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Answer on Real Analysis Question for Junel

Question #4229
Let X be a nonempty set , and let f and g defined on X and have bounded ranges in R. show that
Sup{f(x) + g(x):x element X}<= Sup {f(x):x element X} + Sup {g(x):x element X}
and that,
inf{f(x):x element X} + inf {g(x): x element X} <= inf { f(x) + g(x): x element X}.
also give example to show that each of these inequalities can be either equalities or strict inequalities.
Expert's answer
Proof of (*).
for simplicity denote
& sup(f+g) := sup{f(x) + g(x):x
element X}
& sup(f) := sup{f(x):x element X}
& sup(g) := sup{g(x):x
element X}
& inf(f+g) := inf{f(x) + g(x):x element X}
& inf(f) :=
inf{f(x):x element X}
& inf(g) := inf{g(x):x element X}

We have to
show that
sup(f+g) <= sup(f) + sup(g)

For any x from X we have
that
f(x) <= sup f,&
g(x) <= sup g,&
whence

f(x)+g(x) <= sup(f) + sup(g).
Therefore
sup(f+g) <= sup(f) +
sup(g).



The inequality (**) follows from (8) and the fact that

(***) sup(f) = -inf(-f).

Indeed,
&
& inf(f+g)& = -
sup(-f-g) by (*)
gt;= - sup(-f) -sup(-g) by
(***)
=& inf(f) + inf(g) by (***)


Let us give an
example when (*) and (**) are equalities:
Let
X = [-1,1],
f(x) =
x^2,
g(x) = x^4
f(x)+g(x) = x^2 + x^4

Then
sup(f) =
sup(g)=1
inf(f) = inf(g)=0
sup(f+g)= 2 = sup(f) + sup(g)
inf(f+g)= 0 =
inf(f) + inf(g)
So in this example (*) and (**) are
equalities.


Let us give an example when (*) and (**) are strict
inequalities:
Let
X = [-1,1],
f(x) = x^2,
g(x) =
-x^2
f(x)+g(x) = x^2-x^2=0

Then
sup(f) = 1 inf(f) = 0
sup(g)
= 0 inf(g) = -1
sup(f+g)= 0 < sup(f) + sup(g) = 1+0=1
inf(f+g)= 0
> inf(f) + inf(g) = 0-1=-1

So in this example (*) and (**) are
inequalities.

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