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Suppose X' and Y are Banach spaces and X is a linear subspace dense in X'. Let T be a continuous map of X to Y satisfying:

(1) ||T(x+y)|| is less than or equal to ||T(x)||+||T(y)||.

Please prove whether or not in general it is true that T has a continuous extension to X'.

Please also answer the same question for the case where we assume in addition that

(2) ||T(cx)||=|c|*||T(x)|| for all c real.

If so far, the answer has been "no," then assume that Y is an L^p space on some sigma finite measure space, with p>1. (Let me know if it's true for p=1 too.) If instead of using the p norm in the definitions (1) of subadditivity or (1) and (2) of sublinearity, we have pointwise almost everywhere inequalities of absolute values, then can T be extended?

To be completely explicit, in this case we would have

|T(x+y)| is less than or equal to |T(x)|+|T(y)| and |T(cx)|=|c|*|T(x)| for all c real pointwise almost everywhere on the measure space.

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    $\begingroup$ Do you really want assume continuity of $T$ in your second sentence? $\endgroup$
    – Yemon Choi
    Apr 18, 2011 at 23:52
  • $\begingroup$ As stated, there are trivial counterexamples even when both spaces are the real line: Take $X$ to be a dense subset such that $x+y$ is not in $X$ for all $x$, $y$ in $X4 to make (1) irrelevant. $\endgroup$
    – Bill Johnson
    Apr 19, 2011 at 1:03
  • $\begingroup$ Yes. The purpose of proving the theorem above is to be able to use the Marcinkiewicz Interpolation theorem. The interpolation theorem states that if T is subadditive, and operates on functions with support of finite measure, then inequalities concerning infinity norms and weak L1 norms imply inequalities concerning p norms for 1<p<infinity. The problem is getting something useful out of this theorem for functions that are not bounded, supported on sets of finite measure. To still have the bounds would be nice, since it would imply a maximal inequality for the Hardy Littlewood function. $\endgroup$
    – Jeffrey
    Apr 19, 2011 at 1:20
  • $\begingroup$ My last comment was a response to Yemon, but this one is a response to Bill: X is a subspace. That is, it is a normed linear space. Hence, it is closed under addition. Sorry if I was not clear and you thought I meant topological space. $\endgroup$
    – Jeffrey
    Apr 19, 2011 at 1:22
  • $\begingroup$ edited to show this $\endgroup$
    – Gerald Edgar
    Apr 19, 2011 at 4:08

2 Answers 2

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To 1) and 2). Let $X^{'}=Y=c_{0}$ , and let $X=c_{00}$. Take some $p\in X^{'}\smallsetminus X$, and define $T:X\rightarrow Y$ by $Tx:=\left\Vert x\right\Vert \cdot\left(\sin\left(\left\Vert x-p\right\Vert ^{-1}\right),1,0,0,0,...\right)$.

Then $T$ satisfies your conditions, yet it is not continuously extendable.

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Yes, in your situation, you can continuously extend $T$ and the extension will also be a sublinear operator (the extension is also unique). The reason you can do this is because if $T$ is a sublinear operator which is continuous on $X$, then it is uniformly continuous on $X$. Moreover, $Y$ is complete, so it is possible to continuously extend $T$.

The example $T:c_{00} \to c_{0}: x \mapsto \|x \| (\sin( \|x-p\|^{-1}, 1, 0, \ldots)$ for $p \in c_0 \setminus c_{00}$ fixed is not a sublinear operator; it cannot be continuously extended because it is not Cauchy continuous.

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