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If f is differentiable at a, then f is Lipschitz of order 1 at a.


lim_x->a {(f(x)-f(a))/(x-a)} = f'(a).

hmm if I can somehow change this to f'(a)(x-a)=f(x)-f(a). And let C be a constant such that C > f'(a) then I would be done...

Any help?

Side-Note: Last time I asked something about Lipschitz at a point, I was met with comments asking what this meant, so I'm assuming this isn't standard usage? Anyways what I meant by "f is Lipschitz of order 1 at a": f is Lipschitz continuous at a if there exists a neighborhood of a such that |f(y)−f(x)|< C|y−x|.

EDIT: I have understood that lim_x->a {(f(x)-f(a))/(x-a)} = f'(a) can be changed to f'(a)(x-a)=f(x)-f(a) in a nhbd of a. And we're looking for Lipschitz at point a, so we only consider the nbhd of a anyways. So I think that should complete the proof (coupled with user15464's idea. thanks for the help!

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Since $\frac{f(x) - f(a)}{x-a}$ converges, we can find a neighborhood $U$ of $a$ such that $\frac{|f(x) - f(a)|}{|x-a|} < |f'(a)| + 1$ for all $x \in U$. Now multiply both sides by $|x-a|$. – user15464 Nov 7 '11 at 1:22
user15464's idea looks good. Alternatively, you might consider an argument using the Mean Value Theorem. – Jesse Madnick Nov 7 '11 at 1:25
Your terminology is standard, or at least it should be. The issue is that many people are more familiar with the concept of "uniformly Lipschitz," which (confusingly) is often abbreviated to "Lipschitz." A function is uniformly Lipschitz if there is one constant $C$ that works for all $a$ in the domain. Note that this is a stronger condition than saying that the function is (pointwise) Lipschitz at every point in the domain. The concept you are using is (pointwise) Lipschitz at a point. – Jesse Madnick Nov 7 '11 at 1:29
@Jesse Madnick the MVT would work if this were an interval, but not a point... or am I wrong? – MathMathCookie Nov 7 '11 at 1:37
@MathMathCookie: Ah, good point. – Jesse Madnick Nov 8 '11 at 8:16
up vote 2 down vote accepted

Choose an exponent $u$ and define a function $f$ by $f(0)=0$ and, for $x\ne0$, $$ \color{red}{f(x)=x^u\cos(\pi x^{-1})}. $$ Then $f$ is continuous at $0$ (and everywhere else) if $u>0$ and $f$ is differentiable at $0$ (and everywhere else) if $u>1$. But $x=(2n)^{-1}$ and $y=(2n+1)^{-1}$ for any positive integer $n$ yields $|x-y|=x-y=\Theta(n^{-2})$ while $\cos(\pi x^{-1})=+1$ and $\cos(\pi y^{-1})=-1$ hence $|f(x)-f(y)|=x^u+y^u=\Theta(n^{-u})$.

Finally, for every exponent $u$ in $(1,2)$, $f$ is differentiable everywhere and is not what you call Lipschitz continuous at $0$.

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