# Question on inner products between functions

by mnb96
Tags: functions, products
 P: 626 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?
P: 3,319
 Quote by mnb96 , 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.
P: 626
Hi Stephen!

 Quote by Stephen Tashi 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.

 Quote by Stephen Tashi 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.

P: 3,319
Question on inner products between functions

 Quote by mnb96 , 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.
P: 626
 Quote by Stephen Tashi 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.
 Emeritus Sci Advisor PF Gold P: 9,543 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".
 P: 626 ok! got it. Thanks Fredrik and Stephen.
P: 3,319
 Quote by mnb96 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?.
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.

You need to clarify the context of your question. On one hand, I can imagine you reading a paper by someone who made mathematical errors in symbolic manipulations and is trying to take "the usual" inner product between two functions on different domains without stating any special definitions that would make that a sensible operation. I'd agree that this is an error since, as far as I know, there is no standard definition for that operation that we can assume he is using.

On the other hand, you might be trying to invent some new mathematics and you might be searching for a way to define an inner product of functions on two different domains. I'm not necesarily optimistic about the usefulness of that idea, but I wouldn't say you should automatically give up.

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