Bump Function

[measure]

Consider the function from R to R defined piecewise as follows:

Code: Select all

a = x/(1-x)
b = 1/a
f(x) = 0         | x ≤ 0
f(x) = a^b * b^a | 0 < x < 1
f(x) = 0         | x ≥ 1

I believe that this function is infinitely differentiable at x=0 and x=1. How would I go about proving this?

[Qaanol]

At x=0, can you prove that your function…
• is continuous?
• is differentiable?
• is twice-differentiable?

What do you imagine a proof that it is infinitely-differentiable would look like?

[measure]

At x=0, can you prove that your function…
• is continuous?
• is differentiable?
• is twice-differentiable?

What do you imagine a proof that it is infinitely-differentiable would look like?

Iff it is continuous, the limits from the left and right sides of the transition points will exist and be identical (zero in this case). As x→0+ the function looks like ε^(1/ε) * (1/ε)^ε. As epsilon decreases, the left term vanishes and the right term approaches unity, so the product is zero. Likewise as x→1-.
WolframAlpha gives me first, second, and third derivatives at least, which are all equal to zero in the limit as x approaches the boundaries from the inside. I could calculate the derivatives myself, but the repeated power, product, and quotient rule applications cause the expressions to become very large.

A potential proof might show by induction that if some derivative is continuous, that the next is also continuous, or it could show that all the derivatives have some common structure that implies the relevant limits are zero, but I don't know how to generalize over the large expressions. I am also not very familiar with this sort of proof in general, and it's not especially important to me that I am able to construct the proof myself. I just want to know if my guess is correct.

[Demki]

This may or may not be useful:
by simply taking the derivative, we get that
f'(x) = f(x) * g₁(x) (for some g₁)
therefore
f''(x) = f'(x) * g₁(x) + f(x) * g₁'(x) = f(x) * (g₁(x))^(2) + f(x) * g₁'(x) = f(x) * g₂(x) (for some g₂)
This can be extended to f⁽ⁿ⁾(x) = f(x) * g_n(x) for some g_n, for all natural n by simple induction.
If you can prove that all g_i(x) are differentiable then you can show that f is infinitely differentiable.

[MostlyHarmless]

It’s also worth noting that your title doesn’t match your question, which is good because the title is impossible. A bump function can be infinitely differentiate, but it cannot be analytic.

[measure]

It’s also worth noting that your title doesn’t match your question, which is good because the title is impossible. A bump function can be infinitely differentiate, but it cannot be analytic.

When I wrote the title, I thought "analytic" just meant that it consisted of compositions of "ordinary" functions. I didn't realize that the piecewise definition excluded it. Thanks! (Fixed)

[Qaanol]

Okay, try this on for size:

Can you prove that, for every positive integer n there exists a positive real number a_n such that, if x is within distance a_n of 0 then your function’s magnitude is less than the magnitude of xⁿ?

In other words, by choosing x “close enough” to zero, can you show that your function is bounded by ±xⁿ?

It’s also worth noting that your title doesn’t match your question, which is good because the title is impossible. A bump function can be infinitely differentiate, but it cannot be analytic.

When I wrote the title, I thought "analytic" just meant that it consisted of compositions of "ordinary" functions. I didn't realize that the piecewise definition excluded it. Thanks! (Fixed)

Just to state what you’ve probably already learned, for a function to be analytic means that it can be locally represented by a power series at every point in its domain.

At the boundary of the support of a bump function, the only power series that converges to it on the outside is the zero function, which cannot possibly converge to it on the inside. So at a point on the boundary there is no power series which locally converges that function.

[measure]

Update:

I am just looking at the function y = x^1/x for now. This function should have the same limit behavior near zero since for small x, a ≃ x and b^a ≃ 1.
The first derivative of this function y' = y*(1 - ln(x))/x^2. Let s = 1 - ln(x), and rearrange y' = y*[1/x² * s].
As Demki helpfully pointed out, further derivatives will always be in the form y^k = y*g_k(x,s).
Specifically, the function y^k = y*[a₀ + a₁*s + a₂*s² + a₃*s³ + ...] can be rearranged as y^k = y*a₀ + y*a₁*s + y*a₂*s² + y*a₃*s³ + ....
Differentiate each term using the product rule as (y*a₁*sⁿ)' = y'*a₁*s + y*a₁'*s + y*a₁*(sⁿ)'.
Since y' = y*[1/x² * s] and (sⁿ)' = -n/x*s^n-1, substitute to get (y*a₁*sⁿ)' = y*[1/x² * s]*a₁*s + y*a₁'*s + y*a₁*(-n/x)*s^n-1.
Factor out y and get (y*a₁*sⁿ)' = y*[(a₁/x²)*sⁿ⁺¹ + a₁'*sⁿ - (a₁n/x)*s^n-1].
Combine and collect like terms to get y^k+1 = y*[b₀ + b₁*s + b₂*s² + b₃*s³ + ...].

I need to check whether for arbitrarily large k, the x^1/x factor in front with overwhelm each inner term in the limit as x→0⁺.
The largest term in y^k will be y*(1/x^2k)s^k. Substitute for y to get x^{1/x - 2k} * s^k.
Rearrange as x^1/x-3k * (x*s)^k, and substitute for s to get x^1/x-3k * (x - x*ln(x))^k.
As x goes to zero, the left factor goes to zero because the 1/x part of the exponent will overwhelm the large-but-finite 3k, leaving ε^1/ε→0.
On the right side, x*ln(x) goes to zero, so the right factor goes to (±ε)^k, which also goes to zero.
The whole expression becomes 0*0, so the limit is zero.
Each other term in the expression also goes to zero by a similar argument.

Thus, for any finite k, the limit as x→0⁺ of y^k will be zero. QED

Please let me know if I missed anything.