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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0This includes all functions satisfying the [intermediate value property](http://en.wikipedia.org/wiki/Intermediate_value_property#Darboux_function). The intermediate value property is more stringent since $z \in [x,y]$ in that case. – 2011-03-26
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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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0Thank you. Now suppose a following extension: for each $A$ that is a subset of $[0,1]$, there exists some $z$ such that $\int_{x \in A} f(x) dx = f(z)$. Can each function of $f \in C[0,1]$ satisfy this property? – 2011-03-26
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0The answer is no. In fact take $f(x)=1+2x$, $A=[0,\frac{1}{10}]$: simple computations yield $\int_A f =\frac{11}{100}<1$, therefore no $z$ can exist s.t. inequality $\int_A f=f(z)$ holds, because the range of $f$ is $[1,3]$. The problem here seems to be that the integral function $\phi (A):=\int_A f$ (defined on measurable $A\subseteq [0,1]$) can take arbitrarily small values even if $f$ is bounded away from zero. – 2011-03-26
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0@Pacciu Why do bijections from $[0,1]$ to $(0,1)$ have the modified intermediate values property? – 2011-03-26
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0@Didier: Because they are surjective. Fix $x,y\in [0,1]$; then $0
$f$ is $]0,1[$, hence $f$ takes each value in $[\min \{ f(x),f(y)\} ,\max \{ f(x),f(y)\}] \subset ]0,1[$... The problem is that you are requiring $z$ to be a point in $[0,1]$, not in $[\min \{ x,y\},\max \{x,y\}]$. – 2011-03-26 -
0Make sense. Thank you. – 2011-03-26
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0Oh, I realized that the modified i.v.p. is totally equivalent to the classical i.v.p.. Let $R$ be the range of $f$; then $f$ satisfies the modified i.v.p. iff $R$ is convex; on the other hand $f$ has the i.v.p. iff its range is an interval. Now the equivalence follows, for a set $A\subseteq \mathbb{R}$ is convex iff it's an interval. – 2011-03-26
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0@Pacciu Thanks (for your answer to my question). I misread the property. – 2011-03-26