7
$\begingroup$

I was discussing with a friend of mine about her research and I came across this problem.

The problem essentially boils down to this.

$f(x)$ is a function defined in $[0,1]$ such that $f(x) + f(1-x) = f(1)$. I want to find a condition on $f(x)$ so that I can conclude $f(x) = f(1)x$.

Clearly, $f \in C^{0}[0,1]$ alone is insufficient to conclude $f(x) = f(1)x$.

My hunch is if $f(x) \in C^{\infty}[0,1]$, then $f(x) = f(1)x$. However, I am unable to prove it. Further, is there a weaker condition with which I can conclude $f(x) = f(1)x$?

This problem closely resembles another problem:

If $f(x+y) = f(x) + f(y)$, $\forall x,y \in \mathbb{R}$ and if $f(x)$ is continuous at atleast one point, then $f(x) = f(1)x$.

I know how to prove this statement, but I am unable to see whether this will help me with the original problem.

Though these details might not be of much importance to her, I am curious to know.

EDIT:

As Qiaochu pointed out, I need stronger conditions on $f$ to come up with some reasonable answer.

Here is something which I know that $f$ has to satisfy the following:

$\forall n \in \mathbb{Z}^{+}\backslash \{1\}$, $f(x_1) + f(x_2) + \cdots + f(x_n) = f(1)$, where $\displaystyle \sum_{k=1}^{n} x_k = 1$, $x_i \geq 0$.

Note that $n=2$ boils down to what I had written earlier.

  • 1
    Do you require $f$ to be continuous?2011-01-08
  • 1
    @Asaf: Seems like "$f\in C^{\infty}[0,1]$" answers that pretty handedly.2011-01-08
  • 1
    @Arturo: I can say that I want some function $f$ to hold some property which can be satisfied perhaps by a subset of bounded functions, or something like that. I might conjecture it is true for infinitely smooth functions - doesn't mean it's true. He first stated the property, then conjectured it probably in $C^\infty$, doesn't mean it **has** to be continuous or bounded to have this sort of property (well, it might. I haven't checked that, but my point remains valid - this is how generalizations are made).2011-01-08
  • 6
    @Asaf: Fair enough; it seems to me, based on the sytax and organization of the post, that Sivaram was looking for sufficient, possibly necessary-and-sufficient, conditions on an $f$ for the identity $f(x)+f(1-x)=f(1)$ to imply $f(x)=f(1)x$ for all $x$; continuity by itself was not enough, so he wanted to *strengthen* that property (rather than replace it wholesale by something entirely different).2011-01-08
  • 1
    @Arturo: This makes a lot of sense. I guess you're probably right.2011-01-08
  • 0
    @Arturo: Yes...2011-01-08

5 Answers 5

0

In fact this functional equation belongs to the form of http://eqworld.ipmnet.ru/en/solutions/fe/fe1116.pdf.

The general solution is $f(x)=\dfrac{f(1)}{2}+C(x,1-x)$ , where $C(u,v)$ is any antisymmetric function.

8

Let $g(x) = f(x + 1/2)$, which is defined on $[-1/2, 1/2]$ and which satisfies $g(x) + g(-x) = g(1/2)$. Now, any function $g$ on $[-1/2, 1/2]$ can be written

$$g(x) = \frac{g(x) + g(-x)}{2} + \frac{g(x) - g(-x)}{2} = \frac{g(1/2)}{2} + \frac{g(x) - g(-x)}{2}$$

where the first term is the even part and the second term is the odd part, both of which can be chosen arbitrarily. The problem conditions stipulate the even part and the odd part $h(x)$ is subject to the condition $h(1/2) = \frac{g(1/2)}{2}$. So we can pick an arbitrary odd function for $h$ and then everything else is determined, e.g. if $h(x) = x^3$ then $g(x) = \frac{1}{8} + x^3$.

So the meta-problem is "what conditions on an odd function are strong enough to make it equal to $h(x) = x$?" and I don't think this is a reasonable or interesting question to ask unless you have something very specific in mind.

This problem is not actually much like the Cauchy functional equation because there is only one free parameter instead of two.

