Showing a function is Strictly Decreasing and find the right hand limit

Question

$g(r)=e^{(1/r)(ln( \sum_{i = 1}^{n} |x_i|^r))}$, where all $x_k$ are non-zero. So I have to show that it's strictly decreasing on (0, $\infty$)

$g'(r)= (e^{(1/r)(ln( \sum_{i = 1}^{n} |x_i|^r))})((-1/p^2)ln(\sum_{i = 1}^{n} |x_i|^r)+(1/p)\sum_{i = 1}^{n} |x_i|^r ln(x_i))$

To show decreasing the first derivative must be less than 0 for all $x=(x_1,...,x_n)$, so the e part could be cancelled out, and then I would have

$(1/p^2)ln(\sum_{i = 1}^{n} |x_i|^r) > (1/p)ln(\sum_{i = 1}^{n} |x_i|^r ln(x_i))$,

Cancel out a p from both sides, then I get stuck

Maybe I did it all wrong so far, thanks in advance

Edit: I'm completely lost on right hand limit, although I tried using the definition and it didn't really work out at all

Answer 1

Observe that $g(r)$ is simply the $r$-norm of $x=(x_1,...,x_n)\in\mathbb{R}^n$: $g(r)=e^{\ln({\sum|x_i|^r})^{1/r}}=(\sum|x_i|^r)^{1/r}$. Then it suffices to show that the $r$-norm is decreasing in $r$; and that $\lim_{r\to\infty}g(r)=g(\infty)$, where, with consistent notation, $g(\infty):=\max_i |x_i|$ is the $\infty$-norm.

P.S.: These are well-known results, and both of them have proofs even here on Math SE (e.g. see here and here). If you can't figure them out, I'll record them here as well.

---

Edit: Sorry for the late completion, here is the proof done in the way your teacher wanted. Let $\forall r\in]0,\infty[: g(r):\mathbb{R^n}\setminus\{0\}\to\mathbb{R^n}\setminus\{0\}, g(r)((x_1,...,x_n)):=e^{\frac{1}{r}\ln{\sum_{i=1}^n}|x_i|^r}=\sum_{i=1}^n |x_i|^{r})^{1/r}$. A few remarks are in order: First, $\forall r\in]0,\infty[: g(r)$ is a well-defined function (of $x$), but for $r<0$, it fails to be a norm. Needless to say, you could also consider $g:]0,\infty[\times(\mathbb{R^n}\setminus\{0\})\to\mathbb{R^n}\setminus\{0\}$. Second, $g$ need not be strictly decreasing in $r$, for instance for $n:=1,x_1:=1$, $g$ is even constant in $r$.

$g$ is a differentiable function of $r$, since it is a composition of such functions. Then we may differentiate:

$g'= g\left[\dfrac{1}{r}\ln{(\sum|a_i|^r)}\right]'= g\left[\dfrac{-\ln{(\sum|a_i|^r)}}{r^2}+\dfrac{\sum(|a_i|^r\ln{|a_i|)}}{r\sum|a_i|^r}\right]= \dfrac{g}{r\sum|a_i|^r}\left[\sum{(|a_i|^r\ln{|a_i|})}-\dfrac{1}{r}\ln{(\sum|a_i|^r)}\sum|a_i|^r\right].$

Observe that $\dfrac{1}{r}\ln{(\sum|a_i|^r)}$ does not depend on $i$, so that we can take it inside the sum:

$\dfrac{g}{r\sum|a_i|^r}\left[\sum{(|a_i|^r\ln{|a_i|})}-\dfrac{1}{r}\ln{(\sum|a_i|^r)}\sum|a_i|^r\right]= \dfrac{g}{r\sum|a_i|^r}\left[\sum{(|a_i|^r\ln{|a_i|})}-\sum{\left(|a_i|^r \dfrac{1}{r}\ln{(\sum|a_i|^r)}\right)}\right]= \dfrac{g}{r\sum|a_i|^r}\left[\sum{|a_i|^r \left(\ln{|a_i|}-\dfrac{1}{r}\ln{(\sum|a_i|^r)}\right) }\right]= \dfrac{g}{r\sum|a_i|^r}\left[\sum{|a_i|^r \ln{\dfrac{|a_i|}{(\sum|a_i|^r)^{1/r}}} }\right].$

Note that $\forall i: |a_i|^r\leq \sum |a_i|^r \implies |a_i|\leq(\sum|a_i|^r)^{1/r} \implies \dfrac{|a_i|}{ (\sum|a_i|^r)^{1/r} }\leq 1 \implies \ln{ \dfrac{|a_i|}{ (\sum|a_i|^r)^{1/r} }} \leq0.$ Then any term inside the bracketed sum is $\leq0$, so that the bracketed sum itself is nonpositive. As the first functional factors are nonnegative, the result follows.

Note: I think you can also prove the same result by induction on the dimension $n$ of $\mathbb{R^n}$.