a consequence of a linear transformations $\mathcal{L} (X, Y) $ between two real Banach spaces $X$ and $Y$ I read a following statement in an academic paper from Journal of Mathematical Analysis and Applications. Please refer to Lemma 3 in https://ac.els-cdn.com/S0022247X05001897/1-s2.0-S0022247X05001897-main.pdf?_tid=75c416d8-c0c8-11e7-9e29-00000aacb362&acdnat=1509735466_ce1f132e2c3285a21b65e184e2630ecd.

Let $E$ be a real Banach space endowed with complete norm $\| \cdot \|$ and $P$ be a total cone of $E$.
Suppose $B \colon P \to P$ is a bounded linear operator.
Therefore this operator $B$ can be uniquely extended to a bounded linear operator on
$\overline{B} \colon \overline{P-P} = E \to E$ such that
$\| \overline{B} \| = \| B\|$.

Since there is no proof or any comments regarding this statement in that paper, I did not get why it is true. I was thinking that this statement might be a consequence of the Hahn Banach theorem for linear transformations $\mathcal{L} (X, Y) $ between two real Banach spaces $X$ and $Y$.
In fact, the precondition for such consequences may require the space $Y$ having the extensible property, please refer to Section 10 in this note
http://www-personal.umich.edu/~romanv/teaching/2009-10/602/short-history-of-analysis.pdf .
However, regarding the statement I wrote here, they only assumed that $E$ is a real Banach space with a total positive cone $P$. I did not get why is that.
So, could anyone please help me out and explain it? or could anyone please prove the statement I wrote above?
Any idea or suggestion would be much appreciated! Thanks in advance!
 A: I wrote something for a special case before, but I was being a little dense.
Let $P$ be a cone in $E$ such that $E_0 = P - P$ is dense in $E$.  Let $B: P \to P$ be as linear as it can be,  namely $B(s u + t v) = s B(u) + t B(v)$ for $u, v \in P$ and $s, t \ge 0$.  Try to define a linear extension of $B$ to $E_0$ by $\bar B(u-v) = B(u) - B(v)$, for $u, v \in P$.  Is this well defined?  If $u_1 - v_1 = u_2 - v_2$, then $u_1 + v_2 = u_2 + v_1 \in P$. Thus $B(u_1) + B(v_2) = B(u_1 + v_2) = B(u_2 + v_1) = B(u_2) + B(v_1)$.  Therefore, $B(u_1) - B(v_1) = B(u_2) - B(v_2)$.  Thus $\bar B$ is a well defined extension to $E_0$. Linearity is easy to check.
Now what about boundedness?  If the linear extension $B$ is bounded, it extends uniquely to $E$ by continuity.  
It is here that you might need some extra hypotheses on $E$ and $P$.  The minimum that you need is the existence of a decomposition $f = f_+ - f_{-}$ with $||f_{\pm}|| \le k ||f||$. For then
$$
\begin{aligned}
||\bar B(f)|| &= ||\bar B(f_+ - f_{-})|| \\
&= ||B(f_+) - B(f_{-})|| \le 
||B(f_+)|| + ||B(f_{-})|| \\
&\le K(||f_+|| + ||f_{-}  ||)  \le 2 K k ||f||.
\end{aligned}
$$
You would expect better estimates in special cases.
A: The statement is false. One reason it is false is because it turns out continuity of a linear map on a normed cone is not equivalent to Lipschitz-continuity of this map, in strong contrast to the situation of a linear map acting on a normed vector space. Thus the linear extension of a continuous linear map on $P$ to $\mathrm{span}(P)$ will fail to be continuous if it was not Lipschitz. If it was Lipschitz then continuity will still remain satisfied.
Let $e_1,e_2$ be the standard basis of $\Bbb R^2$ and let $e^*_1, e_2^*$ be the dual basis. Let $\Bbb R^2$ have euclidean norm. Denote with
$$P_\theta = \{ \alpha\, (\cos\theta\, e_1+\sin\theta \, e_2)+\beta\,e_1\mid \alpha,\beta\in\Bbb R_{≥0}\}$$
the "cone of angle $\theta$". The idea is that for small $\theta$ you will need to subtract really big elements of $P_\theta$ in order to reach $e_2$, as the cone starts collapsing. 
