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convergence of unconstraintunconstrained convex optimization

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convergence of unconstraint convex optimization

I encounter an optimization problem. The simplified version is like following:

Denote function $F(x):\mathbf{R}^n\rightarrow\mathbf{R}$, where $F(x)$ is a smooth lower bounded convex function (i.e. $F(x)>0$). The hessian matrix of $F(x)$ is positive definite. However $F(x)$ is not a strongly convex function, namely we can not assume the hessian matrix $H(x)\succ m I_n$.

Suppose for the sequence $\{\mathbf{x}_k\}_{k=1}^\infty$, the gradients of the $F(x)$ converge to zero: \begin{equation} \lim_{k\rightarrow\infty} \|\mathbf{g}_k\|_2 = 0,\text{ and }F(x_{k+1})\le F(x_{k}), k=0,1,... \end{equation} where $\mathbf{g}_k=\frac{\partial F}{\partial x} |_{x=x_k}$. Do we have the following result \begin{equation} \lim_{k\rightarrow\infty} F(x_{k}) = \inf _{x\in\mathbf{R}^n} F(x)? \end{equation}

I expect $\lim_{k\rightarrow\infty} F(x_k) = \inf_{x\in\mathbf{R}^n} F(x)$. However, I can not find proofs for this result. There are some results about this problem. But unfortunately, either assume the optimum point of $F(x)$ is attainable, namely, there exists $x^*\in\mathbf{R}^n$, such that $F(x^*) = \inf F(x)$ or there exists $x^*\in\mathbf{R}^n$, such that $\lim_{k\rightarrow\infty} x_k=x^*$.

Could any one help me to solve this problem or give me some comments? Thanks for your help.