This is a beautiful sample problem from ISI MStat 2016 PSB Problem 2. This is based on application of continuity and integration.
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 \( f(x)=x \text { for all } x \in[-1,1] \).
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 )
\(f:(-1,1) \to (-1,1)\) be continuous function such that \(f(x)=f(x^2) \) for every x and \( f(0)=\frac{1}{2} \)
Then find f(x) .
This is a beautiful sample problem from ISI MStat 2016 PSB Problem 2. This is based on application of continuity and integration.
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 \( f(x)=x \text { for all } x \in[-1,1] \).
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 )
\(f:(-1,1) \to (-1,1)\) be continuous function such that \(f(x)=f(x^2) \) for every x and \( f(0)=\frac{1}{2} \)
Then find f(x) .
