# Inner Product of two functions

The inner product of two vectors $$\vec a$$ and $$\vec b$$ of n dimensions, is given by, $$(\vec a \,, \vec{b}) =a_1b_1 + a_2b_2 + a_3b_3 + \,\,...\,\,+a_nb_n$$

If a function is considered to be a vector of infinite dimensions, for example, $$f(x) = \begin{bmatrix} \vdots \\ f(0.05) \\ f(0.051) \\ f(0.052) \\ \vdots \end{bmatrix}$$ Then, the inner product of two functions say $$f(x), g(x)$$ is given by $$\mathbf {\bigl(} f(x),g(x)\mathbf {\bigr)}=\int_{-\infty}^\infty f(x)g(x) \,\,\, \mathbf{dx}$$ My Question is:

How did we get the $$\mathbf {dx}$$ in the inner product (or)$$\mathbf{why\,\, is}\,\,\,\,\ \sum_{-\infty}^\infty f(x)g(x) = \int_{-\infty}^\infty f(x)g(x) \,\,\, \mathbf{dx}$$ In a Riemann sum , the $$\Delta x$$ term becomes $$dx$$ during the limiting process, but in this sum, there doesn't seem to be a $$\Delta x$$ term. I think the answer to this question would probably answer my another question,

Laplace Transform: Continuous analogue of Power series

The $$\Delta x$$ term is just $$1$$, as the sum is over the integers. As you shorten the range, it is clear that a $$\Delta x$$ term is necessary.

• Wow, that was simple :) Does this answer the linked question? I mean I get this, but then there seems to be an elaborate answer to the linked question, please go through the comments section, I really didn't understand his argument, does this relate to his argument? Commented Apr 17, 2019 at 5:29

We use choices of inner product that (a) indeed satisfy the axioms defining an inner product and (b) prove useful for our purposes. Under appropriate regularisation conditions, $$\sum_{n=-\infty}^\infty f(n)g(n)$$ (the sum implied to run over integers only, since you can't sum uncountably many terms) and $$\int_{-\infty}^\infty f(x)g(x)dx$$ are both inner products. But the latter is of greater interest when studying $$\Bbb R\mapsto\Bbb R$$ functions, since it runs over all values the arguments of $$f,\,g$$ can take.

In particular, (c) this integral is an inner product (proof is an exercise) and (d) you can't have an integral without the $$dx$$ part. What you can do, however, is replace $$dx$$ with the more general $$dh(x)=h^\prime(x) dx$$ to obtain $$\int_{-\infty}^\infty f(x)h^\prime(x)g(x) dx$$, which is an inner product provided $$h^\prime(x)>0$$ for all $$x\in\Bbb R$$. (Similarly, $$\sum_n f(n)j(n)g(n)$$ is an inner product if $$j(n)>0$$ for $$n\in\Bbb Z$$).

So, "why $$dx$$?" really means "why take $$h(x)=x$$ for all $$x$$?" To which we can only answer, "use an $$h$$ that's convenient". For example, you'd use a very different $$h$$ if studying these, whereas in studying these you'd take $$h(x)=x$$ but you'd restrict the integration range $$[-1,\,1]$$ (which still gives an inner product, just as in the discrete case one can restrict to summing over fewer $$n$$).

• What different effect does adding a general $dh(x)$ has? is it like, instead of having double product, we make it a triple product and then get a $dx$ term? But the answer by Eric Miller is also intuitive, in what way does it differ form your answer? I mean How can we arrive at his answer with yours or vice versa Commented Apr 17, 2019 at 6:21
• @AravindhVasu Oh no, it's not a "triple product". Just as a positive-definite matrix $M$ conformable with column vectors $u,\,v$ gives an inner product $(u^TMv)_{11}$, a suitable two-variable $M(x,\,y)$ makes $\int_{-\infty}^\infty\int_{-\infty}f(x)M(x,\,y)g(y)dxdy$ an even more general inner product, so e.g. we can take $M(x,\,y)=h^\prime (x)\delta(y)$.
– J.G.
Commented Apr 17, 2019 at 6:33
• Sorry, to be honest, I didn't get that comment at all, Can you explain it a bit more Commented Apr 17, 2019 at 6:36
• @AravindhVasu You can think of the integral $fg$ as analogous to the usual dot product $u\cdot v$ on $\Bbb R^n$. What the $h^\prime$ factor does is replace that, in this analogy, with $u\cdot Mv$, with $M$ a diagonal matrix whose diagonal entries are all positive. Those diagonal entries may look like a vector, but that's not the most helpful way to consider it.
– J.G.
Commented Apr 17, 2019 at 6:45