Weighted inner product of functions

Question

I am facing difficulty understanding the requirement of the weighted inner product. As per my understanding, the inner or dot product is just the projection of one vector onto another and also gives the angle between them for vector spaces, or it is the integral of the product of two arbitrary functions continuous and differentiable within the interval (a,b) for the function spaces. And analogically I can relate the meaning of orthogonality of functions with the vectors when the integral is zero.$$<f,g>=\int_a^b f(x)g(x)dx$$

But why do we need the integral of the form $$I=\int_a^b w(x)f(x)g(x)dx$$for the inner product of the functions? As per my understanding, I think of it as something similar to the weighted integral statement in Finite Element Analysis where we provide weight to one function (i.e. scale the function values at different 'x' with w(x) ) and then integrate.

But neither am I able to interpret the meaning of the weighted inner product physically nor do I understand why the former inner product is not enough that we needed to define weighted inner product? Any help would be appreciated.

Also,let's say that the functions f(x) and g(x) are orthogonal with respect to weight function w(x) i.e. integral$$I=0$$then f(x) and g(x) are not actually orthogonal (or linearly independent if I speak in analogy with orthogonal eigenvector) right? Then how and why can we express any function as the linear combination of f & g which are orthogonal with respect to w(x) only? Why does f and g still form a basis?

Answer 1

The weight function has large values for those $x$ which are significant for the application you have in mind. Taking $x$ to be "frequency" for example, you might want to determine how similar two signals with spectra (fourier transform) $f$ and $g$ are, but your application requires that for some region of $[a,b]$ the similarity is more important and should be weighted higher. Say the importance is decaying as you move away from a certain point $c.$ Then you might use $$ \frac{\int_{a}^b w(x) f(x) g(x) \,dx}{\int_{a}^b w(x) \,dx} $$ with $w(x)$ (say) being defined as $$ w(x)=e^{-k|x-c|},\quad k>0. $$

Answer 2

Okay, I think I see. I can roughly explain how this fits in with the idea of inner product of functions. The idea is that, suppose we have a vector space and choose a basis $\{e_i\}$ then the dot product of two functions depend on the inner product matrix:

$$ v \cdot w= (v^1 e_1 +v^2 e_2 ) \cdot (w^1 e_1 + w^2 e_) = w^1 v^1 e_1 \cdot e_1 + w^2 v^2 e_2 \cdot e_2 + w^1 \cdot v^2 e_1 \cdot e_2 + w^2 \cdot v^1 e_2 \cdot e_1$$

Typically in Euclidean space, we choose $e_1 \cdot e_1 = e_2 \cdot e_1 = 1$ and $e_2 \cdot e_1 = e_1 \cdot e_2 = 0$

In matrix notation,

$$ v \cdot w = v^t M w$$

Where $M$ is inner product matrix, but how is this consistent with the abstract axiomatic definition of inner product space? A function of two vectors is an inner product only if such a matrix (with some additional properties) actually exist. See here.

Now, here is a set of symbolic manipulation which may help you see what is going on, we can write. I use the Riemann deftn $\int_0^b f(ih)= \sum_{i=0}^{\infty} f(ih) h$, we have:

$$ \left(\sum_{i=0}^{\infty} f(ih) h \right) \cdot_{\text{dot} } \left( \sum_{i=0}^{\infty} g(ih) h \right)= \sum_{i=0}^{\infty} f(ih) g(ih) w(ih) h$$

The idea is you split the interval $[0,a]$ into a set of equally paritioned point, to each point, we attach a number (similar to how we attach a number to a basis), so the values of functions of the function at points could in some sense be thought as components of vector with a basis. So, the dot product would be pointwise multiplication of these components weighted by the analogue of the inner product matrix here the $h$ comes to normalize for the width of partitions we take.

then f(x) and g(x) are not actually orthogonal

I think you may have skipped too many pages in your linear algebra textbook. Inner product determines orthogonality, not the other way around.