show $\int_0^1 f(t)g(t) dt$ is a non-degenerate scalar product Let $C[0,1]$ be the vector space of continuous real-valued functions on the interval [0,1]. The following mapping is defined:
$$
\langle  \bullet ,  \bullet \rangle : C[0,1]^2 \rightarrow \mathbb{R} : (f,g) \mapsto \int_0^1f(x)g(x)dx
$$
Firstly, this map is well defined since the product of two continuous function is continuous, and a continuous function is Riemann integrable.
I can show that $\langle  \bullet ,  \bullet \rangle$ is a scalar product:


*

*Linearity follows from the linearity of the Riemann integral

*Symmetry follows from the commutativity of functions inside the Riemann integral


I would like help showing that $\langle  \bullet ,  \bullet \rangle$ is non-degenerate, that is:
$$
If \ \ f \in C[0,1],\ \ and \ \ \langle  f ,  g \rangle = 0 \ \ \forall \ \ g \in C[0,1], \ \ then \ \ f \equiv 0. 
$$
Ordinarily, if we were working over a finite-dimensional vector space, I would evaluate $g$ at each of the standard basis vectors and hope to show that the corresponding component in $f$ is zero - and so $f$ is the zero vector. However I doubt this method works for $C[0,1]$ since it is infinite dimensional. 
I have also tried to derive a contradiction by assuming: $\exists \ \ c \in [0,1] s.t. f(c)\not= 0$, but to no success.
I'd appreciate any help given. Also, is there a name for this scalar product?

Reference
This was Chapter VI, Section 1, Example 2 taken from 
Lang, S., Linear Algebra (2nd ed.), Addison-Wesley
 A: Thanks to Clement C. for you're comment.
Take $g \equiv f$ and assume, for contradiction, $\exists \ \ c \in [0,1] s.t. f(c)\not= 0$. 
Consider the case $f(c) > 0$ (the case $f(c) < 0$ being similar):
By continuity of $f$,
$$
\forall \ \ \epsilon >0, \exists \ \ \delta >0, \left( \vert c-y \vert < \delta \Rightarrow \vert f(c) - f(y) \vert < \epsilon \right) 
$$
Taking $\epsilon = \frac{f(c)}{2} $  yields,
$$
 \exists \ \ \delta >0, \left( y \in (c-\delta,c+\delta) \Rightarrow f(y) \in \left(\frac{f(c)}{2},\frac{3f(c)}{2}) \right) \right)
$$
Take $\delta^* = min\{c,1-c,\delta\}$ to ensure $(c-\delta^*,c+\delta^*) \subseteq (c-\delta,c+\delta) $ and $(c-\delta^*,c+\delta^*)\subset [0,1]$
Next, define the following subdivision $\Delta = (0,c-\delta^*,c+\delta^*,1)$.
Consider the lower Riemann sum:
$$
\begin{align}
L(f^2,\Delta) 
&\geq \inf\{f^2(t):t \in [c-\delta^*,c+\delta^*]\}\vert 2\delta^*\vert \\
&\geq f(c) \vert \delta^*\vert \\
&> 0
\end{align}
$$
Finally,
$$
0 < L(f^2,\Delta) \leq \int_0^1f^2(t)dt =0
$$
which is a contradiction.
