Kvant Math Problem 563
Assume, seeking a contradiction, that
Verified: yes
Verdicts: PASS + PASS
Solve time: 8m47s
Source on kvant.digital
Problem
The function $f$ is defined on the interval $[a;b]$ of length 4 and has a continuous derivative $f'$ on it. Prove that there exists a point $x$ inside the interval $[a;b]$ for which $$f'(x)-(f(x))^2\lt1.$$
S. V. Fomin
Exploration
Assume, seeking a contradiction, that
$$f'(x)-f(x)^2\ge 1 \qquad\text{for all }x\in[a,b].$$
Then
$$f'(x)\ge 1+f(x)^2.$$
The expression $1+f(x)^2$ suggests dividing by it, because
$$\frac{d}{dx}\bigl(\arctan f(x)\bigr) = \frac{f'(x)}{1+f(x)^2}.$$
Since $1+f(x)^2>0$, the differential inequality can be integrated directly. The interval length $4$ is larger than $\pi$, while the range of the function $\arctan$ has width $\pi$. This mismatch is the source of the contradiction.
Problem Understanding
The goal is to prove the existence of a point $x\in(a,b)$ satisfying
$$f'(x)-f(x)^2<1.$$
A contradiction argument is natural. If the opposite inequality held everywhere, then
$$f'(x)\ge 1+f(x)^2$$
throughout the interval. After dividing by $1+f(x)^2$ and integrating, one obtains a lower bound for the variation of $\arctan f(x)$. Since $\arctan$ always takes values in the interval $(-\pi/2,\pi/2)$, its total variation is strictly less than $\pi$. The interval length is $4>\pi$, which makes the assumed inequality impossible.
Proof Architecture
Assume that
$$f'(x)-f(x)^2\ge1$$
for every $x\in[a,b]$.
From this derive
$$\frac{f'(x)}{1+f(x)^2}\ge1.$$
Integrate over $[a,b]$ and use
$$\frac{d}{dx}\arctan f(x) = \frac{f'(x)}{1+f(x)^2}.$$
This yields
$$\arctan f(b)-\arctan f(a)\ge b-a=4.$$
On the other hand,
$$-\frac{\pi}{2}<\arctan f(t)<\frac{\pi}{2} \qquad\text{for all }t,$$
hence
$$\arctan f(b)-\arctan f(a)<\pi.$$
Combining the two inequalities gives
$$4<\pi,$$
a contradiction.
Solution
Suppose that the conclusion is false. Then
$$f'(x)-f(x)^2\ge1 \qquad\text{for every }x\in[a,b].$$
Equivalently,
$$f'(x)\ge1+f(x)^2.$$
Since $1+f(x)^2>0$, division gives
$$\frac{f'(x)}{1+f(x)^2}\ge1.$$
Because
$$\frac{d}{dx}\bigl(\arctan f(x)\bigr) = \frac{f'(x)}{1+f(x)^2},$$
integrating over $[a,b]$ yields
$$\arctan f(b)-\arctan f(a) = \int_a^b \frac{f'(x)}{1+f(x)^2},dx \ge \int_a^b 1,dx = b-a.$$
The interval has length $4$, so
$$\arctan f(b)-\arctan f(a)\ge4.$$
However, every value of $\arctan$ lies in the interval $(-\pi/2,\pi/2)$. Consequently,
$$\arctan f(b)-\arctan f(a)<\pi.$$
Hence
$$4\le \arctan f(b)-\arctan f(a)<\pi.$$
This implies $4<\pi$, which is false.
The assumption was impossible. Therefore there exists a point $x\in[a,b]$ such that
$$f'(x)-f(x)^2<1.$$
Since the assumed opposite inequality failed everywhere, this point lies in the interval, and the required existence statement follows. ∎
Verification of Key Steps
The only substantive step is the integration of the differential inequality. The continuity of $f'$ guarantees that
$$\frac{f'(x)}{1+f(x)^2}$$
is continuous, so the integral is legitimate. The identity
$$\frac{d}{dx}\arctan f(x) = \frac{f'(x)}{1+f(x)^2}$$
comes from the chain rule. The contradiction arises because the image of $\arctan$ has width $\pi$, while the interval length is $4>\pi$. Thus the assumption $f'(x)-f(x)^2\ge1$ on the whole interval cannot hold.