Square-integrable functions as a vector space

Yes, you have already shown that the set of all normalised functions is ##\textit{not}## a vector space when you 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##.And yes, it is even easier to show that the set of all normalised functions is ##\textit{not}## a vector space by considering the case
  • #1
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Homework Statement



(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.

Homework 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}##

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?
 
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  • #2
The last two terms together are real. For any complex number ##z##, ##{\rm Re}(z) \leq |z|##. Then apply the Schwarz inequality.
 
  • #3
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?
 
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  • #4
failexam said:
How can this help to show that the set of all normalised functions is a vector space?

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
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
Ah right, I missed the "normalised". It is trivial to shot that it is not a vector space.
 
  • #7
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
Almost. You need to insert some values. But that is not the easiest way to prove it. The space has no zero element.
 
  • #9
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##.
 
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  • #10
failexam said:
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##.

Yes, that is fine apart from that the ##|\alpha + \beta|^2## = 0.01 should be an inequality.
 
  • #11
failexam said:
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##??

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

Related to Square-integrable functions as a vector space

1. What is a square-integrable function?

A square-integrable function is a function that is defined on an interval and has a finite integral when squared over that interval. This means that the area under the curve of the function squared is finite.

2. How are square-integrable functions used in mathematics?

Square-integrable functions are often used in functional analysis, which is a branch of mathematics that studies vector spaces of functions. They are also commonly used in probability theory, signal processing, and quantum mechanics.

3. What is the significance of square-integrable functions as a vector space?

Square-integrable functions form a vector space, which means they have properties such as closure under addition and scalar multiplication. This allows for mathematical operations to be performed on square-integrable functions, making them a useful tool in many areas of mathematics and physics.

4. How do square-integrable functions relate to the concept of inner product spaces?

Square-integrable functions can be used as the basis for inner product spaces, where the inner product is defined as the integral of the product of two functions over a given interval. This allows for the calculation of distances, angles, and other geometric properties of functions.

5. Can any function be considered square-integrable?

No, not all functions are square-integrable. A function must meet certain criteria, such as being bounded and having a finite integral, in order to be considered square-integrable. Examples of square-integrable functions include polynomials, trigonometric functions, and exponential functions.

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