I am studying for my real analysis class, and one question involves two increasing functions $f$ and $g$. I have already proven that if $f$ and $g$ are increasing, then so is ($f+g$), and that the product ($fg$) is not necessarily increasing.
The third part that I am working on asks about the composition ($f∘g$). My thoughts are that plugging one increasing function into another increasing function would mean their composition is also increasing. Here is what I have:
Let $a<b$. It follows that $g(a) < g(b)$. Applying $f$ to both sides, we have $f(g(a))<f(g(b))$.
Is this proof sufficient?
I apologize if this question has been already answered as I imagine it has - but I could not find the proof. Thanks in advance!
Yes, the proof is good, let me recap, maybe it helps a bit.
We must prove that if $a<b$ then $(f\circ g) (a) < (f\circ g)(b)$.
Using the definition of composition this is equivalent to $f(g(a))< f(g(b))$.
We have: $a<b \implies g(a)< g(b)$ because $g$ is increasing, and we have $g(a)< g(b)\implies f(g(a))<f(g(b))$ because $f$ is increasing.
Notice that we did not use the fact that $f$ or $g$ are continuous. Although if we used it we could have proved that $f\circ g$ is continuous.