I wanted to use Jensen's Inequality to prove that $f(x)=xlog(x)$ is convex. I know how to do it by simply showing that the second derivative is positive, but I got stuck trying to prove it differently by showing that Jensen's Inequality always holds.
My solution was to use an arbitrary uniform distribution to show that the inequality holds in one case therefore it holds in all cases. Is this logic correct? In other words, if: $$ E[f(X)]\geq fE[X] $$ And the equality holds for one distribution, does this imply it holds for all distributions thus proving $f(x)$ is convex?
The theorem says the if $X$ is integrable and real valued, and $\phi$ a convex function then $E[\phi(X)] \geq \phi(E[X])$. Its not an equivalence only an implication, hence the inequality itself says nothing about convexity.