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.

If the linear functional $f \mapsto f(0)$ were of the form $f \mapsto \langle f, g\rangle$ for some $g \in C_\mathbb{R}[-1,1]$, then in particular it would be bounded:
$$|f(0)| = \left|\langle f, g\rangle\right| \le \|f\|_2\|g\|_2$$
However, $f \mapsto f(0)$ is not bounded. Consider $f_n \in C_\mathbb{R}[-1,1]$ given by $$f_n(x) = \begin{cases} 1+nx, & \text{ if $x \in \left[-\frac1n, 0\right]$} \\
1-nx, & \text{ if $x \in \left[0, \frac1n\right]$}\\
0, & \text{ otherwise }\end{cases}$$
Clearly $f_n(0) = 1$ for all $n \in \mathbb{N}$ but
$$\|f_n\|_2^2 = \int_{[-1,1]}|f_n|^2 = 2\int_{\left[-\frac1n, 0\right]} (1-nx)^2\,dx = \frac2n \int_{[0,1]} u^2\,du = \frac{2}{3n}$$
So $$\|g\|_2 \ge \frac{|f_n(0)|}{\|f_n\|_2} = \frac{1}{\sqrt{\frac2{3n}}} \xrightarrow{n\to\infty} \infty$$
which is a contradiction.
