# Sums and products of bounded functions

I have been on this one for hours, cant figure out how to write this in the proper format/wording.

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

1. If $f$ and $g$ are bounded, then $f + g$ is bounded.
2. If $f$ and $g$ are bounded, then $fg$ is bounded.
3. If both $f + g$ and $fg$ are bounded, then $f$ and $g$ are bounded.

Any help is appreciated, thanks :)

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a) is trivial. For example, let the upper bound of $f$ be $A$ and the upper bound of $g$ be $B$, so the upper bound of $f+g$ is less then $A+B$. same for b) – picakhu Jan 24 '11 at 4:40
For 2) it's worth clarifying whether the question is about 'bounded in absolute value' or just 'bounded above'; the answer for the latter is different than the answer (given below) for the former, in particular if both $f$ and $g$ can go hugely negative. – Steven Stadnicki Jan 24 '11 at 20:36

a) That $f$ is bounded means that $|f(x)| \leq C$ for all $x$. Similarly $|g(x)| \leq D$ for all $x$. Therefore the triangle inequality gives $|f(x) + g(x)| \leq |f(x)| + |g(x)| \leq C + D$ for all $x$ and hence $f+g$ is bounded by $C + D$.

b) is similar but easier

c) $f^2 + g^2 = |f^2 + g^2| = |(f + g)^{2} - 2fg| \leq |f + g|^2 + 2|fg|$ and the last term is bounded by assumption and a) and b).

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Nice, but im a little confused about (c), i understand what you did but cant draw conclusions to it. Is it possible to explain a little more the last part? Thanks – 1337holiday Jan 24 '11 at 14:15
@1337holiday: we have $0 \leq |f(x)|^2 = f^2 \leq f^2 + g^2$, so $|f|^2$ is bounded. If $|f(x)| \geq 1$ then $|f(x)| \leq |f(x)|^{2}$ otherwise $|f(x)| \leq 1$. So together this gives $|f(x)| \leq \max{\{|f(x)|^2, 1\}} \leq \max{\{1,|f(x) + g(x)|^2 + 2|f(x)g(x)|\}} \leq C'$ for all $x$. – t.b. Jan 24 '11 at 15:12

Remember the definition of bounded: A function $h\colon\mathbb{R}\to\mathbb{R}$ is bounded if and only if there exist number $M,N\gt 0$ such that for all $r\in\mathbb{R}$, $-N \leq f(r) \leq M$.

Intuitively, the graph of $f$ goes neither "too high" nor "too low".

So, suppose that $f$ and $g$ are both bounded. That means that there the values of $f$ don't get bigger than some $M_1\gt 0$, and the values of $g$ don't get bigger than some $M_2\gt 0$. When you take $(f+g)(r) = f(r)+g(r)$, how big can the number $f(r)+g(r)$ be, given that $f(r)\leq M_1$ and $g(r)\leq M_2$?

How about discussing how small they can be, given that you know there are numbers $N_1$ and $N_2$ such that $f(r)\geq N_1$ and $g(r)\geq N_2$?

How about the product? That's a bit harder because of the signs, so try working with $|fg|$, $|f|$, and $|g|$. Notice that if $-N\leq f(r)\leq M$, then $0\leq f(r) \leq \max\{M,N\}$ (verify this!).

Number $3$ is a bit trickier... first question: can the sum of two functions, each of which is not bounded, be bounded? Yes: consider for instance $f(x) = x$, which is not bounded, and $g(x) = 1-x$, which is also not bounded. What is $f+g$? Of course, this example does not satisfy that $fg$ is bounded (here, $fg(x) = x-x^2$, which is unbounded). And it's easy to come up with unbounded examples with their product bounded (just make sure that whenever $f$ gets out of control, $g$ "cancels it out").

Perhaps you may want to check that if $ff$ (the product of $f$ with itself) is bounded, then so is $f$, and then consider Theo Buehler's answer.

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Great answer, though im a little confused about the absolute value of the products. How did you come to assume you can do that and what does the max(M, N) mean? Thanks a lot – 1337holiday Jan 24 '11 at 14:12
@1337holiday: Although I used different numbers for the upper and lower bound, you don't have to: $\max\{M,N\}$ means "the largest of $M$ and $N$". It's not hard to check that if $-N\leq f(r)\leq M$ for all $r$, you can replace $M$ by anything larger than $M$, and $N$ by anything larger than $N$, and the inequalities still hold; so you can just take $S=\max\{M,N\}$, and then you will have $-S\leq f(r)\leq S$. This is exactly equivalent to $|f(r)|\leq S$. In essence, you can define bounded using the absolute value instead, and it comes out to be the same. – Arturo Magidin Jan 24 '11 at 14:36

For a and b, you have that $f(x)\le f_{max}$ and similarly for g. What does that tell you about $f(x)+g(x)$ and $f(x)g(x)$? c is harder-it is easy for one of the conditions like $f(x)=-g(x)$ for the sum and $f(x)=0$ for the product, but for both together?

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