How to prove the existence of a subinterval where f is Lipschitz?

Question

The result below is not very known:

Let f:I -> R be differentiable on the open (non empty) interval I. Then, there exists a subinterval of I where f is Lipschitz.

Actually I think this is true under a less rigid assumption. I think the conclusion remains true if, instead of full differentiabilty on I, we assume the existence of the Dini derivatives, that is, suppose that, for every x of I, f is differentiable at least at x+ or x-, not necessarily at both.

Nealjking: There's a subtle detail. It is NOT supposed f' is continuous, so the simple fact that f' exists on every closed subinterval doesn't ensure f' is bounded there.

I'll think carefully about your answers, than you Nealjking and Themathemagician.

For the case of full differentiabilty, I came up with a proof which is not that intuitive and probably most people will hate it, but it works:

We know every derivative f' defined on an interval I is the limit of a sequence g_n of continuous functions defined on I.

Since the functions g_n are continuous and I is a Baire space, it follows from the properties of continuous functions defined on Baire spaces and values in R or the complexes that there is a subinterval J of I where g_n is uniformly bounded by some M >0. Therefore f' is bounded by M on J.

Since f is differentiable on J and f' is bounded, it follows f' is Lipschitz on J.

I'm working on a proof where only the existence of the Dini derivatives exist. I'll send it to you, for your comments.

Answer 1

UPDATED:

If the function is differentiable on the open interval I, then it is also differentiable on any closed sub-interval S. Since S is closed, and the derivative exists on the entire set, there is a maximum value of the derivative, which we can denote as K. This value will serve as the upper limit of the ratio of
df to dx, which defines Lipschitz continuity.

Yes, it should also work if the Dini derivatives are defined on the closed sub-interval: there is a maximum value of the two, which will serve as the K for the Lipschitz condition. But both must exist.

Update:
Hi Steiner, I don't think continuity is needed for my point. The sub-interval S is closed and bounded. If there were no upper bound, you could find a sequence of points x_n, such that If'(x_n)I > n, for n = 1, 2, 3,.. This sequence, in a bounded set, must have a point of accumulation, y; and since S is closed, it contains all its points of accumulation. But then f'(y) = infinity, which is impossible. So, at the very least, If'I must be bounded on S, so I can choose any finite upper bound as my value for K in the argument preceding.


That is already enough to establish the main point under discussion, I believe. As to the subsidiary question, Does f' attain a maximum value on S?, I think the answer is still yes. Certainly, if f' exists on S, it has a least upper bound. If that value is attained at a point z in S, then we are done. If it is not, still there must be infinitely points in S with f' value arbitrarily close to that LUB, so we must have a point of accumulation. Ah! I see that if f' is not continuous, its value at the point of accumulation need not be the LUB.

So:
- You are right that f' may not actually attain a maximum on S.
- But I can still claim that f' is bounded on S, and that is good enough to establish the Lipschitz behavior.


Further update: on the Dini condition
- I had misunderstood what you were saying: I thought the idea was that both the right-hand and the left-hand derivatives existed, but might differ in value. If instead you mean that one or the other might exist, but not necessarily both, my argument wouldn't apply.
- But in that case, I have a counter-example: Let f(x) be defined on [-1, 1] :
f(x) = 0 for x in [-1,0]
f(x) = sqrt(x) for x in [0,1]
Then f is differentiable everywhere except x = 0, where it has only the left-hand Dini derivative of 0. Going to the right, sqrt(x) does NOT have Lipschitz behavior, as the difference-quotient goes like 1/sqrt(x), which is unbounded near x = 0.

Answer 2

EDIT: I believe the following solution may be incorrect. I stated that monotone functions are differentiable from the left and from the right, which may not be correct--I only know that the limit of the function itself exists from the left and from the right. Sorry.

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This is not true under the more rigid assumption.

I will construct a counterexample as the sum of a sequence {f_n} of functions on the interval [0, 1].

Construct a sequence of functions {f_n}, n >=1, defined on [0, 1], as follows:

f_1(x) is:
Equal to 1/2 identically.

f_2(x) is:
Equal to 1/4 if 0 <= x <= 1/2;
Equal to 1/4 + 1/8 if 1/2 < x <= 1

f_3(x) is:
Equal to 1/16 if 0 <= x <= 1/4;
Equal to 1/16 + 1/32 if 1/4 < x <= 3/4;
Equal to 1/16 + 1/32 + 1/64 if 3/4 < x <= 1.

f_4(x) is:
Equal to 1/128 if 0 <= x <= 1/8;
Equal to 1/128 + 1/256 if 1/8 < x <= 3/8;
Equal to 1/128 + 1/256 + 1/512 if 3/8 < x <= 5/8;
Equal to 1/128 + 1/256 + 1/512 + 1/1024 if 5/8 < x <= 7/8;
Equal to 1/128 + 1/256 + 1/512 + 1/1024 + 1/2048 if 7/8 < x <= 1.

and so on in this manner.

Now define f(x) = f_1(x) + f_2(x) + f_3(x) + ...

This is a valid definition because, for any x, the sequence {f(x)} is increasing and bounded above by 1 (because f_k(x) <= f_k(1), and f_1(1) + f_2(1) + f_3(1) + ... = 1.) So this function is defined everywhere on [0, 1].

The function f is also monotone increasing (this should be obvious). Since f is monotone, then the left and right derivatives exist everywhere in (0, 1).

**Note that f has a jump discontinuity at every rational number which can be written with its denominator a power of 2, and that no two jump discontinuities are the same size.**

I've shown that f has left and right derivatives on (0, 1). Now I will show that there is no subinterval on which f is Lipschitz.

Let [a, b] be a subinterval of [0, 1], and let c be a positive real number. I will show that there exist x and y in [a, b] such that |f(x) - f(y)| > c|x - y|.

Let k be the smallest integer such that there is a jump of 1/2^k at some point in (a, b); suppose that this jump occurs at the point x0. Let e > 0, and consider x0 - e and x0 + e. Since there is a jump of size 1/2^k at x0, then we have |f(x) - f(y)| > 1/2^k.

But since this is true no matter how small e is, then if we take e = 1/(c * 2^{k + 1}), we have

c|(x0 - e) - (x0 + e)| = 2ce = 2c/(c * 2^{k + 1}) = 1/2^k < |f(x0 - e) - f(x0 + e)|.

So this x0 - e and x0 + e fail the specified Lipschitz condition.

Since c and [a, b] were arbitrary, then f is not Lipschitz on any subinterval of (0, 1).