# Square-integrable functions as a vector space

• spaghetti3451
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
spaghetti3451

## 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?

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

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?

Abhishek11235
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.

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)?

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

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##??

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

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:
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.

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.

## 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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