Prove bounded function

Question

if f+g is bounded, then f and g are bounded

Counter-example:

if $f(x) = (x-2)$ and $g(x) = (-x+3) |f(x) + g(x)| = 1$. How, do i prove that f and g are bounded or not? Also |f(x) +g(x)| bounded?

Answer 1

It's just the triangle inequality. If $|f(x)| \le M$ for all $x$ and $|g(x)| \le K$ for all $x$, then $|f(x) + g(x)| \le |f(x)| + |g(x)| \le K+M < \infty$.

In your counterexample, $f$ and $g$ aren't bounded functions on $\mathbb{R}$. So what you've deduced is that the set of functions which aren't bounded is not closed under addition.

Edit: I originally read the question (and it may have been posed as such, I don't know whether it was edited or not) as proving $f+g$ is bounded if $f, g$ are bounded. However, it is not true that $f+g$ being bounded implies that $f$ and $g$ are bounded. You provided an example yourself, and there's an even more trivial one: take $f(x) = x, g(x) =-x$. Then neither $f$ nor $g$ is bounded -- do you see why?

$f(x) =x$ gets arbitrarily large. We can't ever say there's a finite number $M$ such that $x \le M$ for all $x$. Same with $g(x) =-x$. But $f(x) +g(x) = x -x =0$ for all $x$. You'd agree that a constant function is bounded, surely. So here we have shown that $f+g$ being bounded doesn't imply that $f, g$ are bounded. Hope this helps!

Answer 2

David Bowman actually proved the converse (If F and G are bounded, then F+G is bounded). What we want to show is that If (F+G)(x) is bounded, then F and G are bounded. You are correct that this is false.

Your counterexample is on the right track in that we know |F(x) + G(x)| $\le$ 1, so we know that (F+G)(x) is bounded, so we just want to show the F(x) and G(x) are not bounded. (This is obvious, right? They're lines!)

To show F(x) and G(x) are not bounded, we want to show that for all m,n $\in$ R, there exists an x such that F(x) > m or G(x) > n . Since the original statement claims that both of them will be bounded, we need only to show that one of them isn't.

So let's use F(x)=x-2. Since "bounded" implies that f(x) is bounded above and below, we need to show that one of those doesn't hold. Traditionally, this means showing that F(x) isn't bounded above.

So let's assume F(x) $\le$ M where M $\le$ 0. Choose x=3, so we have F(x) = 3-2=1 > 0 $\ge$ M. Contradiction.

Now Assume F(x) $\le$ M where M > 0. Choose x= M+3, so we have F(x) = M+3 -2 = M+1 > M (M is positive). Another contradiction.

So for any M you choose, F(x) is not bounded above, so it will not be bounded.