# Logical conclusion from a function being neither increasing nor decreasing.

calculusfunctionsinequalitylogicmonotone-functions

A function $$g:[a,b]→\mathbb{R}$$ is increasing if $$∀x_1{<}x_2$$ in $$[a,b]$$, $$g(x_1)≤g(x_2).\quad$$ Similar is the definition of a decreasing function.

So, if a function $$f:[a,b]→\mathbb{R}$$ satisfies property $$p:\quad f$$ is neither increasing nor decreasing, then:

• My conclusion ($$c_1$$):

$$∃x_1{<}x_2$$ and $$x_3{<}x_4$$ in $$[a,b]$$ which satisfy $$f(x_1) and f$$(x_3)>f(x_4)$$ (or $$f(x_1)>f(x_2)$$ and $$f(x_3)). That is, $$f$$ 'increases' for some $$x_1$$,$$x_2$$ and 'decreases' for some $$x_3,x_4$$ (or vice versa).

• My textbook's and instructor's conclusion ($$c_2$$):

$$∃x_1{<}x_2{<}x_3$$ in $$[a,b]$$ s.t.$$f(x_1) and f$$(x_2)>f(x_3)$$ (or vice-versa).

My questions are:

1. I think that $$c_1$$ can be logically concluded from $$p$$, but not $$c_2.$$

Which conclusion can be derived immediately from $$p$$ by logic?

2. Are the $$c_1$$ and $$c_2$$ identical? If yes, how can we prove it?

There are a few theorems and proofs that can be easily proved from $$c_2$$, not $$c_1$$. But I think that the proofs which use $$c_2$$ are not complete, but neither I can prove those theorems using $$c_1$$, nor prove $$c_2$$ from $$c_1$$, so I need some help.

I suspect you're missing the "(or vice versa)" clause at the end of the conclusion $$(c_2)$$. Without that clause, $$(c_2)$$ certainly wouldn't follow from property $$(p)$$.
However, with that clause, since $$\ f\$$ is not decreasing over $$\ [a,b]\$$, there must exist $$\ y_1,y_2\in[a,b]\$$ such that $$\ y_1 and $$\ f\big(y_1\big)<\,f\big(y_2\big)\$$, and since $$\ f\$$ is not increasing, one of the following four conditions must hold:
• there exists $$\ y_3>y_2\$$ such that $$\ f\big(y_ 2\big)> f\big(y_ 3\big)\$$;
• there exists $$\ y_3\in\big(y_1,y_2\big)\$$ such that $$\ f\big(y_ 3\big)> f\big(y_ 2\big)\$$;
• there exists $$\ y_3\in\big(y_1,y_2\big)\$$ such that $$\ f\big(y_ 1\big)> f\big(y_3\big)\$$; or
• There exists $$\ y_3\le y_1$$ and $$\ y_4 with $$\ f\big(y_3\big)\le f\big(y_1\big)\$$ and $$\ f\big(y_ 4\big)> f\big(y_3\big)\$$.
In the first case, the stated condition of $$(c_2)$$ holds with $$x_i=y_i\$$, and in the second, it holds with $$\ x_1=y_1,\ x_2=y_3\$$ and $$\ x_3=y_2\$$. In the third case the vice-versa condition holds with $$\ x_1=y_1, \ x_2=y_3\$$ and $$\ x_3=y_2\$$. That is, $$\ f\big(x_1\big)>\, f\big(x_2\big)\$$ and $$\ f\big(x_2\big)<\, f\big(x_3\big)\$$ with $$\ x_1. In the fourth case, the same vice-versa condition holds with $$\ x_1=y_4,\ x_2=y_3\$$ and $$\ x_3=y_2\$$.