Proving boundedness of sums and products of functions

[Jamin2112]

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 a proof; if false, provide a counterexample.

a) If f and g are bounded, then f + g is bounded
b) If f and g are founded, then fg is bounded
c) If f+g is bounded, then f and g are bounded
d) If fg is bounded, then f and g are bounded

Homework Equations

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The Attempt at a Solution

"Bounded" just means in the real-numbered set S there is a real number M such that |x|≤M for all x in S.

So, say F is the max for f and G is the max for G.

For example, say f(x)=5-x² and g(x)=6-x². F=5, S=6.

f(x) + g(x) = 11-2x².

Still bounded, of course. But how do I give proofs of all these? Give me an example or two.

 

[Tedjn]

Exactly as you said. Consider (a). If f is bounded by F and g is bounded by G, that means for all x in R, |f(x)| ≤ F and |g(x)| ≤ G. So, |(f+g)(x)| = |f(x)+g(x)| ≤ ? ≤ M. What is M and what goes in the question mark? That is how such a proof would go.

 

[Jamin2112]

Exactly as you said. Consider (a). If f is bounded by F and g is bounded by G, that means for all x in R, |f(x)| ≤ F and |g(x)| ≤ G. So, |(f+g)(x)| = |f(x)+g(x)| ≤ ? ≤ M. What is M and what goes in the question mark? That is how such a proof would go.

M = G + F

Not sure about the question mark

 

[Tedjn]

True, M = G + F, but you cannot just directly conclude |f(x)+g(x)| ≤ G + F. There needs to be an additional step in there, which is the question mark. Remember the definition of F and G.

 

[HallsofIvy]

No, if f(x) is bounded then there exist M such that |f(x)|< M. If g(x) is bounded, there exist N such that |g(x)|< N.

Now, $|f(x)+ g(x)|\le |f(x)|+|g(x)|<$ what?

$|f(x)g(x)|\le |f(x)||g(x)|<$ what?