# $f(x) + f(1-x) = f(1)$

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.

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Do you require $f$ to be continuous? –  Asaf Karagila Jan 8 '11 at 22:51
@Asaf: Seems like "$f\in C^{\infty}[0,1]$" answers that pretty handedly. –  Arturo Magidin Jan 8 '11 at 22:57
@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). –  Asaf Karagila Jan 8 '11 at 23:04
@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). –  Arturo Magidin Jan 8 '11 at 23:06
@Arturo: This makes a lot of sense. I guess you're probably right. –  Asaf Karagila Jan 8 '11 at 23:14

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 $\displaystyle [0,1]$.

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Yes. Nice one... –  user17762 Jan 8 '11 at 23:12
@Sivaram: Your edit is a much stronger restriction, see my revised answer. –  Aryabhata Jan 8 '11 at 23:43
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.) –  Qiaochu Yuan Jan 8 '11 at 23:46
Thanks. My friend was dealing with a two component system and hence I was stuck thinking about f(x)+f(1−x)=f(1). –  user17762 Jan 8 '11 at 23:51

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.

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@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? –  PEV Jan 8 '11 at 23:06
@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. –  Qiaochu Yuan Jan 8 '11 at 23:07
@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 $f(x) = kx$. –  user17762 Jan 8 '11 at 23:11
@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 $h(1/2) = \frac{g(1/2)}{2}$ $\Longrightarrow$ $f(x) = xf(1)$? –  PEV Jan 8 '11 at 23:14
@Sivaram: like I said, I don't think this is an interesting question without much more specific guidelines. –  Qiaochu Yuan Jan 8 '11 at 23:15