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Jul 6, 2019 at 19:10 answer added LinearOperator32 timeline score: 0
May 12, 2014 at 21:14 history edited Ricardo Andrade
edited tags; replaced deprecated tag 'topology'
May 2, 2011 at 1:08 answer added Ady timeline score: 1
Apr 20, 2011 at 3:02 comment added Yemon Choi When you say: $T:X\to Y$ is continuous - what topology is one putting on $X$? The subspace topology inherited from $X'$ ?
Apr 19, 2011 at 4:09 history edited Gerald Edgar CC BY-SA 3.0
added 9 characters in body
Apr 19, 2011 at 4:08 comment added Gerald Edgar edited to show this
Apr 19, 2011 at 1:22 comment added Jeffrey 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.
Apr 19, 2011 at 1:20 comment added Jeffrey 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.
Apr 19, 2011 at 1:03 comment added Bill Johnson 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.
Apr 18, 2011 at 23:52 comment added Yemon Choi Do you really want assume continuity of $T$ in your second sentence?
Apr 18, 2011 at 22:47 history asked Jeffrey CC BY-SA 3.0