To prove that on $C([0,1])$, the integral $\int_{0}^{1} f(x)g(x)dx $ defines a scalar product. Now, for the given operation to be a scalar product, I know I need to check four conditions. Here's what I have done so far:


*

*$\langle f,g\rangle $ = $\int_{0}^{1} f(x)g(x)dx$ = $\int_{0}^{1} g(x)f(x)dx $ = $\langle g,f\rangle $

*$\langle f+g,h\rangle $ = $\int_{0}^{1} (f+g)(x)h(x)dx$ = $\int_{0}^{1} f(x)h(x)dx$ + $\int_{0}^{1} g(x)h(x)dx$ = $\langle f,h\rangle $ + $\langle g,h\rangle $

*$\langle cf,g\rangle $ = $\int_{0}^{1} (cf)(x)g(x)dx$ = c$\int_{0}^{1} f(x)g(x)dx$ = c$\langle f,g\rangle $

*$\langle f,f\rangle $ = $\int_{0}^{1} f(x)f(x)dx$ = $\int_{0}^{1} f(x)^2dx\ \geq 0$ since, $f(x)^2 \geq 0$  $\forall x\in[0,1]$.
But, now how do I show that $\langle f,f\rangle  = 0 $ if and only if $f$ is the constant function zero?


Any help would be greatly appreciated.

EDIT is it necessary to explicitly demonstrate that $\langle f,f\rangle  = 0 $ if and only if $f$ is the constant function zero?
EDIT
here's how I approached the problem based on the replies I got here.
Assuming $f \neq 0,$ we know that $f(a)^2 = c \geq o$ $ a \in [0,1] $
Then we have by definition: for every $\epsilon > 0  \exists  \delta > 0 $ 
 Such that $|x-a| < \delta \implies |f(x)-f(a)| < \epsilon$
and so we can get $|f(x)|> 1/2 f(a)$ or $f(x)^2> 1/4 f(a)^2$ 
Thus, $\int_{0}^{1} f(x)^2dx $ $\geq$ $\int_{a-\delta}^{a+\delta} f(x)^2dx $ $\geq$ $\int_{a-\delta}^{a+\delta} 1/4f(a)^2dx $ $\geq$ $1/4f(a)^2$ $\int_{a-\delta}^{a+\delta}dx$ $\geq$ $1/4f(a)^2\delta$ > 0 
(Since $1/4f(a)^2 >0$ and $\delta$ > 0)
 A: Of course, if $f$ is the zero function $0$, we have
$$\color{#bf0000}{\langle 0, 0 \rangle} = \int_0^1 0^2 \,dx \color{#bf0000}{ = 0 }.$$
On the other hand, if $f$ is not the zero function, at some point $x_0 \in [0, 1]$ we have $f(x_0) \neq 0$, and by continuity there is some $\delta > 0$ such that $$|f(x)| \geq \frac{1}{2} f(x_0)$$ on the interval $I := [x_0 - \delta, x_0 + \delta] \cap [0, 1]$ (which has length at least $\delta$); in particular, if we denote the (positive) r.h.s. by $S$, we have $f^2(x) \geq S^2$ on that interval. Now, since $f^2$ is nonnegative, we have
$$\color{#bf0000}{\langle f, f \rangle} = \int_0^1 f^2 \,dx \geq \int_I S^2 \,dx = S^2 \int_I dx \geq S^2 \delta \color{#bf0000}{ > 0}$$ as desired.
A: Hint: If $g$ is integrable, with $g(x)\geq0,$ and
$$\int_0^{1}g(x)dx=0,$$
then $g(x)=0$ in almost every point. (This is a theorem of Real Analysis.) 
A: Assume $f \neq 0$, so that $f(x_0) =a>0$ (since we are integrating $f^2$. By continuity, there is an interval $(x_0- \epsilon, x_0+ \epsilon); \epsilon >0$ where $f^2>a/2$. Then  $\int f^2  \geq \int_ {x_0 -\epsilon}^{x_0+ \epsilon} f^2 = 2\epsilon a >0$
