# Weak derivative of a Lipschitz function

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 $f$.

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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.
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