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I know that $p$-norm of $xinBbb{R}^n$ is defined as, for all $pge1$,$$Vert{x}Vert_p=left(sum_{i=1}^{n} vert{x_i}vert^pright)^{1/p}.$$

The textbook refers to "Every norm is convex" for an example of convex functions.

I failed to prove $f(x)=Vert{x}Vert_p$ for all $pge1$, then tried to find the proof on the internet but I cannot find it.

Can someone let me understand why $p$-norm is convex for all $pge1$.

The Definition of a norm is:

Be V a Vectorspace, $|cdot|: V rightarrow mathbb{R} $ is a norm $:Leftrightarrow $

1. 
   $forall v in V: |v|ge0$ and $|v| =0 Leftrightarrow v=0$ (positive/definite)


2. 
   $forall vin V, lambdain mathbb{R}: |lambda||v| =|lambda v|$ (absolutely scaleable)


3. 
   $forall v,win V : |v+w| le |v|+|w|$ (Triangle inequality)

The Definition of convex is:

$f:Vrightarrowmathbb{R}$ is convex $:Leftrightarrow$
$forall v,w in V, lambda in [0,1]: f(lambda v+(1-lambda )w)le lambda f(v) +(1-lambda)f(w)$

So using the Triangle inequality and the fact that the norm is absolutely scaleable, you can see that every Norm is convex:
$$|lambda v+(1-lambda )w|le|lambda v|+|(1-lambda)w| = lambda|v|+(1- lambda)|w|$$

So by definition every norm is convex. What is left to show is, that the p-norm is in fact a norm.The first two Requirements are pretty easy to show, the third is hard. That is why it has its own name: the Minkowski Inequality which is a result of the Hölder inequality and shows that the triangle inequality holds for every p-norm (if p>1) and thus that it is a norm.

EDIT: Since this seems to be somewhat popular, I thought I would add a sketch of the proof of the minkowski inequality.

1. You show Young's Inequality: $xyle frac{x^p}{p}+frac{y^q}{q}quad forall q,p>1 text{ with } frac{1}{p}+frac{1}{q}=1, forall x,yge 0$.

You can do that by looking at the function $f(x)=frac{x^p}{p}+frac{y^q}{q}-xy$ find the extremum, show it is a minimum and is greater zero (derivatives).

2. You show the Hölder Inequality: $|fg|_1 le |f|_p|g|_q quad forall q,p>1 text{ with } frac{1}{p}+frac{1}{q}=1$
   

You can do that by setting $x=frac{|f|}{||f||_p}$ and $y=frac{|g|}{||g||_q}$ and plug them into young's inequality.
You get
begin{align}
&&frac{|fg|}{|f|_p|g|_q}&le frac{|f|^p}{p|f|_p^p} + frac{|g|^q}{q|g|_q^q}
\
Rightarrow &&int frac{|fg|}{|f|_p|g|_q} dmu &le int frac{|f|^p}{p|f|_p^p}dmu + int frac{|g|^q}{q|g|_q^q}dmu
\
Rightarrow &&frac{|fg|_1}{|f|_p|g|_q}&le frac{1}{p}+frac{1}{q}=1
end{align}
It works just the same for sequences or $mathbb{R}^n$, you just use young's inequality for every index and then sum over it instead of using the integral.

3. And last the Minkowski Inequality: $|x+y|_ple|x|_p+|y|_p quad forall p>1$
   

Set $q=frac{p}{p-1}$ thus $q(p-1)=p$ and $frac{1}{p}+frac{1}{q}=1$. Then:
begin{align}
|x+y|_p^p&=int |x+y|^pdmuleint |x+y|^{p-1}|x|dmu+ int |x+y|^{p-1}|y|dmu
\
&le left(int|x+y|^{q(p-1)}dmuright)^{1/q}left(int|x|^pdmuright)^{1/p} + left(int|x+y|^{q(p-1)}dmuright)^{1/q}left(int|y|^pdmuright)^{1/p}
\
&=left(int|x+y|^{p}dmuright)^{frac{1}{p}frac{p}{q}}(|x|_p+|y|_p)
=|x+y|_p^{p/q}(|x|_p+|y|_p)
end{align}

If you realize that $p-frac{p}{q}=p(1-frac{1}{q})=1$ you are done.