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If $p$ is convex, thanthen $P(x)=\inf_{t>0}t^{-1}p(tx)$ is sublinear, isn't it? Also, if a linear functional is dominated by $p$, it is also dominated by $P$. Finally, $P\le p$. So there is no non-trivial gain in generality whatsoever unless you start talking about extending non-linear functionals but then you should restate the question accordingly.

If $p$ is convex, than $P(x)=\inf_{t>0}t^{-1}p(tx)$ is sublinear, isn't it? Also, if a linear functional is dominated by $p$, it is also dominated by $P$. Finally, $P\le p$. So there is no non-trivial gain in generality whatsoever unless you start talking about extending non-linear functionals but then you should restate the question accordingly.

If $p$ is convex, then $P(x)=\inf_{t>0}t^{-1}p(tx)$ is sublinear, isn't it? Also, if a linear functional is dominated by $p$, it is also dominated by $P$. Finally, $P\le p$. So there is no non-trivial gain in generality whatsoever unless you start talking about extending non-linear functionals but then you should restate the question accordingly.

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fedja
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If $p$ is convex, than $P(x)=\inf_{t>0}t^{-1}p(tx)$ is sublinear, isn't it? Also, if a linear functional is dominated by $p$, it is also dominated by $P$. Finally, $P\le p$. So there is no non-trivial gain in generality whatsoever unless you start talking about extending non-linear functionals but then you should restate the question accordingly.