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Let $f:\mathbb{R} \to \mathbb{R}$ be a function with the intermediate value property: that is, $f$ maps intervals to intervals. Let $x \in \mathbb{R}$. Suppose to each sequence $ (x_n) $ converging to $x$ there exists a constant $M$ such that $$|f(x) - f(x_n)| ≤M \sup _{n,m}|f(x_n) - f(x_m)|$$ Then show that $f$ is continuous at $x$.

How can I solve this problem? Can anyone help me please. I have the basic idea of real analysis but could not crack this problem.

vonbrand
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damini
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1 Answers1

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Hint: If $f$ is not continuous at $x$, use the Intermediate Value Property of $f$ to find a sequence $\langle x_n \rangle_{n=1}^\infty$ converging to $x$ such that $f(x_n) = f(x) \pm \epsilon$ for some fixed $\epsilon > 0$. Note that either $f(x_n) = f(x) + \epsilon$ for infinitely many $n$, or $f(x_n) = f(x) - \epsilon$ for infinitely many $n$. Derive a contradiction to the stated property of all sequences converging to $x$.

user642796
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  • some more hints please.I can not find such a sequence. – damini Mar 19 '13 at 16:32
  • @user67496: Recall that if $f$ is not continuous at $x$ then there is an $\epsilon > 0$ such that for each $\delta > 0$ there is a $y$ such that $| y - x | < \delta$ and $| f(y) - f(x) | \geq \epsilon$. Pick appropriate $\delta$'s, and then use the IVP of $f$. – user642796 Mar 19 '13 at 16:56