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As i know that linear combination of monotonic increasing is monotonic increasing.Since y=x is monotonic increasing and another function y=[x](integral part of x) is also monotonic increasing Now x-[x]={x} .{x} is fractional part of x which is not monotonic.please help me .IF i am incorrect please tell breifly .I m very confused. THANK YOU

Pawan Thakur
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    Many people regard the use of all capitals as shouting, and therefore find it rude. – lulu Jun 05 '20 at 20:57
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    Anyway, it clearly isn't true that $\textit {any}$ linear combination of such functions is another such. $x$ and $2x$ are both montonically increasing but $x-2x=-x$ is not. – lulu Jun 05 '20 at 20:58
  • https://math.stackexchange.com/a/1501551/775831 – Pawan Thakur Jun 05 '20 at 21:01
  • You are misreading that question. Saying that a sum of good terms is good is not the same as saying that an arbitrary linear combination of good terms is good. Please check my example (or yours, for that matter). – lulu Jun 05 '20 at 21:03
  • I m really confused sir . – Pawan Thakur Jun 05 '20 at 21:07
  • They form a convex cone, but not a subspace. – copper.hat Jun 05 '20 at 21:09
  • @lulu From the link which i have given in comment box..it is clear that sum of two monotonic increasing function is monotonic increasing ...and you are doing same x-2x ..but where is difference..please explain – Pawan Thakur Jun 05 '20 at 21:13

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$-f(x)$ is a linear combination of $f(x)$, but does not share its "increasingness".

Anyway, if

$$u<v\implies f(u)\le f(v)\land g(u)\le g(v)$$

we do have $$u<v\implies af(u)\le af(v)\land bg(u)\le bg(v)$$

and

$$u<v\implies af(u)+bg(u)\le af(v)+bg(v)$$

provided $a,b\ge 0$ !