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Does there exists a differentiable convex function $f: \mathbb{R}^n \rightarrow \mathbb{R}$ with a unique global minimum which we denote by $x^*$ such that there exists a sequence $x_k$, not converging to $x^*$, such that $||f'(x_k)|| \rightarrow 0$?

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No. Let's normalize the function so that $u^*=0$ and $f(0)=0$. I claim that for every $r>0$ the infimum $\inf_{|x|> r}\| f'\|$ is positive; this obviously precludes the scenario described in your post.

Let $m=\min_{|x|=r} f(x)$; this is a positive number because $f$ attains its minimum on the sphere, and the minimum is not zero. Given $x$ with $|x|>r$, consider the one-variable function $\varphi(t)=f(tx/|x|)$. Clearly, $\varphi$ is convex, $\varphi(0)=0$ and $\varphi(r)\ge m$. By the mean value theorem, $\varphi'(\xi)\ge m/r$ for some $\xi\in (0,r)$. Since $\varphi'$ is increasing, it follows that $\varphi'(|x|)\ge m/r$. Thus $||f'(x)||\ge m/r$, proving the claim.

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