Question

Let D be a nonempty set and suppose that f : D → R and g : D → R. Define the function f +g : D → R by (f + g)(x) = f(x) + g(x)

If f(D) and g(D) are bounded above, then prove that (f + g)(D) is bounded above and
sup[(f+g)(D)]≤supf(D)+supg(D)

Answer 1

The sum function of f and g, when both are bounded above, is itself bounded above. Their sum function's upper bound is at most the sum of the individual functions' upper bounds.

Step-by-step explanation:

Mathematics concepts often build upon one another to demonstrate more complex theorems. In this instance, we are looking at the properties of bounded functions and how they interact when combined. When given two real-valued functions f and g on a nonempty set D, both of which are bounded above, what can be said about the bounded nature of their sum, (f + g)?

Now, considering the sum function (f + g)(x) = f(x) + g(x), for every x in D, it follows that (f + g)(x) ≤ M + N. This inequality holds because the addition of two numbers respects the commutative property (A + B = B + A), as mentioned in the associative theory of addition.

Therefore, we can conclude that the set (f+g)(D) is bounded above and the supremum of (f+g)(D) is at most M + N, since sup(f(D)) ≤ M and sup(g(D)) ≤ N. Thus, sup(f+g)(D) ≤ sup(f(D)) + sup(g(D)), establishing our desired proof that the sum of two functions bounded above is itself bounded above.