Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider $C^1([0,1])$ the functions with continuous derivative on $[0,1]$ (one-sided derivatives at each end), and $\operatorname{Lip}([0,1])$ the Lipschitz functions on $[0,1]$. The mean value theorem of course shows that $C^1([0,1]) \subseteq \operatorname{Lip}([0,1])$.

Given $f\in \operatorname{Lip}([0,1])$ with Lipschitz constant $\leq 1$, and $\epsilon>0$, can I find $g\in C^1([0,1])$ with $\|g'\|_\infty\leq 1$ (i.e. $g$ also has Lipschitz constant $\leq 1$) and with $\|f-g\|_\infty \leq \epsilon$?

(E.g. one way to find $f$ is to let $f(x) = \int_0^x F(t) \ dt$ for some $F\in L^\infty$ with $\|F\|_\infty\leq 1$. Then to find a suitable $g$, I can set $g(x)=\int_0^x a(t) \ dt$ for some $a\in C([0,1])$ with $\|a\|_\infty\leq 1$ and $\| F-a\|_1\leq\epsilon$. I can find $a$ by convolving $F$ with a bump function. But not every Lipschitz $f$ arises in this way.)

share|cite|improve this question
Do you mean dense with respect to the $L^\infty$ norm (as written) or the Lipschitz norm? – robjohn Mar 2 '12 at 17:27
Well, the title is ambiguous; but the question (I think) makes it clear I mean the infinity norm. So I think your answer is fine (but let me think!) – Matthew Daws Mar 2 '12 at 17:54
up vote 2 down vote accepted

Let $\varphi$ be a positive $C_c^\infty$ function so that $\int_\mathbb{R}\varphi(x)\;\mathrm{d}x=1$.

$f\ast\varphi_\epsilon$ is $C^\infty$ because $\varphi$ is, and it has Lipschitz constant $\le1$ because $f$ does. This means that $(f\ast\varphi_\epsilon)^\prime\le1$. Since $f$ is continuous on $[0,1]$, it is uniformly continuous, and therefore, $f\ast\varphi_\epsilon\to f$ uniformly. That is, $$ \|f\ast\varphi_\epsilon-f\|_\infty\to0\tag{1} $$

However, if we use the standard Lipschitz norm, $$ \|f\|_\mathrm{Lip}=\|f\|_\infty+\sup_{x\not=y}\,\left|\frac{f(x)-f(y)}{x-y}\right| $$ the same cannot be said. Take for instance $f(x)=|x|$, the absolute value function. Any $C^1$ function that matches the slope of $f$ just below $0$, will fail to match the slope of $f$ by $2$ just above $0$. Thus, the Lipschitz norm of the difference between a $C^1$ function and $f$ is at least $1$.

share|cite|improve this answer
Yes, this seems to work. I was a bit worried about using convolution on functions only defined on $[0,1]$. If you just zero extend $f$ to a function on $\mathbb R$, it's not Lipschitz any more. But I think you can just extend it to be constant on $(-1,0]$ and $[1,2)$ say, and then work on the bigger interval $[-1/2,3/2]$ or something. Thanks! – Matthew Daws Mar 2 '12 at 18:22
I was thinking of extending $f$ outside $[0,1]$ by $$f(x)=\left\{\begin{array}{}f(0)&\text{for }x<0\\f(1)&\text{for }x>1\end{array}\right.$$ then restricting the convolution back to $[0,1]$. – robjohn Mar 2 '12 at 18:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.