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$f:X \rightarrow \mathbb{R} $ for which there is $N \ge 0$ such that $f(x)-f(y)\le N|x-y|$

How may I show that f is continuous?

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    That's what I meant. The inequality implies |f'(x)| < N. I knew that it wouldn't be true everywhere as you say, but at least "almost everywhere" for a well behaved function. Thanks for the info on Lipschitz, I'll read about this a little further.2012-02-05

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The question in the comment of Jonas Meyer 'What is $X$' is very important. Since you use $|x-y|$ for elements of $X$, it would mean that

  • $X$ is a linear space ( you can add, substract, multiply with scalars)

  • there is a norm on $X$.

Since you use the same notation as the absolute value, you maybe mean $X=\Bbb{R}$ or $X=\Bbb{R}^p$. Anyway, the proof remains the same for every metric space.

From your condition it follows easily that $|f(x)-f(y)| \leq N|x-y|$. $f$ is continuous at $x_0 \in X$ if for every $\varepsilon >0$ there exists $\delta>0$ such that $|f(x_0)-f(y)| < \varepsilon$ for every $y$ with $|x_0-y|<\delta$.

Pick $x_0 \in X$ and $\varepsilon >0$. Then $|f(x_0)-f(y)|\leq N|x_0-y|$. If you pick $\delta=\displaystyle \frac{\varepsilon}{N}$ then whenever $|x_0-y|<\delta$, you get $|f(x_0)-f(y)| \leq N |x_0-y|<\displaystyle N\frac{\varepsilon}{N}=\varepsilon$. Therefore $f$ is continuous in $x_0$. Since $x_0$ was chosen arbitrarily it follows that $f$ is continuous.

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    Thank you. I understood the question a while after I had posted. But this is very informative. I shall accept the answer. Thanks!2012-02-02