The uniqueness of the continuation of the functional. I have a continuous linear functional $f_0 \in c_0^*$ where 
$$c_0 = \lbrace{x: (x_1, \dots, x_n, \dots)|\lim_{n\rightarrow\infty}x_n = 0\rbrace} \subset m = \lbrace{x: (x_1, \dots, x_n, \dots)|\sup_{n\in\mathbb{N}}|x_n| < \infty\rbrace}$$
My question:

Prove that $f_0$ have a uniqueness continuation $f$ on space $m$ with respecting the norm, i.e
  $$\forall x \in c_0,\ f(x) = f_0(x)\quad \text{and}\quad ||f_0|| = ||f||.$$

The Hahn-Banach theorem guarantees the existence of such a continuation, but not its uniqueness. I tried to figure out some properties of the space conjugate to $c_0$. I proved that $c_0^* \cong l^1$. Having proved this, I saw how functionals on $c_0$ looks like. Then I proved that $(l^1)^* \cong m$ and saw how functionals on $l^1$ looks like. This is good, because in these spaces I cannot use the Riesz theorem, but, as I said, in the process of the proof I obtained the general form of functionals on these spaces.
But further, I do not know what to do. How to use this information? Or maybe there are some tricky lemmas or theorems that give information about the uniqueness of the continuation?
Could you tell me what to do next? Thanks in advance!
 A: $f_0$ can be written as $x\mapsto \sum a_nx_n$ for $x\in c_0.$ We can define an extension $f\in m^*$ by $f(x)=\sum a_nx_n$ for $x\in m.$
Consider a $g\in m^*$ such that $g(x)=f(x)=f_0(x)$ for $x\in c_0.$ Suppose $g(x)>f(x)$ for some $x\in m$ with $\|x\|_\infty=1.$ Define sequences $y^n\in m$ by $y^n_t=x_t$ for $t>n$ and $y^n_t=\mathrm{sgn}(a_t)$ for $t\leq n.$ Then $\|y^n\|=1$ and $f(y^n)\to \sum |a_n|=\|f\|.$ But $x-y^n\in c_0,$ so $g(y^n)=f(y^n)+g(x)-f(x)\to \|f\|+g(x)-f(x)>\|f\|.$ If $\|g\|=\|f\|$ then $g(x)\leq f(x)$ for all $x\in m$; applying this to $-x$ as well gives $g(x)=f(x).$

It is possible to use the Riesz representation theorem. Bounded functions on $\mathbb N$ extend to continuous functions on the Stone–Čech compactification $\beta\mathbb N.$ We can split any measure on $\beta \mathbb N$ into a measure on the open subset $\mathbb N$ and a measure on the closed subset $\beta\mathbb N\setminus N.$ We can formalize this as follows.
For a locally compact topological space $X,$
let $C(X)$ denote the Banach space of continuous functions on $X$ with sup norm, and let $C_0(X)\subseteq C(X)$ denote the subspace of functions $f$ vanishing at infinity: $\{x\in X:|f(x)|\geq \epsilon\}$ is compact for each $\epsilon>0.$
Let $\mathrm{Meas}(X)$ denote the Banach space of finite regular Borel measures on $X,$ with total variation norm. The "Riesz-Markov theorem" says
$$C_0(X)^*\cong \mathrm{Meas}(X)$$
So
$$c_0^*\cong C_0(\mathbb N)\cong \mathrm{Meas}(\mathbb N)$$
$$m^*\cong C(\mathbb N)\cong C(\beta\mathbb N)\cong C_0(\beta\mathbb N)\cong \mathrm{Meas}(\beta\mathbb N)$$
A measure $\mu\in\mathrm{Meas}(X)$ is specified by a function $\mathcal B(X)\to\mathbb R$ where $\mathcal B(X)$ is the set of Borel sets in $X.$ Given $\mu\in\mathrm{Meas}(\beta\mathbb N),$ restricting the domain to $\mathcal B(\mathbb N)$ gives a measure $\mu'$ on $\mathbb N.$ Similarly, restricting the domain to $\mathcal B(\beta\mathbb N\setminus\mathbb N)$ gives a measure $\mu''$ on $\beta\mathbb N\setminus \mathbb N.$ And $\mu$ can be recovered from these restrictions: $\mu(B)=\mu'(B\cap\mathbb N)+\mu''(B\setminus\mathbb N).$ I think it is easy to show these measures are regular if $\mu$ is. In this way we get an isomorphism
$$\mathrm{Meas}(\beta\mathbb N)\cong \mathrm{Meas}(\mathbb N)\oplus_1\mathrm{Meas}(\beta\mathbb N\setminus \mathbb N)$$
where $\oplus_1$ denotes $\ell_1$ direct sum of Banach spaces.
Combining these isomorphisms gives $$m^*\cong c_0^*\oplus_1\mathrm{Meas}(\beta\mathbb N\setminus \mathbb N)$$
The first component $m^*\to c_0^*$ of this isomorphism is the same as the usual map (given by precomposition by the inclusion $c_0\to m$). So $f\in m^*$ is an extension of $f_0\in c_0^*$ if and only if $f$ gets mapped to $(f_0,\nu)$ under this isomorphism, for some $\nu.$
Note $\|f\|=\|(f_0,\nu)\|=\|f_0\|+\|\nu\|.$ So $\|f\|=\|f_0\|$ if and only if $\|\nu\|=0.$ This means there is a unique extension $f$ of $f_0$ with $\|f\|=\|f_0\|.$
