to prove the Cauchy Schwarz inequality given positive semi-definiteness right? Since you look at the discriminant and conclude it must be real.

Negative, so imaginary, not real - non positive to be exact.

I'm confused as to why you're allowed to use the inequality in the first place. I am trying to prove that something is a vector space. Given that you need a vector space to define an inner product (unless you don't?) the Cauchy Schwarz inequality doesn't seem to be a reasonable theorem to apply. Of course, we can argue that we could pull f(x) and g(x) from the vector space of all functions, but then the inner product as defined by ∫_a^b f*(x)g(x) dx isn't necessarily finite. But does the inner product have to be finite?

I have a vector space of functions and an inner product. If two of them are square integrable, so is it's sum, which is why they form a subspace.

 

[EquationOfMotion]

Negative, so imaginary, not real - non positive to be exact.

I have a vector space of functions and an inner product. If two of them are square integrable, so is it's sum, which is why they form a subspace.

Woops. But should be to the same effect right?

I have a vector space of functions and an inner product. If two of them are square integrable, so is it's sum, which is why they form a subspace.

I'm trying to prove that the set of functions I have (∫_a^b |f(x)|^2 dx is finite) is indeed a vector space. This requires closure under addition. Are you saying we can take two functions which are part of the vector space of all functions with the inner product defined as ∫_a^b f*(x)g(x) dx so we can use Cauchy Schwarz? Which again leads to my question: does an inner product have to be finite? It doesn't seem to be the case, but an infinite inner product seems a little strange.

 

[fresh_42]

An inner product maps a pair of vectors to a scalar. There is no infinity anywhere near. The question is at most, whether a specific definition is an inner product or not.

 

[pasmith]

I can see why you may be concerned about using a statement about an inner product on a vector space to prove closure of the vector space in question.

The answer is that in this context the Cauchy-Schwartz Inequality arises as a special case of the Rogers-Hölder Inequality, which states that if 1 &lt; p &lt; \infty and 1/p + 1/q = 1 and both \int_a^b |f|^p\,dx and \int_a^b |g|^q\,dx are finite then <br /> \left| \int_a^b fg\,dx \right| \leq \int_a^b |fg|\,dx \leq \left(\int_a^b |f|^p\,dx \right)^{1/p}\left(\int_a^b |g|^q\,dx \right)^{1/q}.<br />
This is a statement about the values of integrals of complex-valued functions, not about inner products of vectors. The Cauchy-Schwartz Inequality follows by setting p = 2.

(You need to resort to other propositions of measure theory to conclude that \int_a^b |f + g|^2\,dx actually exists if both \int_a^b |f|^2\,dx and \int_a^b |g|^2\,dx exist, because for non-continuous f and g that isn't obvious. Once you know it exists the Cauchy-Schwartz Inequality will guarantee finiteness.)

 

[pasmith]

Alternatively one can prove <br /> \int_a^b |fg|\,dx \leq \left(\int_a^b |f|^2\,dx\right)^{1/2}\left(\int_a^b |g|^2\,dx\right)^{1/2} as follows.

For convenience, define <br /> \|f\| = \left(\int_a^b |f|^2\,dx\right)^{1/2}. Now if \|f\| = 0 then f = 0 almost everywhere, so |fg| = 0 is zero almost everywhere and the inequality holds.

Assume then that \|f\| &gt; 0 and \|g\| &gt; 0. The arithmetic-geometric mean inequality states that, for positive A and B,<br /> (AB)^{1/2} \leq \frac{A + B}2. For each x \in [a,b] we can apply this with A = |f|^2/\|f\|^2 and B = |g|^2/\|g\|^2 to obtain <br /> \frac{|fg|}{\|f\|\|g\|} \leq \frac 12 \left( \frac{|f|^2}{\|f\|^2} + \frac{|g|^2}{\|g\|^2}\right). Integrating yields <br /> \frac{1}{\|f\|\|g\|}\int_a^b |fg|\,dx \leq \frac12 \left( \frac{\|f\|^2}{\|f\|^2} + \frac{\|g\|^2}{\|g\|^2}\right) = 1 and the result follows on multiplication by \|f\|\|g\|.

It then follows that <br /> \int_a^b |f + g|^2\,dx \leq \int_a^b |f|^2 + 2|fg| + |g|^2\,dx \leq \|f\|^2 + 2\|f\|\|g\| + \|g\|^2 = (\|f\| + \|g\|)^2 is bounded as required.