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I removed the "model theory" tag and added the "continuity". While this is a question about building a model but there is no clear connection between this question and what a model theorist normally deals with. We shouldn't add a model theory tag for every set theory question about building models.
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If a real-valued function $f$ over reals satisfies $$ (1) \; \; \; f({x+y\over2})\le {f(x)+f(y)\over2}, $$and it is continuous, then it is not hard to see that $f$ is indeed convex. On the other hand, a discontinuous additive function, which exists by the axiom of choice, satisfies (1) but is not convex.

My question is, can we find real-valued functions over reals satisfying (1) which are not convex, without using the axiom of choice? Or, are there models of set theory in which all functions satisfying (1) are convex (continuous)?

If a function $f$ satisfies $$ (1) \; \; \; f({x+y\over2})\le {f(x)+f(y)\over2}, $$and it is continuous, then it is not hard to see that $f$ is indeed convex. On the other hand, a discontinuous additive function, which exists by the axiom of choice, satisfies (1) but is not convex.

My question is, can we find functions satisfying (1) which are not convex, without using the axiom of choice? Or, are there models of set theory in which all functions satisfying (1) are convex (continuous)?

If a real-valued function $f$ over reals satisfies $$ (1) \; \; \; f({x+y\over2})\le {f(x)+f(y)\over2}, $$and it is continuous, then it is not hard to see that $f$ is indeed convex. On the other hand, a discontinuous additive function, which exists by the axiom of choice, satisfies (1) but is not convex.

My question is, can we find real-valued functions over reals satisfying (1) which are not convex, without using the axiom of choice? Or, are there models of set theory in which all functions satisfying (1) are convex (continuous)?

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Finding non convex functions satisfying a weak form of convexity, without the axiom of choice

If a function $f$ satisfies $$ (1) \; \; \; f({x+y\over2})\le {f(x)+f(y)\over2}, $$and it is continuous, then it is not hard to see that $f$ is indeed convex. On the other hand, a discontinuous additive function, which exists by the axiom of choice, satisfies (1) but is not convex.

My question is, can we find functions satisfying (1) which are not convex, without using the axiom of choice? Or, are there models of set theory in which all functions satisfying (1) are convex (continuous)?