# Understand the problem

True or False? There exists a function $$f: \Bbb R \to \Bbb R$$ satisfying $$f(-1)=-1, f(1)=1$$ and $$|f(x)-f(y)| \leq |x-y|^{\frac{3}{2}} \forall x, y \in \Bbb R$$
##### Source of the problem
TIFR GS 2017 Entrance Examination paper
Real Analysis
Moderate
##### Suggested Book
Real Analysis by S.K. Mapa

Do you really need a hint? Try it first!

Can you use $$|x-y |^{\frac{3}{2}} \forall x, y \in \Bbb R$$ to find something about the functon $$f(x)$$.
$$|x-y |^{\frac{3}{2}} \forall x, y \in \Bbb R$$ $$\Rightarrow -(x-y)^{\frac{3}{2}} \leq (f(x)-f(y)) \leq (x-y)^{\frac{3}{2}}$$ $$\Rightarrow -(x-y)^{\frac{1}{2}} \leq \frac{f(x)-f(y)}{x-y} \leq (x-y)^{\frac{1}{2}}$$. Now taking limit on both sides as $$x \to y$$, we can get $$f'(x)=0$$. What else can you say about the function over the real line?
We can say that $$f'(x) =0 \forall x \in \Bbb R$$. What does this imply about the function $$f(x)$$?
We can conclude that the function $$f(x)$$ is a constant function. Now look at the other two conditions $$f(-1)=-1, f(1)=1$$. What do you think about the function?
One can say that the function is not possible as it is constant at every point on the real line but goes to -1 at -1 and to 1 at 1. This is not possible. Hence, the statement is false.

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