How to show the k-th derivative of f is bounded, for k =1,2,3....?

Question

Showing the statement below seems to be a cute problem (f^k denotes the k-th derivative of f). I don't have the proof yet, but I was assured by a top dog that the statement is true.

Suppose that, for some positive integer n, the function f: R->R is (n +1) times differentiable and that both f and f^(n+1) are bounded on R. Then, for every k=1,2,3.... , f^k is bounded on R

Clarifying: The given conditions don't imply f has derivatives of all orders on R. What we have to prove is that, for eack k <= n+1, f^k is bounded on R

Answer 1

(Nov 2 2007: Edited in some details, motivated by nealjking's response.)

I will prove the statement by contradiction.

Assume for some 0
Conclusion: If f^(k) is unbounded for some k, then either all higher-order derivatives (that exist on R) are unbounded, or all lower-order derivatives are unbounded. Hence, if both f and f^(n+1) are bounded on R, so are f^(k), k=1,...,n.

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I will here show that f^(k-1),f^(k-2),...,f are unbounded under the assumption that f^(k) is unbounded and f^(k+1) is bounded.

Let N>0, and let |f^(k+1)| be bounded by M.

Since f^(k) is unbounded, it is then possible to find an interval of arbitrary width, say the interval J_k=[a, a+(2^k)*(k+1)] such that |f^(k)| is larger than N on the whole interval, for some a. (Just choose an x where |f^(k)|>N*(1 + M*2^(k-1)*(k+1)) and expand the interval in both directions.)

Now, there must exist a subinterval J_{k-1} in J_k of width at least (2^(k-1))*((k-1)+1) where |f^(k-1)| is larger than N. (This is easily understood by drawing a sketch of a graph on the interval J_k with an positive (or negative) derivative of at least N on the whole interval.)

This argument can be repeated where the width of each interval J_s, s=k-2, k-3,...,0, where |f^(s)|>N, is essentially halved each time.

With the above procedure, we are finally left with an interval J_0 of width 1 where |f|>N is guaranteed.

Since this N>0 was arbitrary, it has been shown that f^(k-1),f^(k-2),...,f are unbounded.

Answer 2

UPDATED, 2nd time:
(1st take: "I am not satisfied with Magee Coe's proposed solution"):

As far as I can tell, it depends critically on the idea that the integral of an unbounded function will itself be unbounded. But this is not true: take
f(x) = 2x*sin(x^2), which oscillates wildly but is unbounded.
It's integral is sin(^2), which is bounded, and also has bounded integrals to all orders, I believe.

This case does not disprove the main question, because that assumed both a bounded f^(k+1) and a bounded f. But I think it illustrates that Magee's proposed proof does not show any logical connection between the boundedness of f^(k+1) and the boundedness of all the derivatives in-between that and f - at least, not a connection explicit enough for me to follow.

Update:
- In answer to Magee: I understand that my example has unbounded derivatives. However, the key issue is that I don't see in your proof how it was that the boundedness of f^(k+1) contributed to the key point that the unboundedness of f^k would be "inherited", in a cogent & convincing way. The basic problem that I see is that there are at least two ways in which a function can be bounded: it can oscillate within bounds, or it can go to a nice limit. The implications for its integral are very different.
- An element that might prove helpful: the Taylor's series with remainder:
f(x) = f(0) + f_1(0)*x + (f_2(0)/2!)*x^2 + ...
...+ (f_n(0)/n!)*x^n + (1/n!) ∫(x-t)^n * f_n+1(t) dt
= f(0) + f_1(0)*x + (f_2(0)/2!)*x^2 + ...
...+ (x^n/n!)*(f_n(0) + ∫(1-(t/x))^n * f_n+1(t) dt)

The fact that f(x) is bounded implies that the coefficient of x^n must => 0 as x => ∞. The Taylor's series for the derivatives f_k have a similar expression for the final coefficient, so perhaps something can be made of this.

2nd Update, based on the revision:

OK, I see the point.

The proof does not dot all the i's and cross all the t's: the fact that the integral of the derivative over an interval can be large does not immediately guarantee that the function will be large over that interval, because the function could have had a large negative value just before the interval started. However, I'm sure this can be fixed up.

Good job!