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Fix constants $0 < m < L$. We say the function $f \in C^1(\mathbb{R}^n;\mathbb{R})$ is bounded in the sector $[m, L]$ if there exists a reference point $y^* \in \mathbb{R}^n$ such that for all $y \in \mathbb{R}^n$,

\begin{equation*} (m(y - y^*) - f(y), L(y- y^*) - f(y)) \le 0, \end{equation*}

where $(\cdot, \cdot)$ denotes the usual inner product on $\mathbb{R}^n$.

Show that the point $y^*$ is the unique global minimizer of $f$.

Here is my attempt so far. Expanding the above inequality and rearranging gives

\begin{equation*} mL \| y - y^* \|^2 + \|\nabla f(y) \|^2 \le (m + L) (y - y^*) \cdot \nabla f(y) \end{equation*}

From this we immediately see that $\| \nabla f(y) \| \neq 0$ for all $y \neq y^* $. Furthermore, continuity gives $\|\nabla f(y^*)\| = 0 $. However, this is not enough to ensure $y^* $ is a global extremum of $f$, see here. Somehow the condition $0 < m < L$ must be used, and here is where I am stuck.

I have tried applying the multidimensional mean value theorem

\begin{equation*} f(y) - f(y^*) = (y - y_*) \cdot \nabla f(w) \end{equation*} for some $w$ on the line segment connecting $y$ to $y^*$. But it seems difficulty to use our above estimate to show $(y - y^*) \cdot \nabla f(w) > 0$.

JZS
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1 Answers1

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The answer is simpler than I expected and seems to rely only on $m > 0$ (that $L$ is strictly greater than $m$ is not needed for a unique minimizer). For any $y \neq y^*$, by the mean value theorem there exists $w$ of the form $y^* + t(y - y^*)$ for some $0 < t < 1$ so that

$$f(y) - f(y^*) = \nabla f(w) \cdot(y - y^*).$$

Then we have

\begin{split} 0 &< \nabla f(w) \cdot (w - y^*)\\ &= t\nabla f(w) \cdot (y - y^*) \\ & = t(f(y) - f(y^*)), \end{split}

where the first line follows because we have assume $f$ is $[m,L]$ sector bounded with $m>0$. Because $t > 0$, we have $f(y) > f(y^*)$ as desired.

JZS
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