# Square-integrable functions as a vector space

1. May 21, 2015

### spaghetti3451

1. The problem statement, all variables and given/known data

(a) Show that the set of all square-integrable functions is a vector space. Is the set of all normalised functions a vector space?

(b) Show that the integral $\int^{a}_{b} f(x)^{*} g(x) dx$ satisfies the conditions for an inner product.

2. Relevant equations

The main problem is to show that the sum of two square-integrable functions is itself square-integrable.

Use Schwarz inequality: $| \int^{a}_{b} f(x)^{*} g(x) dx| \leq \sqrt{ \int^{a}_{b} |f(x)|^{2} dx \int^{a}_{b} |g(x)|^{2} dx}$

3. The attempt at a solution

(a) Let $f(x)$ and $g(x)$ be square-integrable functions. Then,

$\int^{a}_{b} | \alpha f(x) + \beta g(x)|^2 dx$

$= \int^{a}_{b} [\ | \alpha f(x) |^2 + | \beta f(x) |^2 + \alpha^{*} f(x)^{*} \beta g(x) + \alpha f(x) \beta^{*} g(x)^{*} ] dx$

$= |\alpha|^2 \int^{a}_{b} | f(x) |^2 dx + |\beta|^2 \int^{a}_{b} | f(x) |^2 dx + \alpha^{*} \beta \int^{a}_{b} f(x)^{*} g(x) dx + \alpha \beta^{*} \int^{a}_{b} f(x) g(x)^{*} dx$

The first two terms are already square-integrable. So, all I have to do now is to show that the last two terms are square-integrable.

Can you suggest how I can use the Schwarz inequality over the last two terms?

P.S. : The last two terms are complex conjugates of each other, and the sum of two complex conjugates is a real number. Does it somehow prove that the last two terms form a square-integrable combination?

Last edited: May 21, 2015
2. May 21, 2015

### Orodruin

Staff Emeritus
The last two terms together are real. For any complex number $z$, ${\rm Re}(z) \leq |z|$. Then apply the Schwarz inequality.

3. May 21, 2015

### spaghetti3451

Alright, here's my attempt.

Let $f(x)$ and $g(x)$ be square-integrable functions. Then,

$\int^{a}_{b} | \alpha f(x) + \beta g(x)|^2 dx$

$= \int^{a}_{b} [\ | \alpha f(x) |^2 + | \beta f(x) |^2 + \alpha^{*} f(x)^{*} \beta g(x) + \alpha f(x) \beta^{*} g(x)^{*} ] dx$

$= |\alpha|^2 \int^{a}_{b} | f(x) |^2 dx + |\beta|^2 \int^{a}_{b} | f(x) |^2 dx + \alpha^{*} \beta \int^{a}_{b} f(x)^{*} g(x) dx + \alpha \beta^{*} \int^{a}_{b} f(x) g(x)^{*} dx$

$= |\alpha|^2 \int^{a}_{b} | f(x) |^2 dx + |\beta|^2 \int^{a}_{b} | f(x) |^2 dx + \alpha^{*} \beta \int^{a}_{b} f(x)^{*} g(x) dx + ( \alpha^{*} \beta \int^{a}_{b} f(x)^{*} g(x) dx )^*$

$= |\alpha|^2 \int^{a}_{b} | f(x) |^2 dx + |\beta|^2 \int^{a}_{b} | f(x) |^2 dx + 2\ {\rm Re} ( \alpha^{*} \beta \int^{a}_{b} f(x)^{*} g(x) dx )$

$\leq |\alpha|^2 \int^{a}_{b} | f(x) |^2 dx + |\beta|^2 \int^{a}_{b} | f(x) |^2 dx + 2\ | \alpha^{*} \beta \int^{a}_{b} f(x)^{*} g(x) dx |$

$\leq |\alpha|^2 \int^{a}_{b} | f(x) |^2 dx + |\beta|^2 \int^{a}_{b} | f(x) |^2 dx + 2\ | \alpha^{*} \beta | \sqrt{ \int^{a}_{b} |f(x)|^{2} dx \int^{a}_{b} |g(x)|^{2} dx}$

This is a combination of square-integrable functions, so that $\alpha f(x) + \beta g(x)$ is a square-integrable function.

Let $\int^{a}_{b} |f(x)|^2 dx = 1$ and $\int^{a}_{b} |g(x)|^2 dx = 1$. Therefore,

$\int^{a}_{b} | \alpha f(x) + \beta g(x)|^2 dx$

$\leq |\alpha|^2 \int^{a}_{b} | f(x) |^2 dx + |\beta|^2 \int^{a}_{b} | f(x) |^2 dx + 2\ | \alpha^{*} \beta | \sqrt{ \int^{a}_{b} |f(x)|^{2} dx \int^{a}_{b} |g(x)|^{2} dx}$

$\leq |\alpha|^2 + |\beta|^2 + 2\ | \alpha^{*} \beta |$

$\leq |\alpha|^2 + \alpha^{*} \beta + \alpha \beta^{*} + |\beta|^2$

$\leq | \alpha + \beta |^2$.

How can this help to show that the set of all normalised functions is a vector space?

4. May 21, 2015

### Orodruin

Staff Emeritus
This shows you that any (finite) linear combination of functions in the space is also in the space. You should check the other axioms of a vector space as well, but it is rather trivial.

5. May 21, 2015

### spaghetti3451

How exactly can this show any (finite) linear combination of functions in this space is also in this space?

For one thing, I've only shown that, given $\int^{a}_{b} |f(x)|^2 dx = 1$ and $\int^{a}_{b} |g(x)|^2 dx = 1$,

$\int^{a}_{b} | \alpha f(x) + \beta g(x)|^2 dx \leq | \alpha + \beta |^2$, not that $\int^{a}_{b} | \alpha f(x) + \beta g(x)|^2 dx = 1$.

That does not quite prove that the set of all normalised functions form a vector space, does it (skipping the other axioms of a vector space, which are rather trivial to prove for this case)?

6. May 21, 2015

### Orodruin

Staff Emeritus
Ah right, I missed the "normalised". It is trivial to shot that it is not a vector space.

7. May 21, 2015

### spaghetti3451

Haven't I already shown that the set of all normalised functions is $\textit{not}$ a vector space when I proved the following:

if $\int_{a}^{b} |f(x)|^{2} dx = 1$ and $\int_{a}^{b} |g(x)|^{2} dx = 1$, then $\int^{a}_{b} | \alpha f(x) + \beta g(x)|^2 dx \leq | \alpha + \beta |^2$??

8. May 21, 2015

### Orodruin

Staff Emeritus
Almost. You need to insert some values. But that is not the easiest way to prove it. The space has no zero element.

9. May 21, 2015

### spaghetti3451

Right! I can see that the function $f(x) = 0$ is not normalisable.

Therefore, lacking a zero element, the set of all normalised functions do not form a vector space.

I wonder how to complete the proof using my original approach. Is this the correct way?

Let $\alpha = 0.1$ and $\beta = 0.1$, then $|\alpha + \beta|^2$ = 0.01.

So, $\int^{a}_{b} | \alpha f(x) + \beta g(x)|^2 dx \neq 1$.

Last edited: May 21, 2015
10. May 21, 2015

### Orodruin

Staff Emeritus
Yes, that is fine apart from that the $|\alpha + \beta|^2$ = 0.01 should be an inequality.

11. May 21, 2015

### Ray Vickson

Even easier: if $f$ is normalized to 1 and $c$ is a scalar with $|c| \neq1$, then $c f$ is not normalized to 1.