  • 1
    @Qiaochu: Could you define $g(x) = f(x+c)$ for any $c$ on $[-c,c]$ such that $g(x)+g(-x) = g(c)$? Was $\frac{1}{2}$ just arbitrary?2011-01-08
  • 1
    @Trevor: the problem statement says that f is defined on [0, 1] and satisfies a relation symmetric about 1/2. To make this symmetry as easy to see as possible I translated everything to the left by 1/2.2011-01-08
  • 1
    @Qiaochu: +1 Right. So what condition(s) should I further enforce so that I can conclude $f(x) = kx$? I am asking this to figure out if I can translate these conditions into some sort of physical reasoning for the actual problem and give a justification for choosing $f(x) = kx$.2011-01-08
  • 1
    @Qiaochu: So sufficient conditions would be the following: $f(x) \in C^{\infty}[0,1]$, $f(x)+f(1-x) = f(1)$ and the odd part is subjected to $h(1/2) = \frac{g(1/2)}{2}$ $\Longrightarrow$ $f(x) = xf(1)$?2011-01-08
  • 1
    @Sivaram: like I said, I don't think this is an interesting question without much more specific guidelines.2011-01-08
  • 1
    @Trevor: no. The general solution is g(x) = h(1/2) + h(x) where h is an arbitrary odd function.2011-01-08
  • 1
    @Qiaochu: Infact, I know that $f(x)$ satisfies the following (and on the face of it looks like a stronger condition): $\forall n \in \mathbb{Z}^{+}\backslash \{1\}$, $f(x_1) + f(x_2) + \cdots + f(x_n) = f(1)$, where $\displaystyle \sum_{k=1}^{n} x_k = 1$, $x_i \geq 0$. Will this be on any help to draw a stronger conclusion?2011-01-08
  • 1
    @Sivaram: oy. Why didn't you just say that? I think that is a _much_ stronger condition.2011-01-08
  • 1
    @Qiaochu: I realized this only few minutes back. The actual physical problem is from thermodynamics and $x_i$ are the mole fractions. My friend was dealing with a two component system and hence I was stuck thinking about $f(x) + f(1-x) = f(1)$.2011-01-08
6

I presume $\displaystyle f(1) \neq 0$, in which case we can assume $\displaystyle f(1) = 1$.

If $\displaystyle f$ is such a function, then $\displaystyle g(x) = \sin^{2}\left(\frac{\pi f(x)}{2}\right)$ is also such a function.

If $\displaystyle f(1) = 0$, take $\displaystyle g(x) = \sin(f(x))$.

So you will really need much stronger restrictions than infinite differentiability etc.

As to your edit, continuity is enough.

We can first show $\displaystyle f(1/q) = f(1)/q$ for integral $\displaystyle q$.

This can easily be extended to $\displaystyle f(p/q)$ (for instance $\displaystyle f(2/q) + f(1/q) + \dots + f(1/q) = f(1)$) and by continuity, to the whole of $\displaystyle [0,1]$.

  • 1
    Yes. Nice one...2011-01-08
  • 1
    @Sivaram: Your edit is a much stronger restriction, see my revised answer.2011-01-08
  • 1
    Yes, that was my solution to the revised question as well. (Without any continuity assumptions we can use pathological solutions to the Cauchy functional equation.)2011-01-08
  • 1
    Thanks. My friend was dealing with a two component system and hence I was stuck thinking about f(x)+f(1−x)=f(1).2011-01-08
0

Whilst this was a post from a while ago; what you are looking at is a symmetric probability function in some cases (under the definition of Segal 1993)

And I also noticed that your edited solution would work, a while ago. ie Is the Symmetry a bi-conditional injective claim as well, ie as an iff, claim. ie inverse symmetry as well F-1(y)+F-1(1-y)=f-1(1)=1 ? and is it midpoint convex? That might be weaker, as it may give you the continuity via the connection between midpoint convexity and convexity, with F strictly monotonically increasing and bounded.

Where if F is also midpoint convex, inverse symmetric and and Strictly monotonic Increasing;F:[0,1] to [0,1],

Then given your symmetry, condition, along with F(1)=1; , F(0)=0, F(1/2)=1/2; which just fall out.

it should effectively be you jensen's equality Often often a midpoint convex symmetric function, that is strictly monotonic increasing, F(1)=1, F:[0,1] to [0,1] can be only be so, iff it is midpoint concave as well; ie jensen's equality. At least if you have inverse symmetry as well F-1(p)+F-1(1-p)=1

one can get F(2x)=2F(x), and i think F(nx)=n F(x) which generalizes to all integers if not all rationals, numerically, sub-additivity and effectively super-addivitiy and cauchy equation, and jensens equality/or rather conditional cauchy F(x+y)=F(x)+F(y) for all pairs,

Something which will have the same effect under continuity to cauchy equation or to f(x)=x, already I think may be induced by convexity.

Where midpoint convexity given strictly monotone increasing and the other conditions might already give you convexity, and thus continuity etc, F(x)=x already, although I am not sure.

Maybe a mild differentiation or measurability constraint is needed as well.

Although strict monotone increasing generally implies strict quasi convexity and quasi convexity (quasi concave and strict as well), where strict quasi convexity and midpoint convexity very often entails convexity, and thus continuity on [0,1] *lipschitz continuity I believe).

And it would presumably be concave as well in virtue of symmetry.

.

ie, if you have

(1).F:[0,1] to [0,1]

(1.1) and F(1)=1, {F(0)=0, F(1/2)=1/2);

(1.2) symmetry F(1-x)+F(x)=F(1)=1.

and in addition you have (2.a) (2.b) and (3)

(2a.) and F is strictly monotonic increasing function ( and thus, injective function)

(2.b) (inverse symmetry) for all y in range of f, f-1(y)+F-1(1-y)=F-1(1)=1.

