Animated Solution for Mathematics - Differentiation: Let $R$ be the set of real numbers and $f:R\to R$ be such that for all $x$ and $y$ in $R$ $|f(x)-f(y)|\le |x-y{|}^2$. Prove that $f(x)$ is a constant.

Understanding the Inequality

My dear students, today we are going to peel back the layers of a truly elegant problem. We are given a function $f:\mathbb{R}\to \mathbb{R}$ that satisfies the condition $|f(x)-f(y)|\le |x-y{|}^2$ for all real numbers $x$ and $y$.

At first glance, this looks like a simple algebraic constraint, but it is actually a profound statement about the nature of the function. Imagine you are standing at a point $x$ on the real line. The value of the function is $f(x)$.

Now, consider any other point $y$. The inequality tells us that the difference between $f(x)$ and $f(y)$ is bounded by the square of the distance between $x$ and $y$.

Geometrically, this means that if you draw two parabolas, $y=f(x)+(x-y{)}^2$ and $y=f(x)-(x-y{)}^2$, the graph of the function $f$ is trapped within this narrow corridor. As $y$ approaches $x$, this corridor shrinks to a single point. This is the geometric soul of our problem.

Creating the Difference Quotient

To understand how this function behaves, we need to look at its rate of change. In calculus, the rate of change is defined by the derivative.

Let us consider the difference quotient, which represents the slope of the secant line connecting $(x,f(x))$ and $(y,f(y))$. The slope is given by:

We want to see what happens to this slope as $y$ gets closer and closer to $x$. Let us take our original inequality: $|f(x)-f(y)|\le |x-y{|}^2$.

Assuming $x eq y$, we can divide both sides by $|x-y|$. This gives us:

Simplifying the right side, we get:

This is the crucial step. We have successfully bounded the absolute value of the slope of the secant line by the distance between the two points.

Taking the Limit

Now, we apply the power of limits. We want to find the instantaneous rate of change at $x$, which is the derivative $f^{\prime }(x)$.

We take the limit as $y$ approaches $x$ on both sides of our inequality:

On the left side, the limit of the difference quotient is the definition of the derivative, $f^{\prime }(x)$. So, the left side becomes $|f^{\prime }(x)|$.

On the right side, as $y$ approaches $x$, the distance $|y-x|$ approaches $0$. Therefore, we are left with the inequality:

This is the moment of truth. We know that the absolute value of any real number is always non-negative, meaning $|f^{\prime }(x)|\ge 0$. If $|f^{\prime }(x)|$ is both less than or equal to $0$ and greater than or equal to $0$, it must be exactly $0$.

A Constant Reality

We have arrived at the conclusion that $|f^{\prime }(x)|=0$, which implies that $f^{\prime }(x)=0$ for all $x\in \mathbb{R}$.

In the language of calculus, a function whose derivative is zero everywhere is a constant function. The tangent line to the graph of $f$ is horizontal at every single point.

There is no slope, no growth, and no decay. The function is perfectly flat. Thus, we have proven that $f(x)=c$, where $c$ is some constant.

It is a beautiful result, isn't it? A simple inequality, a bit of algebraic manipulation, and the Squeeze Theorem have revealed the hidden, constant nature of the function. Keep this logic in your toolkit, for it is the key to unlocking many more mysteries in the world of analysis.