Measuring nonlinearity and reducing it

How to measure the nonlinearity of a function {f\colon I\to \mathbb R} where {I\subset \mathbb R} is an interval? A natural way is to consider the smallest possible deviation from a line {y=kx+b}, that is {\inf_{k, b}\sup_{x\in I}|f(x)-kx-b|}. It turns out to be convenient to divide this by {|I|}, the length of the interval {I}. So, let {\displaystyle NL(f;I) = \frac{1}{|I|} \inf_{k, b}\sup_{x\in I}|f(x)-kx-b|}. (This is similar to β-numbers of Peter Jones, except the deviation from a line is measured only in the vertical direction.)

NL(f; I) is the maximal vertical distance between red and black, divided by the length of I

Relation with derivatives

The definition of derivative immediately implies that if {f'(a)} exists, then {NL(f;I)\to 0} as {I} shrinks to {a} (that is, gets smaller while containing {a}). A typical construction of a nowhere differentiable continuous function is based on making {NL(f;I)} bounded from below; it is enough to do this for dyadic intervals, and that can be done by adding wiggly terms like {2^{-n}\mathrm{dist}\,(x, 2^{-n}\mathbb Z)}: see the blancmange curve.

The converse is false: if {NL(f; I)\to 0} as {I} shrinks to {a}, the function {f} may still fail to be differentiable at {a}. The reason is that the affine approximation may have different slopes at different scales. An example is {f(x)=x \sin \sqrt{-\log |x|}} in a neighborhood of {0}. Consider a small interval {[-\delta, \delta]}. The line {y = kx} with {k=\sin\sqrt{-\log \delta}} is a good approximation to {f} because {f(x)/x\approx k} on most of the interval except for a very small part near {0}, and on that part {f} is very close to {0} anyway.

Why the root of logarithm? Because {\sin \log |x|} has a fixed amount of change on a fixed proportion of  {[-\delta, \delta]}, independently of {\delta}. We need a function slower than the logarithm, so that as {\delta} decreases, there is a smaller amount of change on a larger part of the interval {[-\delta, \delta]}.

Nonlinearity of Lipschitz functions

Suppose {f} is a Lipschitz function, that is, there exists a constant {L} such that {|f(x)-f(y)|\le L|x-y|} for all {x, y\in I}. It’s easy to see that {NL(f;I)\le L/2}, by taking the mid-range approximation {y=\frac12 (\max_I f + \min_I f)}. But the sharp bound is {NL(f;I)\le L/4} whose proof is not as trivial. The sharpness is shown by {f(x)=|x|} with {I=[-1,1]}.

With the maximum difference 1/2 and the length of interval 2, we get NL(f; [-1, 1]) = 1/4
Proof. Let {k} be the slope of the linear function that agrees with {f} at the endpoints of {I}. Subtracting this linear function from {f} gives us a Lipschitz function {g} such that {-L-k\le g'\le L-k} and {\int_I g'= 0}. Let {A = \int_I (g')^+ = \int_I (g')^-}. Chebyshev’s inequality gives lower bounds for the measures of the sets {g'>0} and {g'<0}: namely, {|g'>0|\ge A/(L-k)} and {|g'<0|\le A/(L+k)}. By adding these, we find that {|I| \ge 2LA/(L^2-k^2)\ge 2A/L}. Since {\max _I g - \min_I g \le A}, the mid-range approximation to {g} has error at most {A/2 \le |I|L/4}. Hence {NL(f; I) = NL(g; I) \le L/4}.

Reducing nonlinearity

Turns out, the graph of every Lipschitz function has relatively large almost-flat pieces.  That is, there are subintervals of nontrivial size where the measure of nonlinearity is much smaller than the Lipschitz constant. This result is a special (one-dimensional) case of Theorem 2.3 in Affine approximation of Lipschitz functions and nonlinear quotients by Bates, Johnson, Lindenstrauss, Preiss, and Schechtman.

Theorem AA (for “affine approximation”): For every {\epsilon>0} there exists {\delta>0} with the following property. If {f\colon I\to \mathbb R} is an {L}-Lipschitz function, then there exists an interval {J\subset I} with {|J|\ge \delta |I|} and {NL(f; J)\le \epsilon L}.

Theorem AA should not be confused with Rademacher’s theorem which says that a Lipschitz function is differentiable almost everywhere. The point here is a lower bound on the size of the interval {J}. Differentiability does not provide that. In fact, if we knew that {f} is smooth, or even a polynomial, the proof of Theorem AA would not become any easier.

Proof of Theorem AA

We may assume {I=[-1, 1]} and {L=1}. For {t\in (0, 2]} let {L(t) = \sup \{|f(x)-f(y)|/|x-y| \colon x, y\in I, \ |x-y|\ge t\}}. That is, {L(t)} is the restricted Lipschitz constant, one that applies for distances at least {t}. It is a decreasing function of {t}, and {L(0+)=1}.

Note that {|f(-1)-f(1)|\le 2L(1)} and that every value of {f} is within {2L(1)} of either {f(-1)} or {f(1)}. Hence, the oscillation of {f} on {I} is at most {6L(1)}. If {L(1) \le \epsilon/3}, then the constant mid-range approximation on {I} gives the desired conclusion, with {J=I}. From now on {L(1) > \epsilon/3}.

The sequence {L_k = L(4^{-k})} is increasing toward {L(0+)=1}, which implies {L_{k+1}\le (1+\epsilon) L_k} for some {k}. Pick an interval {[a, b]\subset I} that realizes {L_k}, that is {b-a\ge 4^{-k}} and {|f(b)-f(a)| = 4^{-k}L_k}. Without loss of generality {f(b)>f(a)} (otherwise consider {-f}). Let {J = [(3a+b)/4, (a+3b)/4]} be the middle half of {[a. b]}. Since each point of {J} is within distance {\ge 4^{-k-1}} of both {a} and {b}, it follows that {\displaystyle f(b) + L_{k+1}(x-b) \le f(x) \le f(a) + L_{k+1}(x-a) } for all {x \in J}.

The green lines have slope L_{k+1} which is close to the slope L_k of the secant line through (a, f(a)) and (b, f(b)). The graph of f is pinched between these green lines, except near a or b.

So far we have pinched {f} between two affine functions of equal slope. Let us consider their difference:
{\displaystyle (f(a) + L_{k+1}(x-a)) - (f(b) + L_{k+1}(x-b)) = (L_{k+1}-L_k) (b-a)}. Recall that {L_{k+1}\le (1+\epsilon) L_k}, which gives a bound of {\epsilon L_k(b-a) \le 2\epsilon L |J|} for the difference. Approximating {f} by the average of the two affine functions we conclude that {NL(f;J)\le \epsilon L} as required.

It remains to consider the size of {J}, about which we only know {|J|\ge 4^{-k}/2} so far. Naturally, we want to take the smallest {k} such that {L_{k+1}\le (1+\epsilon) L_k} holds. Let {m} be this value; then {L_m > (1+\epsilon)^{m} L_0}. Here {L_m\le 1} and {L_0 = L(1)> \epsilon/3 }. The conclusion is that {(1+\epsilon)^m < 3/\epsilon}, hence {m< \log(3/\epsilon)/\log(1+\epsilon)}. This finally yields {\displaystyle \delta = 4^{-\log(3/\epsilon)/\log(1+\epsilon)}/2} as an acceptable choice, completing the proof of Theorem AA.

A large amount of work has been done on quantifying {\delta} in various contexts; for example Heat flow and quantitative differentiation by Hytönen and Naor.

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