Show that there does not exist $g\in C_{\Bbb R}([-1,1])$ such that $f(0)=\langle f,g\rangle$ for every $f\in C_{\Bbb R}([-1,1])$ . Let $C_{\Bbb R}([-1,1])$ be the vector space of continuous real valued functions on the interval $[-1,1]$ with inner product given by $\langle f,g\rangle=\int _{-1}^1f(x)g(x)\,dx$
for $f,g\in  C_{\Bbb R}([-1,1])$ .

Show that there does not exist $g\in C_{\Bbb R}([-1,1])$  such that $f(0)=\langle f,g\rangle$ for every $f\in C_{\Bbb R}([-1,1])$ .

Assume the contrary. Then $\exists g\in C_{\Bbb R}([-1,1])$  such that $f(0)=\langle f,g\rangle$ for every $f\in C_{\Bbb R}([-1,1])$ .
I took $f(x)=$
\begin{cases} 0 & -1\le x\le 0\\mx& 0\le x\le \frac{1}{m}\\1& \frac{1}{m} \le x\le 1\end{cases}
Then $0=f(0)=\int _{-1}^1 fg=\int _0^1 fg=\int _0^{\frac{1}{m}}fg+\int _{\frac{1}{m}}^1fg$
$\implies m|\int _0^{\frac{1}{m}}xg(x)|=|\int _{\frac{1}{m}}^1g(x)|$
Also $g$ is uniformly continuous on $[-1,1]$ .
I am not able to proceed further.
Should I use any other example for $f$?Also how to use the infinite dimensionality of $C_{\Bbb R}([-1,1])$?
NOTE: I took the above since for easy examples of $f$ like polynomials etc. I did not get anything.

Take $f(x)=x^2g(x)$, you get $$\int_0^1(xg(x))^2dx=0.$$
The continuity and nonnegativity of the integrand imply then that $xg(x)=0$ for all $x\in [-1,1]$. Thus $g(x)=0$ for all $x\in [-1,1]\setminus\{0\}$. But, since $g$ is contiuous we conclude that $g\equiv0$. 
Now, testing with the constant function $f\equiv1$ we get the contradiction $1=\int_{-1}^1g(x)dx=0$. So, no such $g$ exist.
