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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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