Let $\psi:\mathbb{R}\to\mathbb{R}$ be a convex function such that $\psi'(0)<0$. Can we say that the point at which $\psi$ attains its minimum is positive? Please help me.
I know the basic results related to second derivative and convexity. Any hint or solution is highly appreciated.
The function need not have a minimum at all: It can be unbounded below (e.g. $\psi(x) = -x$) or bounded below without attaining a minimum (e.g. $\psi(x) = e^{-x}$).
But if $\psi$ is convex and differentiable with $\psi'(0) < 0$ and if the function has a minimum at $x_0$ then $\psi'(x_0) = 0$ implies $x_0 > 0$.
The reason is that the derivative of a convex (differentiable) function is increasing, so that $\psi'(x) \le \psi'(0) < 0$ for all $x \le 0$.
Alternatively one can use that $$ \psi(x) \ge \psi(0) + x \psi'(0) > \psi(0) $$ for $x < 0$, so that $\psi$ cannot have a minimum at $x_0 < 0$.