Consider a class of functions on the real line defined by a following property: for each $\alpha \in [0,1]$, for each $x,y \in [0,1]$, there exists $z \in [0,1]$ such that $ \alpha f(x) + (1-\alpha) f(y)= f(z). $ How general is this class of functions?
Modified intermediate values property - with extensions
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linear-algebra
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0I am not sure. It seems a problem belonging to convex set theory. – 2011-03-26
2 Answers
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It's the set of all functions $f:[0,1]\to{\mathbb R}$ whose image $f\bigl([0,1]\bigr)$ is an interval. E.g., $f$ could map $]0,0.01[$ bijectively onto ${\mathbb R}$ and be anything on the rest of $[0,1]$ .
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Your property is a sort of modified intermediate values property: in fact, you are requiring that, however you choose $x,y\in [0,1]$, the function $f$ takes any value in between $f(x)$ and $f(y)$ (the usual i.v.p. requires that $f$ takes any value in between $\inf_{[x,y]} f$ and $\sup_{[x,y]} f$ in $[x,y]$)
Therefore, each function $f\in C([0,1])$ has your property; but the converse is not true: for example, there are bijections from $[0,1]\to ]0,1[$ (hence they have the modified intermediate values properties) which are not continuous.
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0@Pacciu Thanks (for your answer to my question). I misread the property. – 2011-03-26