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.
 A: First, if $g\ne0$ then $$g(0)=\phi(g)=\int_{-1}^1 g^2(t)\,dt>0.$$ So by continuity there exists an interval $I=(-\delta,\delta)$ with $g>0$ on $I$. 
Take any interval $[a,b]\subset[-\delta,\delta]$ with $0\not\in[a,b]$. Construct a nonzero positive function $f$ with support $[a,b]$ (a "triangle" will do). Then
$$
0=f(0)=\phi(f)=\int_{-1}^1f(t)g(t)\,dt=\int_{a}^bf(t)g(t)\,dt\geq0.
$$
As $g$ is nonnegative and $f$ is positive, we must have $g=0$ on $[a,b]$. As we can do this for any $[a,b]\subset[-1,0)\cup(0,1]$ we obtain that $g$ can only be nonzero at $0$, a contradiction. That is, $g$ cannot exist. 
A: 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.
A: 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.