For any $p\in P_\theta$ we have with $p=\alpha(\cos\theta\, e_1+\sin\theta \, e_2)+\beta\,e_1$ that
$$\|p\|=\sqrt{\alpha^2+\beta^2+2\alpha\beta\cos\theta}≥\alpha, \qquad \left|\frac{e_2^*(p)}{\sin\theta}\right|=\alpha.$$
We will look at a sequence of these cones where the angle $\theta$ collapses to $0$. The functional $\frac{e_2^*}{\sin\theta}$ will remain continuous, because while $\sin\theta\to0$ the cone is also getting thinner with the right speed to guarantee continuity. However the extension to $\Bbb R^2$ will grow unboundedly in norm.
To formalise that look at $P=\sum^+_{k\in\Bbb N}P_{1/k}$ as a cone of $\bigoplus_{k\in\Bbb N}\Bbb R_k^2$. Here $\sum^+$ is supposed to indicate (finite) convex linear combinations, so every element $p\in P$ is the form
$$\sum_{k=1}^N p_k\qquad N\in\Bbb N,\ p_k\in P_{1/k}$$
we give the vector space $\bigoplus_{k\in \Bbb N} \Bbb R^2_k$ the norm
$$\left\|\sum_k x_k\right\|=\sum_{k}\|x_k\|_2$$
where $\|\cdot\|_2$ is the euclidean norm on $\Bbb R^2$, $E$ should be the completion of this space with this norm. Now the counter-example is that the linear functional
$$F:\bigoplus_{k\in\Bbb N}\Bbb R^2_k\to\Bbb R, \qquad \sum_k x_k\mapsto \sum_k \frac{e_2^*(x_k)}{\sin 1/k}$$
is continuous on $P$, but it is not continuous on $\bigoplus_k \Bbb R^2_k=\mathrm{span}(P)$. First we show continuity on $P$:
Let $\sum_k p{k,n}\to \sum_k p_k$ in $P$. It follows that
$$\left\|\sum_k (p_{k,n}-p_k)\right\|=\sum_k\|p_{k,n}-p_k\|\to0$$
now only finitely many of the $p_k$ are allowed to be non-zero, so the expression is of the form
$$\sum_{k>N}\|p_{k,n}\|+\sum_{k=1}^N\|p_{k,n}-p_k\|≥\sum_{k>N}\alpha_{k,n} +\sum_{k=1}^N\|p_{k,n}-p_k\|$$
it follows $\sum_{k>N}\frac{e_2^*(p_{k,n})}{\sin1/k}=\sum_{k>N}\alpha_{k,n}\to0$. On the other hand $\sum_{k=1}^N\frac{e_2^*(p_{k,n}-p_{k,n})}{\sin1/k}$ must converge to zero, since $\sum_{k=1}^N p_{k,n}\to \sum_{k=1}^N p_k$ in $\bigoplus_{k=1}^N\Bbb R_k^2$, which is a finite dimensional vector space and all linear maps on finite dimensional spaces are continuous. Thus
$$F(\sum_k p_{k,n})-F(\sum_k p_k) = \sum_k F(p_{k,n}-p_k)=\sum_{k>N}\alpha_{k,n}+\sum_{k=1}^N \frac{e_2^*(p_{k,n}-p_k)}{\sin1/k}\to0.$$
So continuity of $F$ on $P$ has been verified. $F$ is not continuous $\bigoplus_k \Bbb R^2_k$: Look at the sequence $\sum_k x_{k,n}$ with $x_{k,n}=\sqrt{\sin1/k}\,\delta_{k,n}\cdot e_2$. Now the norm of this sequence is $\|x_{n,n}\|=\sqrt{\sin1/n}$, which converges to $0$. But applying $F$ to this sequence retrieves $\frac1{\sqrt{\sin1/n}}$, which diverges.
