# Differentiability of Lipschitz functions

## Contents

# Differentiability of Lipschitz functions#

The regularity of a Lipschitz function \(f\colon \mathbb R^n\to \mathbb R\) is very interesting. Of course, Lipschitz functions are continuous, but they may not be differentiable everywhere. However, it is quite easy to convince yourself that they cannot be non-differentiable on quite a large set. The question to quantify how large the non-differentiability set of a Lipschitz function can be was one of the motivating questions of Lebesgue’s development of measure theory.

(Total variation of a measure)

Let \(f\colon [a,b]\to \mathbb R\). The *total
variation* of \(f\), \(Vf\colon [a,b]\to [0,\infty]\) is
defined by

where the supremum ranges over all \(a=t_0 < t_1 <\ldots <t_n=b\).

If \(Vf(b)<\infty\), \(f\) is said to have *bounded variation
(BV)*.

(Absolutely continuous function)

A function \(f\colon [a,b]\to \mathbb R\) is
*absolutely continuous (AC)* if for any \(\epsilon>0\) there
exists a \(\delta>0\) such that, for any intervals
\((a_1,b_1),(a_2,b_2)\ldots\subset [a,b]\) with
\(\sum_i |b_i-a_i|<\delta\), we have
\(\sum_i |f(b_i)-f(a_i)|<\epsilon\).

Note that AC functions are BV, and Lipschitz functions are AC (see Example 33). Also, if \(f\) is BV then \(Vf\) and \(Vf-f\) are non-decreasing. If \(f\) is AC then so are \(Vf\) and \(Vf-f\), see Example 34.

(Lebesgue)

Let \(f\colon [a,b]\to \mathbb R\) be absolutely continuous. Then \(f\) is differentiable \(\mathcal L^1\) almost everywhere. Moreover, for any \(x>y \in [a,b]\),

Proof. By Example 34 it suffices to assume that \(f\) is non-decreasing. In this case define a measure \(\mu\) on \([a,b]\) using the Carathéodory construction with \(F\) the set of compact intervals and \(\zeta([c,d])= f(d)-f(c)\). This defines a finite Borel measure such that \(\mu([c,d])=f(d)-f(c)\) for all intervals \([c,d]\subset [a,b]\). Indeed, for any \(\delta>0\), we may cover \([c,d]\) by finitely many intervals \([c,c_1],[c_1,c_2],\ldots,[c_k,d]\) of width \(\delta'\leq\delta\), showing

The reverse inequality holds because \(f\) is non-decreasing.

Note that \(\mu\ll \mathcal L^1\). Indeed, given \(\epsilon>0\), let \(\delta>0\) be given by the definition of \(f\) being absolutely continuous. If \(\mathcal L^1(N)=0\), we may cover \(N\) by countably many closed intervals \(I_i\) such that \(\sum_i\mathcal L^1(I_i)<\delta\). In particular \(\sum_i f(I_i)<\epsilon\) and hence \(\mu(N)<\epsilon\). Therefore,

with \(\, \mathrm{d}\mu/\, \mathrm{d}\mathcal L^1 \in L^1(\mathcal L^1)\).

By the Lebesgue differentiation theorem, for any Lebesgue point \(x\) of \(\, \mathrm{d}\mu/\, \mathrm{d}\mathcal L^1\),

(Rademacher)

Any Lipschitz \(f\colon \mathbb R^n\to \mathbb R\) is differentiable \(\mathcal L^n\) almost everywhere.

Proof. For notational simplicity, we prove the case \(n=2\).

For each \(y\in \mathbb R\), \(x\mapsto f(x,y)\) is a Lipschitz function \(\mathbb R\to\mathbb R\) and so is differentiable \(\mathcal L^1\)-a.e. That is, for every \(y\), \(\partial_1 f(x,y)\) exists for \(\mathcal L^1\)-a.e. \(x\). By Fubini’s theorem, \(\partial_1 f\) exists \(\mathcal L^2\)-a.e. Similarly, \(\partial_2 f\) exists almost everywhere too.

Fix \(\epsilon>0\). For \(D\in \mathbb Q^2\) and \(j\in \mathbb N\) let

These are Borel sets. Further, for \(D\in\mathbb Q^2\), if

then \(x\in X_{D,j}\) for sufficiently large \(j\). That is,

is a set of full measure.

Fix \(D\in \mathbb Q^2\) and \(j\in\mathbb N\). Let \(x\) be a density point of \(X_{D,j}\). Let \(R>0\) such that

for all \(0<r<R\). In particular, for every \(y\in B(x,r)\) there exists \(y'\in X_{D,j}\) with

Now let \(r<\min\{R,1/j\}\) and \(\|x-y\|<r\). Set \(h=y-x\), \(\tilde y = x + \pi_1 y\) and \(\tilde{\tilde y} \in X_{D,j}\) with

Also let \(y',y''\) lie on the same vertical line as \(\tilde{\tilde y}\) such that \(y',\tilde y\) have the same vertical component as do \(y'',\tilde{\tilde y}\). Then, since \(x\in X_{D,j}\),

Since \(f\) is Lipschitz,

Since \(\tilde{\tilde y}\in X_{D,j}\),

Since \(f\) is Lipschitz,

By combining (17), (18) (19) and (20),

This is true for all \(y\) with \(\|x-y\|<r\) and for any density point \(x\) of the full measure set \(X^\epsilon\). That is, for \(\mathcal L^n\)-a.e \(x\). Taking a countable intersection over \(\epsilon \to 0\) concludes the proof.

