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Let the function $f$ be twice differentiable on $[0,1]$ such that $|f''(x)|\leq 1$ for all $x\in [0,1]$. If $f(1)=f(0)$, then show that $|f'((x)|\leq 1$ for all $x\in [0,1]$.

My effort

Applying Lagrange's mean value theorem we have
$f(1)-f(0)=f'(c)$ for some $c\in (0,1)\implies f'(c)=0$.
Next, I don't know how to show the required result.

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4 Answers 4

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We have $f'(c) = 0$ for some $c \in (0,1)$. By the MVT,

$$\left|\frac{f'(x)}{x-c}\right| \leq 1$$

Since $0<|x-c| \leq 1$, the result follows.

Note that this does not assume the integrability of $f''$.

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  • $\begingroup$ I added a little edit. Excellent answer. $\endgroup$ Commented Aug 29, 2016 at 18:58
  • $\begingroup$ @user254665 The edit makes it ugly. Anyone with a brain knows $c \neq x$. $\endgroup$ Commented Aug 29, 2016 at 19:56
  • $\begingroup$ It's sufficient to modify it as $0 < |x - c| \leq 1$. $\endgroup$
    – rubik
    Commented Aug 29, 2016 at 19:59
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    $\begingroup$ @rubik Fair enough. $\endgroup$ Commented Aug 29, 2016 at 20:02
  • $\begingroup$ @MathematicsStudent1122 thank you for your answer(+1) $\endgroup$
    – user356595
    Commented Aug 30, 2016 at 3:12
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By Taylor expansion with Lagrange remainder term, for any $x \in (0, 1)$: \begin{align} & f(0) = f(x) - f'(x)x + \frac{1}{2}f''(\xi)x^2, \tag{1} \\ & f(1) = f(x) + f'(x)(1 - x) + \frac{1}{2}f''(\eta)(1 - x)^2, \tag{2} \end{align} where $\xi \in (0, x)$, $\eta \in (x, 1)$.

Subtract $(2)$ from $(1)$ and use the condition $f(0) = f(1)$ yields $$0 = -f'(x) + \frac{1}{2}f''(\xi)x^2 - \frac{1}{2}f''(\eta)(1 - x)^2,$$ which implies that \begin{align} |f'(x)| = \left|\frac{1}{2}f''(\xi)x^2 - \frac{1}{2}f''(\eta)(1 - x)^2\right| \leq \frac{1}{2} + \frac{1}{2} = 1. \end{align}

To treat the end points case, directly expand $f(0)$ at $1$ yields $$f(0) = f(1) - f'(1)(-1) + \frac{1}{2}f''(\zeta),$$ where $\zeta \in (0, 1)$, hence $$|f'(1)| = |f''(\zeta)/2| \leq 1/2 \leq 1.$$ $|f'(0)|$ can be bounded similarly.

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With $c$ so chosen, $f'(x)$ can only get so far away from $f'(c)$. What I mean is that \begin{equation} f'(c) + m(x-c) \leq f'(x) \leq f'(c) + M(x-c), \end{equation} where $m$ and $M$ are the smallest and largest possible values of $f''(x)$, respectively. Explicitly: \begin{equation} m = \inf_{x\in[0,1]} f''(x) \qquad\text{and}\qquad M=\sup_{x\in[0,1]} f''(x). \end{equation} Given what you know about $f''(x)$ and about $f'(c)$, can you make the desired conclusion about $f'(x)$?

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This conjecture is wrong. Just take the counterexample $$f:x\mapsto y=3\cdot\left(x-\frac12\right)^2.$$ Obviously holds $f(0)=\tfrac34=f(1)$, but with $f^\prime(x)=6\cdot\left(x-\tfrac12\right)$ you have $f^\prime(\tfrac34)=\tfrac32>1$.

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  • $\begingroup$ You missed the second derivative condition. $\endgroup$
    – Ian
    Commented Aug 30, 2016 at 10:34
  • $\begingroup$ Thanks, I just read the headline. $\endgroup$
    – user40471
    Commented Aug 30, 2016 at 10:39

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