True or False? There exists a function \( f: \Bbb R \to \Bbb R \) satisfying \( f(-1)=-1, f(1)=1 \) and \(|f(x)-f(y)| \leq |x-y|^{\frac{3}{2}} \forall x, y \in \Bbb R \)

TIFR GS 2017 Entrance Examination paper

Real Analysis

Moderate

Real Analysis by S.K. Mapa

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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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