Proving $\langle f,f \rangle =0 \implies f=0$

Question

Let $Y=C[0,1]$, prove that $\langle f,g \rangle = \int _0^1 f(x)g(x)dt$ is an inner product.

Just need to talk about the condition on the title. Can we simply say $<f,f>=0$ so $||f||=0$ so by definition of norms, this means $\iff f=0$ which proves the statement?

Answer 1

By properties of integrals, $\langle f,f\rangle = 0$ implies $$ 0 \le \int_{0}^{x}|f(t)|^2dt \le \int_{0}^{x}|f(t)|^2dt+\int_{x}^{1}|f(t)|^2dt = \int_{0}^{1}|f(t)|^2dt = 0. $$ Therefore, $\int_{0}^{x}|f(t)|^2dt=0$ for all $x \in [0,1]$. And $|f(t)|^2$ is continuous on $[0,1]$ because $f(t)$ is continuous. Therefore, by the Fundamental Theorem of Calculus, $$ 0 = \frac{d}{dx}\int_{0}^{x}|f(t)|^2dt = |f(x)|^2,\;\; 0 < x < 1. $$ The right derivative at $0$ is $|f(0)|^2$, which must also be $0$; the left derivative at $1$ is $|f(1)|^2$, which must also be $0$. So $f\equiv 0$.

Answer 2

Ah, but how do you know $||f||=<f,f>$ defines a norm? That requires a proof too!

Suppose $f\ne 0$. Then we can find $x \in [0,1]$ such that $f(x) \ne 0$. So $f(x)>0$ or $f(x)<0$. In either case, $f(x)^{2}>0$. But since $f$ is continuous at $x$, for any $\epsilon>0$ we can choose $\delta$ sufficiently small that if $|y-x|<\delta$, $|f(y)-f(x)|<\epsilon$. In particular, we can keep $f(y)^{2}>0$ on a small interval. Can you bound the integral of $f(y)^{2}$ on this small interval? What does it tell you about the integral over the whole of $[0,1]$?

Answer 3

Note that you need to know that $\|f\|=\sqrt{\langle f,f\rangle}$ is a norm to use it. But to prove that $\|f\|$ is a norm implies to prove that $\|f\|=0\Rightarrow f=0$, and prove this is equivalent to prove what you need to prove. So you are using what you want to prove to prove it! So what you did is not a good idea.

Suppose that $\langle f,f\rangle =0$, that is, $\int_0^1 f^2(x)dx =0$, and let $h=f^2$, so $h(x)\geq 0$ for every $x\in [0,1]$. Suppose that $h(a)> 0$ for some $a\in [0,1]$. As $h$ is continuous, you can find an open interval $I=]a-\epsilon , a+\epsilon[$ containing $a$ such that $h(x)>0$ for every $x\in I\cap [0,1]$. But then, writing $I\cap [0,1] = (\alpha,\beta)$ we have

$$ 0=\int_0^1 h(x)dx \geq \int_\alpha^\beta h(x)dx > 0,$$

an absurd. So $h(x)=0$ for every $x\in [0,1]$ and this implies that $f=0$.