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# ISI MStat 2016 (Sample) Problem 2 | Continuous function | PSB

This is a beautiful sample problem from ISI MStat 2016 PSB Problem 2. This is based on application of continuity and integration.

## Problem- ISI MStat 2016 Problem 2

Let $$f:[-1,1] \rightarrow \mathbb{R}$$ be a continuous function. Suppose that $$f^{\prime}(x)$$ exists and $$f^{\prime}(x) \leq 1$$ for all $$x \in(-1,1)$$ . If $$f(1)=1$$ and $$f(-1)=-1,$$ prove that $$f(x)=x \text { for all } x \in[-1,1]$$.

## Solution

Given function is continuous and $$f^{\prime}(x)$$ exists .

Hence we can integrate $$f^{\prime}(x)$$ .

Another thing is given that $$f^{\prime}(x) \leq 1$$ for all $$x \in(-1,1)$$ ----(1) (say)

Now if we integrate both side of (1) from -1 to y , where $$y \in(-1,1)$$ then we get ,

$$\int^y_{-1} f^{\prime}(x) \,dx \leq \int^y_{-1} 1\,dx$$ , for all $$y \in(-1,1)$$

$$\Rightarrow f(y)-f(-1) \leq y+1 \Rightarrow f(y) \leq y$$ ,for all $$y \in(-1,1)$$ [ since $$f(-1)=-1$$ given ]---(2)

Again if we integrate both side of (1) from y to 1 ,where $$y \in(-1,1)$$ then we get,

$$\int^1_{y} f^{\prime}(x) \,dx \leq \int^1_{y} 1\,dx$$ , for all $$y \in(-1,1)$$

$$\Rightarrow f(1)-f(y) \leq 1-y \Rightarrow f(y) \geq y$$ ,for all $$y \in(-1,1)$$ [ since $$f(1)=1$$ given ] ---(3)

Hence from (2) & (3) we get $$f(y)=y$$ for all $$y \in(-1,1)$$ and $$f(-1)=-1 , f(1)=1$$ .

Therefore , $$f(x)=x$$ for all $$x \in[-1,1]$$ ( proved )

## Challenge Problem

$$f:(-1,1) \to (-1,1)$$ be continuous function such that $$f(x)=f(x^2)$$ for every x and $$f(0)=\frac{1}{2}$$

Then find f(x) .

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