is product of norms convex? Is a function of the form 
$f(x) = \|x\|_1\|x\|_2$
convex in x? I have tried plotting it in wolfram alpha and it appears convex, althought I ahve not been able to show it yet
 A: No, it isn't.  Consider (for functions on $\mathbb R$)
$$ x(t) = \cases{1/n^3 & for $0 \le t \le n^4$\cr 0 & otherwise\cr}, \ y(t) = \cases{n^3 & for $-1/n^4 < t < 0$\cr 0 & otherwise\cr} $$
We have $\|x\|_1 = n$, $\|y\|_1 = 1/n$, $\|x\|_2 = 1/n$, $\|y\|_2 = n$, so $f(x) = f(y) = 1$.  However, $\|(x+y)/2\|_1  > n/2$ and $\|(x+y)/2\|_2 > n/2$,
so $f((x+y)/2) > n^2/4 > 1$ if $n > 2$.
EDIT: it seems that $f$ is convex on ${\mathbb R}^3$ but not on ${\mathbb R}^4$.
Counterexample for ${\mathbb R}^4$: $x = (1,0,0,0)$, $y = (0,1/5,1/5,1/5)$,
$f((x+y)/2) =  .4233202099 < (f(x)+f(y))/2 = .6039230485$.
EDIT: Oops, wrong way.  Try $x = (1,0,0,0)$, $y = (.9,.07,.07,.07)$,
$f((x+y)/2) = 1.004288519  > (f(x)+f(y))/2 = 1.004012120$.
A: Hint: we have $(\lVert x\rVert_1+\lVert y\rVert_1)((\lVert x\rVert_2+\lVert y\rVert_2)\geqslant 0$. Multiply this by $\alpha(1-\alpha)\geqslant 0$. It's the difference between $$\alpha f(x)+(1-\alpha)f(y)\mbox{ and  }\alpha \lVert x\rVert_1+(1-\alpha)\lVert y\rVert_1))(\alpha \lVert x\rVert_2+(1-\alpha)\lVert y\rVert_2).$$
