# Find all continuous functions that satisfy the Jensen inequality(?) $f(\frac{x+y}{2})=\frac{f(x)+f(y)}{2}$ [duplicate]

Find all continuous functions that satisfy the Jensen inequality(I don't know why the problems states ''inequality'') $$f(\frac{x+y}{2})=\frac{f(x)+f(y)}{2}$$

I've checked around stackexchange and google a bit, but none of the questions related to this problem gave me any hints on what I should do. The expression ''find all continuous functions'' confuses me. Am I supposed to determine some kind of a property that applies to all continuous functions that satisfy the property above or? Likewise there is a problem asking to find all continuous functions that satisfy$$f(x+y)=f(x)+f(y)$$ I'm sorry if this is a duplicate question, but even after reading all of those answers I've failed to find an adequate one. Also, another problem is that I'm not allowed to use derivations, and hence anything more complex than that.

## marked as duplicate by Martin Sleziak, Sahiba Arora, JMP, user296602, Claude LeiboviciAug 8 '17 at 6:55

• what is with $$y=x$$? – Dr. Sonnhard Graubner Jan 29 '17 at 10:49
• The Jensen inequality appears when you change $=$ into $\leq$, and once you've done that, it is true for all convex functions. I agree that the wording is a bit unfortunate, but the problem itself seems clear enough. – Arthur Jan 29 '17 at 10:49
• @Dr.SonnhardGraubner what do you mean, what is with? – Collapse Jan 29 '17 at 10:49
• i think all linear functions satisfy your condition – Dr. Sonnhard Graubner Jan 29 '17 at 10:52
• The equality case in Jensen's inequality holds for both convex and concave functions. And a function both convex and concave can be shown to be of the form $y=ax+b$. See convex+concave=affine and its linked posts for more. – StubbornAtom Jan 29 '17 at 13:22

The first question is actually similar to the second question $$f\left(\frac{x+y}{2} \right)=\frac{f(x)+f(y)}{2}$$ Then we set $f(x)-f(0)=g(x)$. Note that $g(x)$ is also a solution to the functional equation. Now, let $x=2a, y=0$. Then notice $$2g(a)=g(2a) \implies g(a)=2g\left(\frac{a}{2}\right)$$ So $$g(x)+g(y)=2g\left(\frac{x+y}{2} \right)=g(x+y)$$ So the solutions to the first question are functions that satisfy the second question, plus some arbitrary constant $c$.