**(3). F is midpoint Convex F(x/2+y/2)<=F(x)/2+F(y)/2

  • 0
    Maybe also 2.(C) if for any arbitrary two points in domain [0,1], x1+x2>1 iffF(x)+F(x1)>1; x1+x2<1,iff F(x)+F(x1)<1; and likwise for the two points in the range for the inverse, y1, y2 in range of F, which is at least in [0,1]y1+y2>1 iffF(y)+F(y1)>1; x1+x2<1,iff F(y)+F(y1)<1 " condition; 2.c.(this is probably redudant given strict monotonic, and inverse symmetric** i think that this, 2.c and 2.b is necessary may be be redundant, given (1), (1.1) (1.2), presumably are implied by (1), (1.2) symmetry, F(1)=1 and strict monotonic increasing(and injective).2017-04-19
  • 0
    They may help in the justification of it function being strictly monotonically increasing, if its modelled on a strictly linear order, that is total, dense/non atomic, separable, and bounded (and min and max elements; where midpoint convexity might give you the solv-ability requirements to show that it limits or attains its min and max, and/or is complete). of course, its already a consequence of F(1)=1 that F(1/2)=1/2 in your def, and comes for free already F(0)=0 and conversely. But I added F-1(1)=F(1)=1, F-1(0)=F(0)=0 and F(1/2)=F-1(1/2)=1/2. To be sure2017-04-19
  • 0
    Do you have midpoint Convexity and that F(1)=1, and thus F(1/2)=1/2 F(0)=0 with F strictly monotonic increasing and injective F:[0,1]to [0,1]? . Does f-1(x)+f-1(1−x)=f-1(1)=F(1). (ie x +y=1 iff F(x)+F(y)=1)? Then Midpoint convexity is sufficient I think. I think if you have further restrictions it can be weakened F(x) +F(y)>1 iff x +y>1 etc. Even the weaker for any given three points x + y +z=1 iff f(x +y +z)=1, x + y +z>1 iff f(x +y +z)>1x + y +z<1 iff f(x +y +z)<1. If you have dual sequence for example2017-05-02
  • 0
    should be sufficient; Given F(0,5)=0.5 and F(1)=1 and F(0)=0; this is related to the anticipated utility functional. If you have a G(0.25) and G(0.5)=0.5 then if G(0.25)=F(0.25) where these x +y +z =1 then F(0.25)+G(0.25)+F(0.5)=1 ; 2F(0.25)=0.5 F(0.25)=0.25 etcwhere one function denote the A outcome, the other decreasing precisely identically (but downward) denoting ~A2017-05-02
0

In fact if F strictly monotonic; with $F(1)=1$, and thus $F(0.5)=0.5$ as a result then midpoint convexity just at at 0 and 1 (mid star convexity at zero or one)

Would be sufficient, if $F:[0,1]\to [0,1]$; or just star convexity, given F(1)=1 and $F(1-x)+F(x)=F(1)=1$ just at $0$ for all real t, at zero alone.

Not sure if strictly monotonic increasing or inverse symmetry or $F^{-1}(1-x)+F^{-1}(x)=F^{-1}(1)=1$ is necessary. The inverse version of your condition essentially, making it into a bi-conditional claim.

with $F$ being$(A)$ 'Locally midpoint Jensen convex', as well as $(B)$ F strict monotone increasing. $(A1)$ Almost approximately convex might work with $(1)\,(2)\,(3)\,(4.1), (4.2)$ it may work.F

$$(1)F:[0,1]\to [0,1]\,\text{where}\,F(1)=1,\text{where F(0)=0, F(0.5) (if not already implicit)}$$ $$(2)\forall (p_1,p_2)\,\in \text{Im};\,(F)F^{-1}(1-p)+F^{-1}(p)=F^{-1}(1)=1\,\text{where p_2=1-p}$$ $$(3)\forall (x_1,x_2)\,\in \text{dom}(F);\,F(1-x)+F(x)=F(1)=1\,\text{where x_2=1-x}$$

$$(4.1)\forall (x_1,x_2)\,\in \text{dom}(F);\,[F(x_1)+F(x_2)>1 \leftrightarrow [x_1 +x_2>1]\,\land\quad\,[F(x_1)+F(x_2)<1] \leftrightarrow \,[x_1 +x_2]<1]$$

$$(4.2)\forall (p_1,p_2) \in \text{Im}(F);\,[F^{-1}(x_1)+F^{-1}(x_2)>1] \leftrightarrow\, p_1 +p_2>1\,\land\,[F^{-1}(p_1)+F^{-1}(p_2)<1] \leftrightarrow [p_1 +p_2]<1]$$ $(5)$ where $F$ strictly pseudo-linear or is twice/thrice differentiable with a non vanishing first derivative. Not sure if log concavity or anything weaker would help replace the convexity condition; mid concavity or local mid concavity (would do the same job) as would star concavity; or super/sub-additivity by itself would work. It may