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Check out convexity of a function $J(u)=cu^r$, $J:[a,b]\rightarrow R$,
$0<a<b<\infty$, depending on values of parameters $c,r\in R$.

I know a definition of a convexity:
"Function J,defined on convex interval U, is convex on U if $J(\alpha u + (1-\alpha) v)\leq \alpha J(u) + (1-\alpha) J(v)$
and I know theorems:

Theorem 1
Let U be non-empty convex set in $R^n$ and $J\in C^1(U)$.
Function $J$ is convex if and only if

$\langle J'(u)-J'(v),u-v\rangle \geq 0$ for all $u,v\in U$

Theorem 2
Let U be convex set in $R^n$ with non-empty interior. Let $J\in C^2(U)$.
Function $J$ is convex on U if and only if
$\langle J''(u) ξ,ξ\rangle \geq 0$ for all $u\in U$ and all $ξ\in R^n$.

But I don't know what to use.
I tried in 3 ways:
Using Theorem 1

$u,v\in [a,b]$.
$\langle J'(u)-J'(v),u-v\rangle = \langle cru^{r-1}-crv^{r-1},u-v\rangle = cr\langle u^{r-1}-v^{r-1},u-v\rangle = cr \sqrt[2] {|u^{r-1}-v^{r-1}|^2+|u-v|^2}$
If this is ok, then J is convex if
1) $c,r\geq 0$
2) $c,r\leq 0$

Using Theorem 2:

$J'(u)=rcu^{r-1}$
$J''(u)=r(r-1)cu^{r-2}$

$\langle J''(u) ξ,ξ\rangle=\langle r(r-1)cu^{r-2} ξ,ξ\rangle$
and I don't know what now.

Using definition

I should show that
$J(\alpha u + (1-\alpha) v)\leq \alpha J(u) + (1-\alpha) J(v)$
for all $u,v\in [a,b]$
So, let $u,v\in [a,b]$.
$J(\alpha u + (1-\alpha) v)=c(\alpha u + (1-\alpha) v)^r$
And I don't know what to do now.

user23709
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  • So, I can write $⟨r(r−1)cu^{r−2}ξ,ξ⟩=r(r−1)cu^{r−2}\langle ξ,ξ⟩$ ? – user23709 Jun 02 '13 at 19:44
  • @copper.hat why second derivative is scalar? – user23709 Jun 11 '13 at 14:08
  • Because the function $J:[a,b] \to \mathbb{R}$ (that is, the domain is a scalar). – copper.hat Jun 11 '13 at 14:45
  • why first derivative is not scalar then? And what about 3rd derivative? It would be $0$ then? – user23709 Jun 11 '13 at 14:55
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    All derivatives are scalars if the domain & range consist of scalars! – copper.hat Jun 11 '13 at 15:02