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Consider a strongly convex function $~f: \mathbb{R}^n \rightarrow \mathbb{R^+}~$ with a unique minimum at the point $x^* \in \mathbb{R}^n$.

I am wondering: if I have another point $y \in \mathbb{R}^n$ such that $\nabla{f}(y)^T e_i = 0$ for some $i = 1 \ldots n$, then is it true that $y_i = x_i^*$.

If so, I was hoping that someone could outline a proof. If not, can someone provide a counter example that is easy to visualize?

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  • $\begingroup$ @StevenTaschuk No just for 1 value of $i \in {1,\ldots,n}$ but not the others. $\endgroup$
    – Elements
    Jul 31 '15 at 0:10
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Let $f(x) = x^T A x$ with $$A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}. $$ Then the minimum is at $x^*=(0,0)$. At $y = (1, -2)^T$, we have that $e_1^T \nabla f(y) = 2(Ay)_1 = 0$. Thus in this case $y_1 \neq x_1^*$.

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  • $\begingroup$ Doesn't work. In 2 dimensions, $\nabla f(x) = x_1^2 + x_2^2$ so $x^* = (0,0)$. At $y = (0,1)$, you do have that $e_1^T \nabla f(y) = 2y_1 = 0$. However, in this case, $y_1=x_1^*$. That being said, your answer was insightful! I think that if the function is $f$ is separable in each dimension, then the property I describe will hold. $\endgroup$
    – Elements
    Jul 31 '15 at 0:02
  • $\begingroup$ @Elements: Fixed the example. $\endgroup$
    – user251257
    Jul 31 '15 at 0:09

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