What is an example of a function which is not lipschitz but satisfies the following weaker notion of linear growth
$$f(x) < K(1+x) \forall x, K > 0$$
along with being continuous
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Maybe $f(x)=-1_{\mathbb Q}(x)$ ? – Hagen von Eitzen Sep 01 '14 at 17:23
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The fractional part of $x$, $f(x)=x-\lfloor x \rfloor$? – hardmath Sep 01 '14 at 17:47
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Related:http://math.stackexchange.com/questions/14735/is-a-uniformly-continuous-function-vanishing-at-0-bounded-by-axc – Jonas Meyer Sep 02 '14 at 02:30
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Since the Lipshitz condition is a strong form of uniform continuity, any discontinous function will not be Lipschitz.
The estimate $f(x) \lt K(1+x)$ tells us nothing about continuity, only about "growth" of the function being limited, at least as viewed for positive $x \gt 0$. So one easily fabricates examples like ones given in Comments above for $x \in \mathbb{R}^+$, and a bit of fudging would make those work on the entire real line.
A formula that works everywhere on $\mathbb{R}$ is discontinuous function $f(x) = \lfloor x \rfloor - \frac{1}{2}$ with $K=1$. Thus $f(x) \lt x \lt 1+x$.
Added in response to change in Question:
A continuous function that satisfies the growth condition with $K=1$, but fails to be Lipshitz because of points with arbitrarily steep slope is:
$$ f(x) = \lfloor x \rfloor + \sqrt{1-(1-\{x\})^2} - 1 $$
where $\{x\} = x - \lfloor x \rfloor$ denotes the fractional part of $x$. Again we have $f(x) \lt 1+x$.
hardmath
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