Prove that $\langle f,g \rangle = \int_0^1 f(t)g(t) dt$ is an inner product on $C([0,1])$

Question

Question: Let $V=C([0,1])$ be the vector space of real-valued continuous functions on $[0,1]$. For $f,g \in V$, define $$\langle f,g \rangle = \int_0^1 f(t)g(t) dt. $$ Prove that $(V,\langle\cdot,\cdot\rangle)$ is an inner product space.

I know that we need to prove according to the $4$ conditions of inner product space.
But I don't understand how to prove the $4$-th condition, i.e. the concept and steps behind it.

$4$-th condition: $\langle u,u \rangle >0$, if $u \neq 0 $, my prof sort of explained how to prove using the concept of continuity.

Answer 1

If $f$ is not identically zero, then there is $x_0\in [0,1]$ such that $f^2(x_0)>0$. By continuity there are $0\leq a<b\leq 1$ such that $x_0\in[a,b]$ and $f^2(x)\geq f^2(x_0)/2$ for all $x\in [a,b]$. Hence $$\langle f,f \rangle=\int_0^1 f^2(x) dx\geq \int_{a}^{b} f^2(x)\, dx\geq \frac{f^2(x_0)}{2}\int_{a}^{b} \, dx=\frac{(b-a)f^2(x_0)}{2}>0.$$

Answer 2

Suppose $\langle u,u\rangle = 0$. Then, for all $0 \le x \le 1$, $$ 0 \le \int_{0}^{x}u(t)^2 dt \le \int_{0}^{1}u(t)^2 dt = 0. $$ Therefore, by the Fundmental Theorem of Calculus, $$ 0=\frac{d}{dx}\int_{0}^{x}u(t)^2dt = u(x)^2,\;\; x \in [0,1]. $$