Talk:Power Function is Convex Real Function

Probably need a better name for this, seeing as it clashes with Power Function on Strictly Positive Base is Convex. Caliburn (talk) 18:38, 6 January 2023 (UTC)

They're effectively the same thing, so they should be merged. Anyone up for it? --prime mover (talk) 20:04, 6 January 2023 (UTC)

They're a bit different, the first is basically about $x \mapsto e^x$, this is about $x \mapsto x^2$, $x \mapsto x^\pi$ and similar. You could compose with a log to get this result, though. Caliburn (talk) 20:09, 6 January 2023 (UTC)

They are all just examples of the same truth: $a \to b^c$ is convex in general. Within certain constraints, of course, which are to be presented in the conventional way. --prime mover (talk) 20:29, 6 January 2023 (UTC)

I don't get the importance of this particular result. Why is the strict convexity of $x^p$ in case of $p>1$ is not mentioned? Why not more generally state Real Function with Positive Second Derivative is Convex? --Usagiop (talk) 22:30, 6 January 2023 (UTC)

Because it hasn't been posted. Well volunteered, that man. --prime mover (talk) 23:10, 6 January 2023 (UTC)

Throwing this up as a lemma for the result that if $\tuple {X, \Sigma, \mu}$ is a finite measure space and $p \ge q \ge 1$ then $\map {\LL^p} {X, \Sigma, \mu} \subseteq \map {\LL^q} {X, \Sigma, \mu}$. Will put this up tomorrow. Also in my PDEs class I always found myself appealing to the fact that $x \mapsto \size x^p$ is convex. Caliburn (talk) 22:57, 6 January 2023 (UTC)

Contains $0^{-1}$

$x^{p-2}$ for $x\geq0$ and $p\geq1$ seems not ok, or am i missing something?

We should indeed probably make a special case for $x = 0$, or something. I'm no mathematician so I haven't a clue about how to properly express the details. --prime mover (talk) 09:09, 21 May 2024 (UTC)

We could use the continuity of the power functions to push the defining estimation to the boundary, which we know by the given method for the interior of our interval. Peter Grabs (talk)

Yeah this is an omission. Isn't convexity with $0$ immediate from $\map f {t x} = \paren {t x}^p = t^p x^p \le t x^p = t \map f x$ for $t \in \closedint 0 1$. Then the proof here shows that $\map f {t x + \paren {1 - t} y} \le t \map f x + \paren {1 - t} \map f y$ for $x, y \in \hointl 0 1$. Caliburn (talk) 10:24, 21 May 2024 (UTC)

very good! Peter Grabs (talk) 12:03, 21 May 2024 (UTC)

Done Caliburn (talk) 16:07, 22 May 2024 (UTC)