Show that $v\mapsto \|v\|^p$ for $1\le p<\infty$ is strictly convex I want to show that $f(v)=\|v\|^p$ for $1\leq p&lt;\infty$ is strictly convex.

In the simplest case when $p=2$, we have:
$$\left\|\frac{v_1+v_2}{2}\right\|^2=\left(\frac{v_1+v_2}{2},\frac{v_1+v_2}{2}\right)=$$
$$\frac{\|v_1^2\|}{4}+\frac{\|v_2^2\|}{4}+\frac{1}{2}(v_1,v_2)$$
$$\leq \frac{\|v_1^2\|}{4}+\frac{\|v_2^2\|}{4}+ \frac{\|v_1^2\|}{4}+\frac{\|v_2^2\|}{4}$$
(by Cauchy inequality). That is convex, but not strictly convex - where did I make a mistake?

Please help me. Thanks.
The $p=2$ case:

If $f(\frac12(u+v))=\frac12(f(u)+f(v))$ then $\frac14\|u\|^2+\frac14\|v\|^2+\frac12\langle u,v\rangle=\frac12\|u\|^2+\frac12\|v\|^2$ by the computation in your post, hence $\|u\|^2+\|v\|^2-2\langle u,v\rangle=0$, that is, $\|u-v\|^2=0$. Thus, for every $u\ne v$, $f(\frac12(u+v))\lt\frac12(f(u)+f(v))$, QED.
The way that you show a function is convex is to draw a line between any two values; the values should always be on the same side.

It relies on a definition of convexity, where a line that strikes it twice is only inside (ie a specified $&lt;$ or $&gt;$) between these two striking points, and outside (the opposite inequality) everywhere else.  

So, for example, one might find $f(a)=p_a$, and $f(b)=p_b$, then $f(ax+by)&lt;x\cdot f(a)+y\cdot f(y)$, where $x+y=1$.  

So, one deconstructs the function as $f(p+dx) &lt; fp(p) + d.s$, where $s=f(p+x)-f(p)$ while $d&lt;1$, then the function is convex.