Riesz representation theorem fails for the pair 1, $\infty$ consider the space of Natural numbers equipped with the counting measure. 
find a bounded linear functional $\Phi \in (l^\infty)^*$ which is not of the form $\Phi=\Phi_g$  for some $g \in l^1$
Not sure where to start looking, any hints?
bump
 A: You will need the Hahn-Banach theorem. In the following $x=(x_k)_{k=1}^\infty$ represents a sequence.


*

*Consider the subspace $c$ of all convergent sequences. Define $\Phi_0\colon c\to \mathbb{R}$ (or $\mathbb{C}$) by $\Phi_0(x)=\lim_{k\to\infty}x_k$.

*Show that $\Phi_0$ is bounded on $c$ with norm equal to $1$.

*Use the Hahn-Banach theorem to extend it to a bounded linear functional on $\ell^\infty$.

*Show that there is no sequence $a\in\ell^1$ such that $\Phi(x)=\sum_{k=1}^\infty a_kx_k$ for all $x\in\ell^\infty$. (Consider $\Phi$ on $c_0$, the space of sequences converging to zero.)

A: To expand on Chris Eagle's suggestion (since it seems to be somewhat confusing to the OP):
An ultrafilter $\mathcal{F}$ on $\mathbb{N}$ is a collection of subsets of $\mathbb{N}$ such that:


*

*$\emptyset\notin \mathcal{F}$;

*If $A\in\mathcal{F}$ and $A\subseteq B\subseteq\mathbb{N}$, then $B\in\mathcal{F}$;

*If $A,B\in\mathcal{F}$, then $A\cap B\in\mathcal{F}$;

*For every $X\subseteq\mathbb{N}$, either $X\in\mathcal{F}$ or $\mathbb{N}-X\in\mathcal{F}$.


For example, $\mathcal{F}=\{A\subseteq \mathbb{N}\mid 0\in A\}$ is an example of an ultrafilter; this is a principal ultrafilter: an ultrafilter that consists of "all subsets that contain" a given element.
Using Zorn's Lemma one can show that there exist ultrafilters on $\mathbb{N}$ that are not principal. These are often useful for defining structure, especially on products indexed by $\mathbb{N}$.
Consider $\ell^0$, the set of all bounded real sequences. Given a sequence $(x_n)$, a point $x$, and an $\epsilon\gt 0$, we can consider the $\epsilon$ ball around $x$, $B(x,(x_n),\epsilon) = \{n\in\mathbb{N}\mid |x_n-x|\lt \epsilon\}$. 
We say that the sequence $(x_n)$ $\mathcal{F}$-converges to $x$ if and only if for every $\epsilon\gt 0$, $B(x,(x_n),\epsilon)\in\mathcal{F}$. 
Then:


*

*If $\mathcal{F}$-limit, if it exists, is unique:
Suppose that $x\neq y$ and $B(x,(x_n),\epsilon)\in\mathcal{F}$ for all $\epsilon\gt 0$. Let $\delta=\frac{1}{2}|x-y|$. If $|x_n-y|\lt\delta$, then $|x_n-x|\geq \delta$, since $2\delta=|x-y|\leq |x-x_n|+|x_n-y| = |x-x_n|+\delta$. Therefore, $B(y,(x_n),\delta) \subseteq \mathbb{N}-B(x,(x_n),\delta))$. In particular, we cannot have $B(y,(x_n),\delta)\in\mathcal{F}$: if $B(y,(x_n),\delta)\in\mathcal{F}$, then its intersection with $B(x,(x_n),\delta)$ would also be in $\mathcal{F}$, which is impossible because the intersection is empty. So $(x_n)$ does not converge to $y$. 

*If $\lim_{n\to\infty}x_n = x$  in the usual sense, then $(x_n)$ will $\mathcal{F}$-converge to $x$.
Let $\epsilon\gt 0$. Then there exists $N$ such that for all $n\geq N$, we have $|x_n-x|\lt \epsilon$. In particular, $\mathbb{N}-B(x,(x_n),\epsilon)\subseteq \{1,2,\ldots,N-1\}$. But $X=\{1,2,\ldots,N-1\}\notin \mathcal{F}$, because no singleton is in $\mathcal{F}$ (since $\mathcal{F}$ is nonprincipal), so no finite subset of $\mathcal{N}$ is in $\mathcal{F}$ (the complement of every singleton is in $\mathcal{F}$ since $\mathcal{F}$ is an ultrafilter, so any finite intersection of complements of singletons is in $\mathcal{F}$; this means that the complement of any finite subset of $\mathbb{N}$ is in $\mathcal{F}$, so no final subset is in $\mathcal{F}$).
Since the complement of $B(x,(x_n),\epsilon)$ is finite, it follows that $B(x,(x_n),\epsilon)\in\mathcal{F}$. This holds for all $\epsilon$, so $(x_n)$ also $\mathcal{F}$-converges to $x$.

*If $(x_n)$ $\mathcal{F}$-converges to $x$, then for any real number $\alpha$, $(\alpha x_n)$ $\mathcal{F}$-converges to $\alpha x$.
This is clear if $\alpha=0$, since then $(\alpha x_n)$ is the constant sequence $0$ which converges to $0$; assume $\alpha\neq 0$. Let $\epsilon\gt 0$. Then 
$$\begin{align*}
B(\alpha x,(\alpha x_n),\epsilon) &= \{n\in\mathbb{N}\mid |\alpha x -\alpha x_n|\lt\epsilon\}\\
 &= \{n\in\mathbb{N}\mid |x-x_n|\lt\epsilon/|\alpha|\}\\ 
&= B(x,(x_n),\epsilon/|\alpha|)\in\mathcal{F},
\end{align*}$$
so $(\alpha x_n)$ will $\mathcal{F}$-converge to $\alpha x$.

*If $(x_n)$ $\mathcal{F}$-converges to $x$ and $(y_n)$ $\mathcal{F}$-converges to $y$, then $(x_n+y_n)$ will $\mathcal{F}$-converge to $x+y$.
Let $\epsilon\gt 0$. Then $B(x,(x_n),\epsilon/2), B(y,(y_n),\epsilon/2)\in\mathcal{F}$, hence their intersection lies in $\mathcal{F}$. If $|x_n-x|\lt\epsilon/2$ and $|y_n-y|\lt\epsilon/2$, then $|(x_n+y_n)-(x+y)|\lt\epsilon$; so 
$$B(x+y, (x_n+y_n),\epsilon) \supseteq B(x,(x_n),\epsilon/2)\cap B(y,(y_n),\epsilon/2) \in \mathcal{F}$$
so $B(x+y, (x_n+y_n),\epsilon)\in\mathcal{F}$. Therefore, $(x_n+y_n)$ $\mathcal{F}$-converges to $x+y$.

*The big nontrivial part which I won't prove: if $(x_n)$ is bounded, then $x_n$ has an $\mathcal{F}$-limit.  I won't be able to prove it here, but it's nonetheless true. (It follows from the fact that if your sequences take values in a compact metric space, then they necessarily have an $\mathcal{F}$-limit).
So, taking these facts, define $\Phi_{\mathcal{F}}\colon\ell^{\infty}\to\mathbb{R}$ by letting $\Phi_{\mathcal{F}}(x_n)$ be the (unique) $\mathcal{F}$-limit of $(x_n)$. This is a linear functional. Now show that it is not equal to $\Phi_g$ for any $g\in\ell^1$. 
