# Question on inner products between functions

Hello,

let's suppose I have two functions $f,g\in L^2(\mathbb{R})$ and I consider the inner product $$\left\langle f,g \right\rangle = \int_\mathbb{R} f(x)g(x)dx$$
If I transform the function f in the following way $f(x) \mapsto f(\phi(u))$, where $\phi:\mathbb{R}\rightarrow \mathbb{R}$ is smooth and bijective, I can still calculate the inner product $$\left\langle f \circ \phi,g \right\rangle = \int_\mathbb{R} f(\phi(u))g(u)du$$ Instead, if $\phi:U\rightarrow \mathbb{R}$ is smooth and bijective but U is not necessarily ℝ, I can't calculuate the inner product $\left\langle f \circ \phi,g \right\rangle$ anymore.

Does this happen because in the first case $\phi$ acted as a mapping $L^2(\mathbb{R}) \rightarrow L^2(\mathbb{R})$ to the same vector space, while in the second case we had a mapping $L^2(\mathbb{R}) \rightarrow L^2(U;\mathbb{R})$ which is a different vector space. Am I right?

Stephen Tashi
, I can't calculuate the inner product $\left\langle f \circ \phi,g \right\rangle$ anymore.

That's a fairly confusing question!

On a given vector space, it may be possible to define more than one inner product. When you have picked a definite "inner product space" then you have picked one particular inner product. There is no rule that the inner product on a vector space of functions must be chosen to be the intergral of their pointwise product. Their isn't even any rule that says the sum of two functions in a vector space of functions must be defined as pointwise sum of the two functions. All that is needed is that there be some sort of operation that satisifes the axioms for vector addition. So when you say "I can't calculate the inner product...", it isn't a precise statement. You have to define exactly what the inner product you are talking about before asserting it can't be calculated. You also need to define exactly what vector space you are talking about.

Hi Stephen!

You also need to define exactly what vector space you are talking about.

As a vector space I am considering L2(ℝ), the square integrable functions of one variable over the reals.

You have to define exactly what the inner product you are talking about before asserting it can't be calculated.

the inner product I am considering over L2(ℝ) is $f\cdot g = \int_\mathbb{R}f(x)g(x)dx$

What I was trying to say is that, with the above definitions of our inner product vector space, when we consider a function $f\circ \phi$, where $\phi:U\rightarrow \mathbb{R}$ is smooth and bijective but U is not necessarily ℝ, it does not make sense to calculate the inner product between $f\circ \phi$ and another function g in L2(ℝ) simply because $f\circ \phi$ is not in L2(ℝ) anymore.
I just wanted to know if this reasoning is correct.

Stephen Tashi
, it does not make sense to calculate the inner product between $f\circ \phi$ and another function g in L2(ℝ) simply because $f\circ \phi$ is not in L2(ℝ) anymore.
I just wanted to know if this reasoning is correct.

It's correct that an inner product of two functions on $U$ is not defined merely by stating the definition of an inner product of two functions on $\mathbb{R}$ and the existence of the transformation $\phi$. It may be possible to come up with a "sensible" definition for an inner product of functions on $U$ that makes reference to the transformation $\phi$ and an inner product of functions on $\mathbb{R}$. It wouldn't be correct to say that creating such a definition is impossible.

It may be possible to come up with a "sensible" definition for an inner product of functions on $U$ that makes reference to the transformation $\phi$ and an inner product of functions on $\mathbb{R}$

I am not sure I understood exactly what you meant here; did you mean something like the following? : $$\left\langle f\circ \phi,\, g\circ\phi) \right\rangle_U = \int_U J_\phi^{-1}f(\phi(u))g(\phi(u))du=\int_\mathbb{R}f(x)g(x)dx = \left\langle f, g \right\rangle$$ where J is the determinant of the Jacobian matrix of the transformation $\phi$.

Here I just defined an inner product for functions on U that makes reference to $\phi$ and to our original inner product on the functions on ℝ. However I don't see how we could make a definition of inner product such that we could take inner products of functions on U with functions on ℝ. Is that possible? It seems to me that such an inner product (if it exsited) would not satisfy the "symmetry"-axiom.

Last edited:
Fredrik
Staff Emeritus
Gold Member
An inner product (or semi-inner product) on a real vector space V is a map from V×V into ℝ (that satisfies some additional properties).

The problem with your (second) ##\phi## and ##f\circ\phi## is just that they're not members of the vector space. So the answer to your question is simply "yes".

ok! got it.
Thanks Fredrik and Stephen.

Stephen Tashi
I'm not saying that I know how to do it but I see no proof that it is impossible. To do it, you would have to define a vector space that contained both functions on $\mathbb{R}$ and functions on $U$ as vectors. You might try defining vectors for such a space as a linear combinations of two such functions.