# Analysis problem for my research

blackbelt5400
This is not a homework problem. I've encountered something in my research in potential theory, and I need to prove the following.

Given:
h(t) = f(t) + g(t),
f '(t) < 0 for all t,
g'(t) > 0 for all t, and g'(t) is monotonically increasing,
f '(0) = 0,
g'(0) > 0,
f '(t) has exactly one minimum, and $$\lim_{t\to\infty}f'(t) = 0$$.

Show that h'(t) has at most three zeros.

Aaron792
This is not a homework problem. I've encountered something in my research in potential theory, and I need to prove the following.

Given:
h(t) = f(t) + g(t),
f '(t) < 0 for all t,
g'(t) > 0 for all t, and g'(t) is monotonically increasing,
f '(0) = 0,
g'(0) > 0,
f '(t) has exactly one minimum, and $$\lim_{t\to\infty}f'(t) = 0$$.

Show that h'(t) has at most three zeros.

Replace it by:
H(t)=F(t)-G(t)
H(t)=h'(t), F(t)=f'(t), G(t)=-g'(t)
F(t)<0, F(0)=0, G(t)<0, G(0)<0, G(t) monotonically decreasing
$$\lim_{t\to\infty}f'(t) = 0$$.
I see at most 2 zeros from graph example.
But I am not a mathmatical professional so can't help u walk it through.
Hope this can help.

Homework Helper
are f,g analytic, or maybe polynomials? if they are only smooth functions, it seems fairly easy to construct counterexamples where h' has infinitely many zeroes.

just let g = e^x, and let the graph of -f be parallel to, even equal to, the graph of g, over some interval like 1 < x < 2. then for x > 2 let the graph of -f have a flex (unique max of -f') and let -f' start to decrease toward zero as x--> infinity.

for x<1, just let graph -f straighten up to the left faster than that of g, so it becoems horizontal at x = 0, to get -f' = 0.

then -f' = g' for 1 < x < 2, so h' = f' + g' = 0 for all x: 1<x<2.

blackbelt5400
are f,g analytic, or maybe polynomials? if they are only smooth functions, it seems fairly easy to construct counterexamples where h' has infinitely many zeroes.

The most I can say is that they are subharmonic.

These are potential functions using the Newtonian kernel. Particularly, I've placed two positive charges on the x-axis at (0,0) and (1,0), with the third positive charge somewhere in the upper half-plane. Choose 0<c<1, and let $$\vec{v}$$ be the line segment originating at c directed into the upper half-plane.

Parameterize the potential on this vector with x(t) = c, y(t) = t.

Then the potential on this vector is given by h(t) = f(t)+g(t), where f(t) is the potential from the two charges on the x-axis, and g(t) is the potential from the third charge in the upper half-plane.

Clearly f '(t)<0, and f '(0)=0. I've proven that f '(t) also has exactly one minimum. Also, g'(0)>0, and if this line segment is chosen to be an altitude of the triangle formed by the three charges, then g(t) is convex and increasing. In this case, I need to show that h'(t) has at most two zeros.

If the vector is not an altitude of the triangle, then you lose the convexity of g(t). It will still be true that g'(t)>0 for all t in [0,y_3) (where (x_3,y_3) is the coordinate of the third charge). I can show that g'(t) is increasing initially. However, g'(y_3)=0, so g takes a max and is therefore concave down after some point.

In this case, the problem changes to showing that h'(t) has at most three zeros.

maze
To create the function mathwonk said, you can create f in 3 separate peices - f in [0,1], in (1,2], and in (2,infty).

At this point it will have "corners" where the peices meet up. Then convolve it with a thin mollifier to smooth out the corners.