Let D be a non-empty subset and suppose that f: D map to R (set of real number).

Question

Let D be a non-empty subset and suppose that f: D map to R (set of real number). Define the function f + g: D map to R by (f+g)(x) = f(x) + g(x). a) If f(D) and g(D) are bounded above, then prove that (f + g)(D) is bounded above and sup(f + g)(D) <= sup f(D) + sup g(D).
b) Find an example to show that a strict inequality in part (a) may occur.
Let D be a non-empty subset and suppose that f: D map to R (set of real number). Define the function f + g: D map to R by (f+g)(x) = f(x) + g(x). a) If f(D) and g(D) are bounded above, then prove that (f + g)(D) is bounded above and sup(f + g)(D) <= sup f(D) + sup g(D).
b) Find an example to show that a strict inequality in part (a) may occur.

Explanation / Answer

a) Since f is bounded above let it be by M, that is,    f(x)<=M for every x in D.    Similarly let N be the upper bound for g i.e., g(x)<=N for every x in D.    Now for x in D, (f+g)(x)=f(x)+g(x) <= M+N    Since x is arbitrary, we have (f+g)(x)<=M+N for every x in D. That is M+N is an upper bound for f+g and hence f+g is bounded above. Now let A=sup(f), B=sup(g). So A is an upper bound for f and B is for g. So by above argument, we have (f+g)(x)<=A+B for every x in D. Since A+B is an upper bound for f+g and supremum of any function is less than or equal to upper bound, we have sup(f+g)<=A+B. Now substitute back A and B. Thus we get sup(f+g)<=sup(f) + sup(g). Hence we are done! b) Let D=[0,1] f: D-->R be defined by f(x)=x. So sup(f)=1 on D. g:D-->R be defined by g(x)=-x. So sup(g)=0 on D. Now, (f+g)(x)= x-x=0 for every x in D. So Sup(f+g)=0.    Hence we have sup(f+g) =0 < 1+0= sup(f) + sup(g).

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.