How would we define an inner product on the space $C[a,b]$ with sup-norm? Sorry for the dumb question.  I'm having some trouble figuring out how to define an inner product on this space.  The space is $C[a, b]$ and the norm here is defined by $\|f\| = \sup_{a \leq x \leq b}|f(x)|$.  
I know that $\|f\|^2 = \langle f, f \rangle$ but what about more generally $\langle f, g \rangle$?  I hope my question makes sense. 
 A: If you take $f(x)=\frac{x-a}{b-a}$ and $g(x)=\frac{b-x}{b-a}$ then $f,g$ take their values in $[0,1]$ and $||f||^2=||g||^2=1$
$f(x)+g(x)=\frac{b-a}{b-a}=1$ and $||f+g||^2=1$.
$f(x)-g(x)=\frac{2x-a-b}{b-a}$ takes its values in $[-1,1]$ and $||f-g||^2=1$
$||f+g||^2+||f-g||^2=2\neq 2(||f||^2+||g||^2)=4$
If $||\cdot||$ was issued from an inner product it would verify the parallelogram identity for any $f,g$, but this isn't the case here.
It doesn't mean that $C^0([a,b])$ can't be equiped with an inner product, just that if there exists one, the norm derived from it won't be equivalent to $||\cdot||_\infty$.
For instance $\displaystyle \langle f,g\rangle=\int_a^b f(t)g(t)dt$ is an inner product (real case, for complex case use $\bar g$ as user1952009 suggested).
Bilinearity, positivity, symetry are immediate, and $\langle f,f\rangle=0$ gives $f^2=0$ almost everywhere, but since $f$ is continuous we actually have $f=0$.
And now considering $h_n(x)=\sqrt{2n+1}\big(\frac{x-a}{b-a}\big)^n$ we have $||h_n||_\infty=\sqrt{2n+1}$ while $||h_n||_2=\langle f,f\rangle^{\frac 12}=\sqrt{b-a}$ so the two norms can't be equivalent since one is bounded and the other not for this sequence.
