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Let $f$ be twice differentiable function on $(0,1)$. Given that for every $x\in(0,1)$, $|f''(x)|\leq M$ where $M$ is a non-negative real number. Prove that $f$ is uniformly continuous on $(0,1)$.

My Approach:

Since $f$ is twice differentiable function on $(0,1)$, it follows that $f'$ is continuous in $(0,1)$. So, by Mean Value Theorem, for any $x,y\in (0,1)$ with $x<y$, there is a $z$ with $x<z<y$ such that $|\frac{f'(x)-f'(y)}{x-y}|=|f''(z)|$. That is, $|f'(x)-f'(y)|\leq M|x-y|$ for all $x,y\in(0,1)$.

Now , I don't know how to proceed further.

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    $\begingroup$ $f''$ is bounded, hence $f'$ is uniformly continuous, hence $f'$ is bounded, hence $f$ is uniformly continuous $\endgroup$ – MathematicsStudent1122 May 12 '17 at 3:44
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We have since $f''$ is bounded $f'$ is Lipschitz and hence uniformly continuous on $(0,1)$ Then $f'$ is bounded because given any $x,y$ in $(0,1)$ we have $\frac{|f'(x)-f'(y)|}{|x-y|}\le M$ and hence $|f'(x)-f'(y)|\le M$. Thus $f'$ is bounded so $f$ is Lipschitz once again and hence uniformly continuous.

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The inequality at which you arrived implies that $|f'(x) - f'(y)| \leq M$ for all $0 < x,y < 1$. If $f'$ is unbounded on $]0,1[$, then for every $B > 0$ there are some $0 < x,y < 1$ such that $|f'(x) - f'(y)| > B$; hence $f'$ is bounded on $]0,1[$ from above by $M$. Then mean-value theorem gives that for every pair of $0 < x,y < 1$ there is some $x < \xi < y$ such that $|f(x)-f(y)| \leq |f'(\xi)||x-y| \leq M|x-y|$. If $\varepsilon > 0$, the number $\delta := \varepsilon/M$ is such that $|x-y| < \delta$ implies $M|x-y| < \varepsilon$; hence $f$ is uniformly continuous on $]0,1[$.

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  • $\begingroup$ Can you explain the proof of boundedness of $f'(x)$? $\endgroup$ – Unknown x May 29 '18 at 15:42

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