Proofing Boundedness: Explaining the Sign Change

[Robb]

Homework Statement

Let f and g be functions from R to R. For the sum and product of f and g, determine which statements below are true. If true, provide proof; if false provide counterexample.

e) If both f + g and fg are bounded, then f and g are bounded.

Homework Equations

I don't understand how to go from abs[(f(x) + g(x))^2 - 2f(x)g(x)] to abs[(f(x) + g(x))^2 +2f(x)g(x)]
Why/how do the signs change?

 

[haruspex]

Why/how do the signs change?

You have not correctly transcribed the step. It is
|(f+g)²-2fg| ≤ (f+g)²+|2fg|
This is just an example of |a+b|≤|a|+|b|

 

[Robb]

ok. Next question, why is (f(x) + g(x))^2 chosen? When I first worked this problem I used this: abs(f(x) + g(x))=< abs(f(x)g(x))=< M+N =< MN, where m,n > 0. Hence, f & g are bounded.

 

[haruspex]

abs(f(x) + g(x))=< abs(f(x)g(x))

Try f=0, g=1.

 

[Robb]

RIght. What leads to thinking (f(x) + g(x))^2? Trial and error? Guessing?

 

[haruspex]

RIght. What leads to thinking (f(x) + g(x))^2? Trial and error? Guessing?

Given a problem involving the sum and product of the same two variables, x and y, it is natural to play around with (x+y)² and (x-y)².

 

[Ray Vickson]

RIght. What leads to thinking (f(x) + g(x))^2? Trial and error? Guessing?

No guessing needed: if $f+g$ is bounded, then $-M \leq f+g \leq M$ for some finite $M > 0$. Thus, $0 \leq (f+g)^2 \leq M^2,$ so $(f+g)^2$ is bounded.

 

[WWGD]

Just note that fg being bounded by itself does not imply boundedness of f,g: f(x)= $x^2+1 , g(x)= \frac {1}{x^2+1}$