## Exercises#

**Exercise 48**. Prove that Lipschitz functions are AC and that AC
functions are BV.

**Exercise 49**. Let \(f\colon [a,b]\to \mathbb R\) be BV. Show
that \(Vf\) and \(Vf-f\) are non-decreasing. If \(f\) is
AC then show that \(Vf\) and \(Vf-f\) are AC.

**Exercise 50**. Show that any monotonic
\(f\colon \mathbb R\to \mathbb R\) is continuous except at
countably many points.

**Exercise 51**. In this exercise we will show that monotonic
functions are differentiable almost everywhere.

Let \(f\colon [a,b]\to \mathbb R\) be non-decreasing. For each \(x\in (a,b)\) let

Observe that the set of \(x\in(a,b)\) where \(f\) is not differentiable at \(x\) is the countable union, over \(p<q\in\mathbb Q\), of the sets

We now fix \(p<q\in\mathbb Q\).

Let

\[\mathcal B =\{[x,x+h]:f(x+h)-f(x) <ph\}.\]Note that \(\mathcal B\) satisfies the hypotheses of the Vitali covering theorem (recall Theorem 43). Let \(\mathcal B'\) be a disjoint sub-cover obtained from the Vitali covering theorem with respect to Lebesgue measure and let \(S=\cup \mathcal B'\). Prove that

\[\mathcal L^1(f(B_{p,q}\cap S)) \leq p \mathcal L^1(B_{p,q}\cap S).\]Note: this is the step where we require \(f\) to be monotonic.

Similarly, prove that \(\mathcal L^1(f(B_{p,q}\cap S)) \geq q \mathcal L^1(B_{p,q} \cap S)\).

Deduce that \(f\) is differentiable almost everywhere.

Deduce that a BV function is differentiable almost everywhere.

However, BV functions do not satisfy the fundamental theorem of calculus:

**Exercise 52**. Recall the definition of the Cantor set from
Exercise 6. Define the *Cantor function*
\(f\colon [0,1] \to [0,1]\) as follows. For each
\(n\in\mathbb N\), define \(f_n \colon [0,1] \to [0,1]\) by

Show that the \(f_n\) converge uniformly on \([0,1]\) to a monotonic, continuous function \(f\). For each \(x\in [0,1]\setminus C\), show that \(f'(x)=0\).

Thus, \(f\) is monotonic and hence BV, has derivative \(0\) almost everywhere, but does not satisfy the fundamental theorem of calculus.

**Exercise 53**. In lectures we proved that the derivative of any AC
function is an absolutely continuous measure. Prove the converse: for
any finite, absolutely continuous measure \(\mu\) on
\([0,\infty)\), show that

defines an absolutely continuous function.

Up to now, we have considered points where functions *are*
differentiable. We now consider points of non-differentiability (which
are much more interesting).

Show that the Cantor function is not differentiable at any point of the Cantor set.

Let \(N\subset [0,1]\) satisfy \(\mathcal L^1(N)=0\).

For each \(n\in\mathbb N\), iteratively construct a countable collection of open intervals \(\mathcal O_n\) such that, for each \(n\in\mathbb N\),

\(N\) is contained in the union of \(\mathcal O_n\);

for every \(I\in \mathcal O_n\) there exists \(J\in\mathcal O_{n-1}\) with \(I\subset J\);

for each \(I\in \mathcal O_{n-1}\),

\[\mathcal L^1(I\cap \cup\{J:J\in \mathcal O_n\}) < 2^{-n} |I|.\]

Let

\[S= \bigcap_{n\in\mathbb N} \bigcup_{m>n} \cup\{J:J\in\mathcal O_m\},\]the “limsup” of the \(\mathcal O_n\) (\(S\) is the set of points that are contained in infinitely many intervals from the \(\mathcal O_n\)). In particular, \(S\supset N\).

For each \(x\in [0,1]\setminus S\), let \(N(x)\) be the largest \(n\) for which there exists \(I\in \mathcal O_n\) with \(x\in I\). Define \(P(x)=1\) if \(N(x)\) is even, \(P(x)=0\) otherwise. Finally, for each \(x\in[0,1]\) define

\[f(x) = \mathcal L^1(\{t\in [0,x] : P(t)=1\}).\]Show that \(f\) is Lipschitz, monotonic, and not differentiable at any point of \(N\). Hint: show that \(\underline D f(x)=0\) and \(\overline Df(x)=1\) for each \(x\in N\).