Slope Equals Derivative at Midpoint

Problem? Find all differentiable functions f:\mathbb{R}\to\mathbb{R} such that


for all x,y\in\mathbb{R}.

Solution? First we prove two lemmas.

Lemma 1. Suppose a differentiable function g satisfies g(x)-g(y)=g'\left(\frac{x+y}2\right)(x-y) for all x,y\in\mathbb{R}. If g'(a)=g'(b)=0 (a< b), then g(a)=g(b) and g'\left(\frac{a+b}2\right)=0.

Proof of lemma 1. For any \epsilon>0, there exists \delta such that \left|\frac{g(x)-g(a)}{x-a}\right|<\epsilon for x\in(a,a+\delta) and \left|\frac{g(x)-g(b)}{x-b}\right|<\epsilon for x\in(b-\delta,b). So, if x\in(a,a+\delta), we have

g(a)-\epsilon(x-a)< g(x)< g(a)+\epsilon(x-a)


g(b)-\epsilon(x-a)< g(a+b-x)< g(b)+\epsilon(x-a).


g(b)-g(a)-2\epsilon(x-a)< g(a+b-x)-g(x)< g(b)-g(a)+2\epsilon(x-a)

g(b)-g(a)-2\epsilon(x-a)< g'\left(\frac{a+b}2\right)(a+b-2x)< g(b)-g(a)+2\epsilon(x-a).

Since g(b)-g(a)=g'\left(\frac{a+b}2\right)(b-a), then

g(b)-g(a)-2\epsilon(x-a)< \frac{g(b)-g(a)}{b-a}(a+b-2x)< g(b)-g(a)+2\epsilon(x-a)



Since we choose \epsilon arbitrarily, then \frac{g(b)-g(a)}{a-b}=0, which implies g(a)=g(b). This further implies g'\left(\frac{a+b}2\right).

Lemma 2. If g'(a)=g'(b)=0 (a< b), then g is constant on [a,b].

Proof of lemma 2. From lemma 1, we know that g(a)=g(b)=:c. Call a number x\in[a,b] good if g(x)=c and g'(x)=0. From lemma 1, if x and y are good, then \frac{x+y}2 is good.

We prove by induction on n that for any positive integer k<2^n, the number x+\frac{k}{2^n}(b-a) is good.

This is clearly true for n=1. Suppose it is true for smaller n. Then a+\frac1{2^{n-1}}(b-a) and a+\frac{k-1}{2^{n-1}}(b-a) are good, which implies that a+\frac{k}{2^n}(b-a) is good, as desired.

The numbers of the form a+\frac{k}{2^n}(b-a) are dense on [a,b]. By continuity, we get that g is constant on [a,b]. Hence, lemma 2 is proved.

Now we will prove that f is a quadratic function. Assume the opposite. Then there exists an interval [a,b] such that f does not coincide with any quadratic function. Let u,v\in\mathbb{R} be such that f'(a)=2ua+v and f'(b)=2ub+v. Let g(x)=f(x)-ux^2-vx. Then g'(a)=g'(b)=0. It is easy to see that g satisies g(x)-g(y)=g'\left(\frac{x+y}2\right)(x-y). By lemma 2, g is constant, say g(x)=m. Thus f(x)=ux^2+vx+m for all x\in[a,b], which is a contradiction to the assumption that f does not coincide with any quadratic function.

Therefore, f must be a quadratic function. Conversely, it is easy to check that any quadratic function satisfies the required properties.

Source? A. Y. Dorogvstev, Mathematical Analysis (Russian)


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