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.

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.

## Comments

## Leave a comment