4

Let $f:[a,b] \rightarrow \mathbb R$ be integrable and satisfies $$ f\left(\frac{x+y}{2}\right)=\frac{1}{y-x} \int_x^y f(t)dt $$ for all $x \neq y$, $x,y \in [a,b]$.

What about $f$? Is it affine function?

Thanks

user84413
  • 27,211
Richard
  • 4,432
  • What your statement says, is that $f$ takes its average value over an interval at the average of the interval endpoints. Interesting – amcalde Oct 08 '14 at 17:37
  • Through your condition, we can't get any information about $f$ at $a,b$, so at least it may not be affine over $[a,b]$ – John Oct 08 '14 at 17:46
  • This implies immediately that $f$ cannot be, say, concave up or down, because this property would fail. – amcalde Oct 08 '14 at 17:50
  • I think that means that it has to be linear, i.e. $f(x) = \mu x$. This is affine. – amcalde Oct 08 '14 at 17:51
  • Fix $x$, since the integral is continuous with respect to y, we know $f$ should be continuous in the interval, then though integral, we know $f$ is differentiable in the interval, then we can get $f^{'}$,$f^{''}$ exist, then use Tylor theorem to approximate $f(t)$, I think we can get it's affine in the interval. – John Oct 08 '14 at 17:58
  • Note that $f$ satisfies the Mean Value Property, and therefore, it is harmonic. The only harmonic functions in one dimension are those with $f''(x)=0$. – robjohn Oct 08 '14 at 19:44

2 Answers2

4

Without loss of generality, $a < 0 < b$. The integral formula tells us that $f$ is continuous. Using the integral formula again, we see that $f$ is differentiable in $(-1,1)$. Then by differentiating the equivalent formulation $(y-x)f(\frac{x+y}{2}) = \int_x^y f(t) dt$, we see that $$ f(y) = f\left(\frac{x+y}{2}\right) + \frac{y-x}{2}f'\left(\frac{x+y}{2}\right) $$ for all $a \le x, y \le b$. Now choose $h$ such that $a \le h, -h \le b$ and set $y = h, x = -h$. It follows that $$ f(h) = f(0) + hf'(0) . $$ Therefore $f$ is affine on any symmetric interval that is contained in $[a,b]$. By shifting the interval, it follows that $f$ is affine everywhere.

robjohn
  • 345,667
Hans Engler
  • 15,439
2

If $f: x \mapsto \lambda x$ for some $\lambda \in \Bbb{R}$, Then $f$ satisfies the above equality and is affine.

Otherwise suppose $f$ is not affine, that is, $f(x) \ne \lambda x$ for any $\lambda \in \Bbb{R}$. Since the graph of $f$ is not a straight line, we can find two points $x$, and $y$ such that $\left.f\right|_{(x,y)}$ lies entirely above (or below) the line segment from $(x,f(x))$ to $(y,f(y))$. Indeed, we can find $(x,y)$ such that the value at the midpoint, $f((x+y)/2)$, is farthest from the straight line from all the other points in $(x,y)$. That is set $g(u)$ as follows:

$$g(u)=\left|\left((1-u) x + u y,f\left((1-u) x + u y\right)\right)-\left(\frac{(1-u)(x,f(x))+(u)(y,f(y))}{2}\right)\right|$$

where $u \in [0,1]$. The first term traces the graph of $f$ the second traces the line segment above.

We can always find some $x$ and $y$ such that the maximum of $g(u)$ occurs exactly at $u=(x+y)/2$. By the equality $f((x+y)/2)$ is the average of $f$ over the interval $(x,y)$ and therefore cannot be farthest from the straight line from $(x,f(x))$ to $(y,f(y))$ since $f$ is continuous (due to the integral) and must therefore continuously approach the line segment at its endpoints. Therefore on average $f$ is closer to the line than it is at its maximum distance. This is a contradiction for this case.

Hence the only option is for $f$ to be affine.

enter image description here

In the figure I try to encapulate this proof. The dotted line is the average of the blue function over the segment. It cannot be as far from the straight line as it's farthest point (the apex of the blue graph).

amcalde
  • 4,674