Can two functions be substantially different if their inner product is close to $1$?

Question

I have met a strange feature. Assume that I have two functions $f(x)$ and $g(x)$ defined at the interval $(-1,1)$. The question is, if the inner product of these two functions is close to $1$, are these functions "close" in some sense to each other? I thought, yes, but then considered an example: $f(x)=(1-x^2)^{1/8}$, $g(x)=(1-x^2)^{-1/8}$. If I normalize both functions and calculate their inner product I get $0.977$. But the functions are obviously very different: $f(x)$ tends to $0$ when $x$ tends to $1$ or $-1$, while $g(x)$ tends to infinity. Can the functions be indeed so different in spite of their inner product being almost $1$, or do I make a fundamental error somewhere?

Answer 1

We have $$ \|f - g\|_2^2 = \|f\|_2^2 + \|g\|_2^2 - 2 (f,g)_2.$$ In your case, this results in $$ \|f - g\|_2 = \sqrt{2(1-0.977)} \approx 0.214. $$ Thus, it seems that the functions are not very close, as you observed.

Answer 2

The functions are only very different near two points in the domain $[-1,1]$. Over most of that domain the two functions are both close to $f(x)=1$.

The behavior of a function near an isolated point (or two) does not have as much effect on its similarity to other functions as its behavior over the rest of its domain, at least if you are taking inner products in a typical way, for example, $$ \langle f, g \rangle = \int_D f(x) g(x) \, \mathrm dx. $$

Also, is $0.977$ really close to $1$ when dealing with inner products? If you had $\cos\theta = 0.977$ in an exercise where $\theta$ was the angle between to vectors in $\mathbb R^3$, the angle between the vectors would be more than $12.3$ degrees.