Slope Equals Derivative at Midpoint

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

$f(x)-f(y)=f'\left(\frac{x+y}2\right)(x-y)$

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

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

and

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

Therefore,

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

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

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

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

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

$-\epsilon<\frac{g(b)-g(a)}{a-b}<\epsilon$

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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Tags: calculus, derivative, real analysis

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