Question

# Let $$f$$ defined on $$[0,1]$$ be twice differentiable such that $$\left| f''(x) \right| \le 1$$ for all $$x\in [0,1]$$. If $$f(0)=f(1)$$, then show that $$\left| f'(x) \right| <1$$ for all $$x\in [0,1]$$

Solution

## Let $$f$$ be defined on $$[0, 1]$$$$|f^{\prime\prime}|\le 1$$ for all $$x\in [0, 1]$$for any $$f(x)=x^2$$$$f^{\prime}(x)=2x$$$$f^{\prime\prime}(x)=2$$At $$x_1=0, f^{\prime \prime}(0)=2$$At $$x_2=1, f^{\prime\prime}(1)=2$$$$f(0)=f(1)$$Prove : $$|f^{\prime}(x)|<1$$for $$f(x)=x$$$$f^{\prime}(x)=1$$for $$x_1=0, f^{\prime }(0)=1$$for $$x_2=1, f^{\prime}(1)=1$$Hence, $$|f^{\prime} (x)|<1$$ for all $$[0, 1]$$Hence proved.Mathematics

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