I was reading Peter Lindqvist's material on the p-laplacian, there he derives a result from a convex based inequality that i have been stuck trying to show, how does one show that
$$
|b|^p \geq |a|^p + p<|a|^{p-2}a,b-a>
$$
based on the fact that for $p>1$ the function $f(x) = x^p$ (or $|x|^p$) is convex.
It seems as a direct application of jensen's inequality, but that leads to some inequalities as
$$
|a+b|^p \leq 2^{p-1} (|a|^p+|b|^p)
$$
and for showing with an inner product it seems that might be some trick lead by going to the $2$-norm, any hints?
Edit 1:
Trying to solve it for a while i think i got closer to it, the RHS of the first inequality can be rewritten as
$$
|a|^p + p<|a|^{p-2}a,b-a> = |a|^p -p |a|^p + p<|a|^{p-2}a,b> \\= (1-p)|a|^p + p<|a|^{p-2}a,b>
$$
using that $(1-p)/p = -1/p^\prime$, and dividing both sides by $p$, we get
$$
\frac{1}{p} |b|^p + \frac{1}{p^\prime} |a|^p \geq ^\ <|a|^{p-2}a,b>
$$
this looks more like a convex combination of the function, as taking $t=1/p$, we have in the LHS $tf(b) + (1-t) f(a)$, so we know that
$$
\frac{1}{p} |b|^p + \frac{1}{p^\prime} |a|^p \geq |\frac{1}{p} b + \frac{1}{p^\prime} a|^p
$$
but i still don't know how to compare both RHS.
Best Answer
Ok so after struggling with this one for a bit, it is a direct use of subgradient inequality for convex functions, that is $$ f(y) \geq f(x) \ + \ f^\prime (x)(y-x) $$ now as $f^\prime (x)(y-x) = p|x|^{p-1} < \frac{x}{|x|},y-x>$, by the chain rule, the result follows directly.
Note that the subgradient inequality requires the function to be differentiable at the point $x$, so our restraint is $a \neq 0$.