This is a question from "Calculus of Variations and Optimal Control Theory" (by Daniel Liberzon).
"Suppose that $f$ is a $C^2$ function and $x^*$ is a point of its domain at which we have $\nabla f(x^*)\cdot d \geq 0$ and $d^T \nabla ^2f(x^*)d>0$ for every nonzero feasible direction $d$. Is $x^*$ necessarily a local minimum of $f$? Prove and give a counterexample."
It seems to me that this is indeed true, but I was not able to prove it. Can anyone prove this or give a counterexample?
The Taylor expansion around $x^*$ is given by $$ f(x^* + d) = f(x^*) + \nabla f(x^*)^T d + d^T \nabla^2 f(x^*) d + o(\lVert d \rVert^2) $$ which means that $$ f(x^* + d) - f(x^*) = d^T \nabla^2 f(x^*) d + o(\lVert d \rVert^2) $$ We use that $ d^T \nabla^2 f(x^*) d \geq \sigma_1 \lVert d \rVert^2$, where $\sigma_1$ is the smallest eigenvalue of $\nabla^2 f(x^*)$ and apply it so that $$ f(x^* + d) - f(x^*) \geq \sigma^1 \lVert d \rVert^2 + o(\lVert d \rVert^2) $$ Now we let $\lVert d \rVert \rightarrow 0$ and use that $$ \underset{\lVert d \rVert \rightarrow 0}{\lim} \frac{o (\lVert d \rVert^2)}{\lVert d \rVert^2} = 0 $$ to show that $$ \underset{\lVert d \rVert \rightarrow 0}{\lim} \frac{f(x^* + d) - f(x^*)}{\lVert d \rVert^2} \geq \sigma^1 $$ which implies that $x^*$ is a local minimum.