why $C_{2}[a,b]$ is an inner product space under the given inner product, while $C[a,b]$ is not? I could not see the problem here How can I prove if this is an inner product? as the calculation of the inner product is 1 with me, and also why $C_{2}[0,1]$ is an inner product space with the inner product defined by
$<f,g>=∫_{0}^{1}f(x)\bar g(x)dx$
, what is the difference?
 A: The linear space $C[a,b]$ of continuous functions on $[a,b]$ is an inner product space under $\langle f,g\rangle = \int_{0}^{1}f(t)\overline{g(t)}dt$. The only property in question is whether $\langle f,f\rangle =0$ iff $f=0$. But this property follows easily from the Fundamental Theorem of Calculus. For suppose
$$
   \int_{0}^{1}|f(t)|^2dt = 0.
$$
Then $\int_{0}^{x}|f(t)|^2dt=0$ for all $0 \le x \le 1$ because
$$
 0 \le \int_{0}^{x}|f(t)|^2dt \le \int_{0}^{x}|f(t)|^2dt + \int_{x}^{1}|f(t)|^2dt = \int_{0}^{1}|(f)|^2dt = 0.
$$
$|f(t)|^2$ is continuous because $f$ is. Hence, $0=\frac{d}{dx}\int_{0}^{x}|f(t)|^2dt=|f(x)|^2$ for all $x$.
The linear space $C_2[0,1]$ is also an inner product space under the same inner product.
Note that neither space is a Hilbert space; the completion of either space is $L^2[0,1]$.
A: The normed space is an inner product space if and only if the parallelogram law holds. This is Jordan-von Neumann Theorem (1937). Check whether or not the parallelogram law holds in $C[0,1]$.
