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Let $f:\mathbb{R}^2 \to \mathbb{R}$ be a convex function which is strictly convex in its first argument, i.e. $x_1\mapsto f(x_1,x_2)$ is strictly convex for every $x_2\in\mathbb{R}$. Does it follow that for every $x,y\in\mathbb{R}^2$ with $x_1\neq y_1$ we have $$ f(tx + (1-t)y) < tf(x) + (1-t)f(y) $$
for every $t\in(0,1)$?

I am struggling to come up with a proof or a counterexample.

msaBU
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  • Geometrically we construct the graph by taking a parabola in one variable, which is convex, and have the base point follow say, another parabola in the other variable with the same base point but opposite direction and orthogonal to the first parabola, which is concave. If fix some $x_1$ or $x_2$ this means we are restricting our attention to a plane parallel to the axes. In one direction this will always be convex but in the other it will always be concave. – CyclotomicField Feb 04 '20 at 01:40
  • But I don't think that this construction gives us a function which is convex in $\mathbb{R}^2$, does it? Maybe a concrete example in form of a function would help to clarify my doubts. – msaBU Feb 04 '20 at 02:02

1 Answers1

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Using the idea of @CyclotomicField leads to $$ f(x_1, x_2) = (x_1 - x_2)^2 .$$

gerw
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