# Differential inequalities

The other thing I was thinking about recently was the idea of differential inequalities I mentioned before.

I think in the case of: x’ > x if x > 0 we can say something about what x(t) must be, as long as x'(t) is continuous.

I think with careful thinking about this there could be used for a proof that there is no infinitely differentiable f(x) such that its integral and derivative are both less than it for all x > 0

If x'(t) is continuous and x'(t) > x(t) for all t > 0, and both x and x’ are always greater than 0 on that range, then set a = x(1), whatever that positive value is.

You can find an exponential function such that it also is a at t = 1. The function would be:

$g(t) = \frac{a}{e} e^t$

Now notice:
x'(1) > a, because x'(t) > x(t) and x(1) = a

And
$g'(x) = D(\frac{a}{e} e^x) = \frac{a}{e} e^x$
$g'(1) = \frac{a}{e} * e = a$

Call x'(1) = b > a. Even if x'(t) from t = 1 onward is dropping back towards g'(t), there still must be some time h before x'(t) = g'(t), because a continuous function can not have a jump discontinuity, it must smoothly merge back into the curve of g'(t).

That is, for some 1 < t < 1 + h, x'(t) > g'(t).

$\int_{1}^{1+h} x'(t) > \int_{1}^{1+h} g'(t)$
$\int_{1}^{1+h} x'(t) = x(1+h) - x(1) > g(1+h) - g(1)$
$x(1+h) - x(1) > g(1+h) - g(1) \rightarrow x(1+h) > g(1+h) - g(1) + x(1)$

But since g(1) and x(1) are both a:

$x(1+h) > g(1+h) - g(1) + x(1) \rightarrow x(1+h) > g(1+h) - a + a$
$x(1+h) > g(1+h) -a + a \rightarrow x(1+h) > g(1+h)$

This relationship is true for anywhere between $1 < t \leq t + h$ because we can use with some other j < h. So throughout this range, x(t) > g(t), or $x(t) > \frac{a}{e} e^x$

However there is no actual h where g'(t) = x'(t) again, because g”(t) = g'(t), x”(t) > x'(t), x'(t) > g'(t), and so x”(t) > g”(t). x is accelerating away from g, so it cannot rejoin the slope of g(t). Similarly we can show that for all t > 1, \latex x^{(n)} > g^{(n)}.

Since x(t) > 0 for all t > 0, we can pick any arbitrary t > 0 and use this same argument that x(t) > c exp(t) for all t > p and some multiple c. Since we can do so for any p > 0, we can show that there is some c such that x(t) is greater than c exp(t) for all t > 0.

For the other direction, x’ < x, we can use the same argument that from some point p, x(t) could not grow as fast or faster than an exponential that has the same value at p.

If x(2) = 1, then $\frac{1}{e^2} e^x$ has the same value at t=2. x’ is less than 1 there, and up to some distance h from there, so it must fall behind in the same way.

Now my question is how far can you take this line of argument, and what are the pitfalls that look like true statements you can make about inequalities, but you can’t.

It seems like the inequalities largely tell us about the growth rates of a function and nothing about its current value or velocity. We could have a function with x(1) = 1000000, but by the argument it still could not grow exponentially, because the derivative must be less than 1000000 there. It might have a connection with O notation.