# Can all integration be thought of as projections?

For example, the integral of the function f(x) could be thought of the projection of f on the function g, where g is identically 1.

Following this logic, can we think of the multiplication of f and g as the area between f and g, no matter how complicated f or g is?

But then this is a bit more if we think of the area between f and g as the projection of f on g, but how do we explain negative area i.e. integrating sin(x).

I don't know if I was taught the right calculus but I've never seen a textbook that introduces the notion of the integral of f and g as the projection of f on g. Why is that? And if possible, can someone please provide me a good notes on this subject? Treating integration from a projection perspective.

-
This is one of the basic ideas behind Riesz representation. – Adam Hughes Jul 31 '14 at 17:36
I won't do that. Integrals are more basic than projections. I mean, if somebody wakes me up at 3 a.m. and abruptly asks me what an integral is, I would probably blabber something about areas under curves, not about projections. This said, it surely is a nice train of ideas that is interesting to explore once. – Giuseppe Negro Jul 31 '14 at 18:07

Just to expand a little, the Riesz representation theorem has these ideas embedded into it. Generically, on a Hilbert space of $\Bbb C$-valued functions, you will see

$$\int f(x)\overline{g}(x)\,d\mu(x)$$

(if they have values in $\Bbb R$ the complex conjugation over $g$ disappears)

as an inner product of two functions, which--you may remember from basic vector calculus--was how you talked about projecting one vector onto another

$$\text{proj}_{\mathbf{v}}(\mathbf{u})={\mathbf{u}\cdot\mathbf{v}\over\lVert\mathbf{v}\rVert^2}\mathbf{v}$$

which (if you go back even further) you may remember you had the inner product come in by drawing triangles with the law of cosines.

Albeit, there is no notion of "area" going on with this sort of thing. It measures more just ideas of "amount of one vector in the direction of another." However, you do have a nice interpretation of negatives, since we have the formula

$$\cos(\theta)={\mathbf{u}\cdot\mathbf{v}\over \lVert\mathbf{u}\rVert\lVert\mathbf{v}\rVert}$$

where $\theta$ is the angle between the two vectors, so that negative numbers just mean the angle is not in the range $-{\pi\over 2}\le\theta\le {\pi\over 2}$.

It's quite natural that you would not have found this in a calculus class, it requires some good amount of linear algebra and proving that that formula gives an inner-product on a space of square-integrable functions is much harder mathematics than just calculus.

Any book on Hilbert spaces which mentions a suitably generalized version of the Riesz representation theorem should be sufficiently satisfying for someone seeking to pursue this line of thought, and any good mathematics library should have $n+1$ books on the subject if you just look in the catalog for "Hilbert spaces." A quick googling of "Hilbert space" yields many results, such as these notes. A proof that $L^2$ of a measure space is a Hilbert space is a classical result in any functional analysis textbook, and follows from the Minkowski inequality.

-