Let $I$ and $J$ be compact intervals.
Let $f:I\times J\to\mathbb R$ be differentiable and strictly convex.
Is the function $g:I\to\mathbb R$ defined by
$$ g(x) = \min_{y\in J} f(x,y) $$
convex?
Remarks:
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I know that minimum of convex functions is in general not convex. However, I can't find a counter example in which $f$ is convex.
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The regularity ensures that the minimizer $y^*(x)$ of $f(x, \cdot)$ is unique.
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Assume $y^*$ as function is convex, $y^*$ maps $I$ into an interval $J^*$, and $f(x, \cdot)$ is increasing on $J^*$ for every $x\in I$. Then, $g$ is convex.
Thanks for any input 🙂
Best Answer
It is convex!
Your first statement that the minimum of convex functions is in general not convex is true, but here you have a lot more structure! In a sense you are projecting onto $x$. In fact, $g$ is also called the inf-projection of $f$. Let $\lambda \in (0,1)$ and $y_1, y_2 \in J$ arbitrary:
$$ \begin{aligned} g(\lambda x_1 + (1-\lambda) x_2) &= \min_{y} f(\lambda x_1 + (1-\lambda)x_2, y) \\ &\leq f(\lambda x_1 + (1-\lambda)x_2, \lambda y_1 + (1-\lambda)y_2)\\ &\leq \lambda f(x_1, y_1) + (1-\lambda) f(x_2,y_2)\\ \end{aligned} $$
Now first minimize with respect to $y_1$, then with respect to $y_2$ to finally get: $$g(\lambda x_1 + (1-\lambda) x_2) \leq \lambda g(x_1) + (1-\lambda) g(x_2)$$
Also notice that you do not need the regularity conditions you imposed onto $f$.