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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).

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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 $$f(x_0)-15|x-x_0|<f(x)<f(x_0)+15|x-x_0|$$ so that means $f(x)$ should be between the cone formed by the lines $f(x_0)+15(x-x_0)$ and $f(x_0)-15(x-x_0)$, as can be seen in the image (for $x_0=0$):

enter image description here

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  • $\begingroup$ Could you give some intutition on how this condition implies continuity? (On a graph, perhaps) $\endgroup$ – The Cryptic Cat Feb 21 '17 at 0:26
  • $\begingroup$ @flytothesurface Thank you. Should there not be a "for all of x" quantifier in the definition of continuous? $\endgroup$ – Dole Feb 21 '17 at 0:54
  • $\begingroup$ @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. $\endgroup$ – A. Salguero-Alarcón Feb 21 '17 at 17:16
  • $\begingroup$ @TheCrypticCat I edited my answer according to your comment. $\endgroup$ – A. Salguero-Alarcón Feb 21 '17 at 17:31
  • $\begingroup$ @flytothesurface Thanks! $\endgroup$ – The Cryptic Cat Feb 22 '17 at 15:10

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