Given a strictly convex function $f(x,y)$, is the following function still strictly convex?
$$ g(x) := \inf_y f(x,y) $$
Is there any counterexample for it?
There is a question here Is the minimum of a parametric convex function convex again? But the question is for the convex case, does this hold for the strictly convex case?
As demonstrated in the other answers, $g$ need not be strictly convex. However, under the additional assumption that the infimum is attained, i.e. $g(x) = \min_y f(x, y)$ for all $x$, one can show that $g$ is strictly convex:
Let $x_1, x_2$ be distinct points in the domain of $g$ and $0 < \lambda < 1$. Choose $y_1, y_2$ such that $g(x_1) = f(x_1, y_1)$ and $g(x_2) = f(x_2, y_2)$. Then $$ \begin{align} g(\lambda x_1 + (1-\lambda) x_2) &= \inf_y f(\lambda x_1 + (1-\lambda) x_2, y)\\ &\le f(\lambda x_1 + (1-\lambda) x_2), \lambda y_1 + (1-\lambda) y_2) \\ &\boxed{<} \lambda f(x_1, y_1) + (1-\lambda)f(x_2, y_2) \\ &= \lambda g(x_1) + (1-\lambda)g(x_2) \, . \end{align} $$
The function $f \colon \mathbb R^2 \to \mathbb R$ given by $$ f(x,y) = \exp(y) + \exp(x+y) $$ is a counterexample, since $g(x) = 0$.
Let $$ f(x,y)=e^{x^2-y}\,. $$ The gradient and Hessian of this function are $$ \nabla f=\begin{pmatrix}2x\\-1 \end{pmatrix}e^{x^2-y}\,,\quad\nabla^2f=\begin{pmatrix}2+4x^2&-2x\\-2x&1\end{pmatrix}e^{x^2-y}\,. $$ Both eigenvalues of the Hessian are real and strictly greater zero. The function $f$ is therefore strictly convex. However, $$ g(x)=\inf_yf(x,y)\equiv 0 $$ is only convex.
