Proving a result of a function on $L^1[-1,1]$ 
For every $\phi\in C[-1,1]$ define $f(\phi)=\phi(0)$.  
  
  
*
  
*Verify that $f\in C[-1,1]^*$ and compute $||f||$
  
*Prove that there is no Function $g\in L^1[-1,1]$ such that $f(\phi)=\int\limits_{-1}^1\phi gdx$ for all $\phi\in C[-1,1]$. 
  
*Identifying $C[-1,1]$ with a closed subspace of $L^{\infty}[-1,1]$ prove that standard embedding of $L^1[-1,1]$  into $L^{\infty}[-1,1]^*$
  is not onto
  

I was able to do the part 1:
Linearity is trivial.
Then by considering the continuity at $0$, we have $|f(\phi)|=|\phi(0)|\leq \epsilon+|\phi(x)|\leq\epsilon+||\phi||$.
Thus it gives the boundedness and $||f||\leq 1$.
And by taking the constant function $\phi=1$ we have $||f||\geq 1$.
Therefore $||f||=1$ 
For part 2:
I guess I can use the functions $\phi_n=1/n$  and contradict the fact that $g\in L^1$. that is $\int|g|<\infty$. Is it correct.  
And for 3, I'm clueless. what I know is that $L^1[-1,1]$ space is the all $g\in[-1,1]$ such that $\int\limits_{-1}^1|g|<\infty$.
And $L^{\infty}[-1,1]$ gives the bounded functions on $[-1,1]$ 
 A: For part three, assume the contrary that it is onto. Then for such an $f\in C[-1,1]^{\ast}$, as $C[-1,1]$ is a closed subspace of $L^{\infty}[-1,1]$, Hahn-Banach gives us some extension $\overline{f}\in L^{\infty}[-1,1]^{\ast}$ of $f$ such that $\overline{f}(\phi)=f(\phi)$ for $\phi\in C[-1,1]$. Then the surjective map gives some $g\in L^{1}[-1,1]$ such that $\overline{f}(\phi)=\displaystyle\int_{-1}^{1}\phi(x)g(x)dx$ for any $\phi\in L^{\infty}[-1,1]$. Realizing $\phi\in C[-1,1]$ and keep in mind that $\overline{f}(\phi)=f(\phi)$, this contradicts the result in part two.
For part two, let $\phi_{n}(x)=n^{2}x+n$ for $-1/n\leq x\leq 0$ and $\phi_{n}(x)=-n^{2}x+n$ for $0\leq x\leq 1/n$ and zero otherwise. Then $f(\phi_{n})=n$. On the other hand, given $\epsilon>0$, since $g$ is continuous at $0$, there is some $\delta>0$ such that $|g(x)-g(0)|<\epsilon$ for $|x|<\delta$. Then for large $n$, 
\begin{align*}
|f(\phi_{n})-g(0)|&=\left|\displaystyle\int_{-1}^{1}\phi_{n}(x)g(x)dx-\int_{-1}^{1}\phi_{n}(x)g(0)dx\right|\\
&\leq\displaystyle\int_{-1/n}^{1/n}|\phi_{n}(x)||g(x)-g(0)|dx\\
&\leq\epsilon\int_{-1/n}^{1/n}\phi_{n}(x)dx\\
&=\epsilon, 
\end{align*}
this means that $f(\phi_{n})\rightarrow g(0)$, this is a contradiction since $f(\phi_{n})\rightarrow\infty$.
Edit:
I overlooked that as if $g$ were continuous, the above argument for part two is only valid for continuous $g$.
