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Let f be a function from real numbers to real numbers. The following holds for the function:

$ |(f(x)-f(y))|\leq 15|x-y|$

How do I prove that the function is continuous?

(Utilizing the limit concept of continuity).

1 Answers 1

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$f$ is continuous in $x_0$ if $$\forall \varepsilon>0 \ \exists \delta : |x-x_0|<\delta \Rightarrow |f(x)-f(x_0)|\leq \varepsilon$$

So, fix $\varepsilon>0$. We have $|f(x)-f(x_0)|\leq15|x-x_0|$, so it's enough to pick $\delta=\frac{\varepsilon}{15}$:

$$|x-x_0|<\delta \Rightarrow |f(x)-f(x_0)|\leq15|x-x_0|\leq 15 \delta=15\frac{\varepsilon}{15}=\varepsilon$$

For some intuition, fix a point $f(x_0)$. This condition says that $|f(x)-f(x_0)|\leq15|x-x_0|$. Eliminating absolute values, we get $$f(x_0)-15|x-x_0|

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    Could you give some intutition on how this condition implies continuity? (On a graph, perhaps) – 2017-02-21
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    @flytothesurface Thank you. Should there not be a "for all of x" quantifier in the definition of continuous? – 2017-02-21
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    @Dole you mean in like $$\forall \varepsilon >0 \ \exists \delta>0: \forall x: |x-x_0|<\delta \Rightarrow |f(x)-f(x_0)|<\varepsilon$$ If it's yes, it's implied in the "$\Rightarrow$", it says that "whenever $|x-x_0|<\delta$, then... But I'm aware some authors write it. – 2017-02-21
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    @TheCrypticCat I edited my answer according to your comment. – 2017-02-21
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    @flytothesurface Thanks! – 2017-02-22