Proofs of Boundedness for Sum and Product of Functions | Homework Question

  • Thread starter Jamin2112
  • Start date
In summary: Exactly as you said. Consider (b). If f and g are founded, then fg is founded. So, |f(x)+g(x)| = |f(x)| + |g(x)| = |f(x)| + |g(x)| = ?. |f(x)| + |g(x)| = ?.This is just a basic algebraic equation. You can solve it for x if you want.No, there is no proof for (c).
  • #1
Jamin2112
986
12

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



?

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.
 
Physics news on Phys.org
  • #2
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.
 
  • #3
Tedjn said:
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
 
  • #4
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.
 
  • #5
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, [itex]|f(x)+ g(x)|\le |f(x)|+|g(x)|< [/itex] what?

[itex]|f(x)g(x)|\le |f(x)||g(x)|< [/itex] what?
 

FAQ: Proofs of Boundedness for Sum and Product of Functions | Homework Question

What is a proof of boundedness for a sum of functions?

A proof of boundedness for a sum of functions involves showing that the sum of two or more functions is always within a certain range or bound. This means that the sum of the functions will never exceed a certain value, regardless of the input values.

How is a proof of boundedness for a product of functions different from a proof for a sum of functions?

A proof of boundedness for a product of functions is different from a proof for a sum of functions because it involves showing that the product of two or more functions is always within a certain range or bound. This means that the product of the functions will never exceed a certain value, regardless of the input values.

What are some common techniques used to prove boundedness for sums and products of functions?

Some common techniques used to prove boundedness for sums and products of functions include using the properties of limits, using the squeeze theorem, and using induction.

Why is it important to prove boundedness for sums and products of functions?

Proving boundedness for sums and products of functions is important because it allows us to understand and analyze the behavior of these functions. It also helps us to determine the range of values that these functions can take, which can be useful in various applications.

Can proofs of boundedness for sums and products of functions be applied to any type of function?

Yes, proofs of boundedness for sums and products of functions can be applied to any type of function as long as the functions involved have well-defined limits and behave within certain bounds.

Similar threads

Back
Top