Solution
= Solution
A <convex function> satisfies $g(tx+(1-t)y)\le tg(x)+(1-t)g(y)$ for all $x,y$ in its domain and $0\le t\le1$. If the second <derivative> exists on an interval, this is equivalent to $g''\ge0$ there. The equivalence follows from the nondecreasing first <derivative>, or from the supporting-line inequalities obtained by integration. Here
$$
f'(x)=\log x+1,\qquad \boxed{f''(x)=1/x>0\quad(x>0).}
$$
Thus it is a <strictly convex function>, and in particular convex.