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.

Any idea or suggestion would be much appreciated! Thanks in advance!

• What is the definition of a total cone? – fredgoodman Nov 2 '17 at 23:31
• Thanks @fredgoodman. A cone $P$ is called reproducing if and only if $E = \text{span} (P)$, that is, any element $w$ in $P$ can be expressed as the form of $w = u - v$ where $u, \, v \in P$ (i.e., $E = P-P = \{ u-v \colon u,\, v \in P \}$). If $\text{span} (P)$ is dense in $E$ (i.e., $E = \overline{P-P}$), then we say that $P$ is total. – Paradiesvogel Nov 2 '17 at 23:32
• What do linearity and boundedness mean if the domain of $B$ is not a vector space? If $B$ is (extended to) a linear and continuous operator on the subspace $P-P$ then you get the unique continuous extension from uniform continuity and this extension is again linear. – Jochen Nov 3 '17 at 8:28
• Thanks @Jochen. Let's treat the domain of $B$ as the positive cone $P$ of a real Banach space $E$. Actually, I'm wondering two things. The first one is that how could we extend this bounded linear operator $B \colon P \to P$ to a bounded linear operator $\overline{B} \colon P-P \to P-P$? Would you mind to explain it in detail please? Thanks a lot:) – Paradiesvogel Nov 3 '17 at 8:59
• Whenever you have a dense subspace $L$ of a Banach space $E$ and a bounded linear operator $B:L\to L$ the unique continuous extension $\overline B: E\to E$ has the same norm: By definition $\overline B(x)=\lim B(x_n)$ for every sequence $L\ni x_n\to x$ and hence $\|\overline B(x)\|= \lim \|B(x_n)\| \le \lim\sup \|B\| \|x_n\| = \|B\| \|x\|$ for every $x\in E$. – Jochen Nov 3 '17 at 9:37

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.

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.

• Thanks so much @fredgoodman. Yes, you are right, the paper mainly focus on the Banach space $E= C(X)$ with the positive cone of $C(X)$ which is solid and hence reproducing (i.e., $E = P-P$). But before stating the lemma 3, they generally denote $E$ be a real Banach space with a total cone $P$. Regarding your answer, I thought it might only work in some Banach spaces like $C(X)$ where $X$ is compact, or $C_b(X)$ continuous bounded functions space, or $L^{\infty} (X, \mu)$ essentially bounded $\mu$-measurable functions space. – Paradiesvogel Nov 4 '17 at 22:40
• However, for such $L^{p} (X, \mu)$ space with a Gaussian distribution $\mu$, we cannot always find such $M \gg 0$. For instance, take $p=1$ and $X = \mathbb{R}$ for simplicity. Let $u(x) := - \exp(x)$, which is in $E= L^{p} (\mathbb{R}, \mu)$, but I didn't see we can find a sufficiently large number $M >0$ such that $- \exp(x) + M \geq 0$ for all $x \in \mathbb{R}$. Am I right? Thanks a lot again:) – Paradiesvogel Nov 4 '17 at 22:43
• Right, my answer will only work in special cases, as you say. I wouldn't necessarily trust that the general case is true without digging into the literature. That something is stated (without proof) in a math paper is only an indication that it might be true. – fredgoodman Nov 4 '17 at 23:18
• Oh, I see. Many thanks again @fredgoodman :) That was a huge help to me. May I ask you one more question please? I thought: since $P$ is a total cone of $E$, which means span$\{P\}=P−P$ is dense in $E$, thus we can do like this: as $P$ is a basis for a vector space $P−P$, then for any linear transformation $B \colon P \to P$, there exists a unique linear map (called it linear extension) $\overline{B} \colon P-P \to P-P$ such that $\overline{B} |_{P} = B$. In this connection, $B$ can be extended linearly, and it's what one usually does in linear algebra. – Paradiesvogel Nov 4 '17 at 23:37
• After that, continuous linear extension theorem may apply and yield the result of the statement. Could you please tell me am I right? If not, which step I go wrong? Thanks:) – Paradiesvogel Nov 4 '17 at 23:38