Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f \in C^\infty_c(\mathbb{R})^*$ be a distribution. How can I show the following: $$f \in C^{0,1}(\mathbb{R}) \Leftrightarrow f \in L^\infty(\mathbb{R}) \text{ and } f' \in L^\infty(\mathbb{R}) \text{.}$$ Here $C^{0,1}(\mathbb{R})$ is the space of bounded Lipschitz functions on $\mathbb{R}$ and $f'$ is the distributional derivative of $f$.

share|improve this question

1 Answer 1

up vote 2 down vote accepted

First, suppose that $f$ is a bounded Lipschitz function (hence in $L^\infty$). Then $f$ is absolutely continuous and you have $f(x) - f(a) = \int_a^x f^{'} (t)dt$. The Lipschitz condition gives that there is a constant $L$ so that $|\int_a^x f^{'} (t)dt| \leq L(x-a)$. Dividing by $(x-a)$ and applying the Lebesgue differentiation theorem gives that $|f^{'} (t)| \leq L$ almost everywhere.

Conversely, suppose that $f \in L^\infty$ and $f^{'} \in L^\infty$ as distributions. It suffices now to show that $f(x)$ differs by a single constant almost everywhere from the Lipschitz function $\int_0^x f^{'} (t)dt$, so view $\int_0^x f^{'} (t)dt$ as another candidate distribution. An elementary exercise (which is pretty standard) shows that two distributions which have the same distributional derivative differ by a constant and so the result follows.

share|improve this answer
    
Does $f' \in L^\infty$ as a distributions mean that there exist $g\in L^\infty$ such that for every test function $\phi \in C^\infty_c$, $(f',\phi)=(g,\phi)$? –  Cantor Nov 19 '12 at 20:31
    
What is the meaning of $\int_0^\infty f'(t) dt$, when I only know $f' \in L^\infty$ as a distribution? –  Cantor Nov 19 '12 at 20:45
    
@Cantor Yes, $f^{'} = g \in L^\infty$ means that there is an actual function so that $ \langle f, \varphi^{'} \rangle = - \langle g, \varphi \rangle$. It doesn't make sense to talk about distributions living in function spaces unless you mean that there is an actual function which agrees with the distribution on all test functions. –  Chris Janjigian Nov 19 '12 at 21:11

Your Answer

 
discard

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.