Inner Product spaces with functions?

Question

I understand inner product space with vectors, but the conversion to functions is throwing me off. Also why do they use an integral here, I've always seen summations. I think I'm missing something with notation here. Any help/hints would be appreciated.

Answer 1

First, remember that an integral is a summation under a limit, if that helps with understanding why that could possibly be an inner product.

Second, remember that $C([0,1])$ is a vector space. The functions are vectors. You can add two functions together, there is a constant zero function, you can multiply functions by scalars. If you wish, think of a function as a vector with infinitely many indices: i.e., $f_x=f(x)$ for all $x$ from $0$ to $1$, rather than just integers.

Third, it is an inner product because of the axioms (Wikipedia). The integral will work out as an inner product for basically the same reason the finite-dimensional version did: they both involve a sum of products of components.

(By the way, the summation sign is a Greek sigma, for sum, and the integral sign is a "long s" because Liebniz thought of it as an infinite sum of infinitesimal summands -- it is the smooth/continuous version of summation.)

Answer 2

They are defining the inner product in this sense. It is true that this is an inner product similar to what you know from basic linear algebra. But put that aside for a moment.

You have probably been given some properties of inner products, e.g. that $\|f\| = \sqrt{\langle f,f\rangle}$. If the thing that has been given to you is in fact an inner product, then it must satisfy these properties.

So let us compute, for instance, $\|f\|$ with the definition we have been handed:

$$\|f\| = \sqrt{\langle f, f\rangle} = \left(\int_0^1 t^2 t^2\, dt\right)^{1/2} = \sqrt{\frac15 \left.t^5 \right|_0^1} = \sqrt{\frac15}.$$

That's it. The rest should follow.

Answer 3

An inner product $V$ on a real vector space $E$ is a form which is symmetric positive definite. This means that:

1. form: $V:E\times E\mapsto\mathbb{R}$,
2. symmetric: for all $x,y\in E$, we have that $V(x,y)=V(y,x)$,
3. positive: for all $x\in E$, we have that $V(x,x)\ge0$,
4. definite: if for some $x\in E$, we have that $V(x,x)=0$, then $x=0$.

If $E=\mathbb{R}^n$, it is easy to see that $V(x,y)=x^Ty$ satisfies all those properties.

We can also see that if we take $E=\ell^2(\mathbb{R})$ which is the space of all $\mathbb{R}$-valued infinite sequences $(x_0,x_1,\ldots)$ such that $\sum_{i=0}^\infty x_i^2<\infty$, then the form $$V(x,y):=\sum_{i=0}^\infty x_iy_i$$ is also an inner-product on $E$.

Now, consider $E=C([0,1],\mathbb{R})$ which is the space of continuous functions from $[0,1]$ to $\mathbb{R}$. Recall also that an integral is also a sum, that is we have that

$$\int_0^1f(t)dt=\lim_{n\to\infty}\dfrac{1}{n}\sum_{i=0}^\infty f(t_i),\ t_i=i/n$$

which comes from the definition of a Riemann integral. So, the integral can be seen as averaged infinite sum.

Therefore, one can see that $$\langle f,g\rangle=\lim_{n\to\infty}\dfrac{1}{n}\sum_{i=0}^\infty f(t_i)g(t_i),\ t_i=i/n$$ which as a very similar structure as the inner-product on $\ell^2$ where $x_i=f(t_i)$ and $y_i=g(t_i)$; the only difference being the factor $1/n$.

One can easily prove that this is indeed an inner product as it verifies all the conditions stated earlier.

So, in the end, an integral is nothing else but a continuous sum that acts on spaces of functions with continuous domain. Those functions can then be seen as vectors with infinitely many (a continuum) of indices.