# Thread: inequality of supremums

1. ## inequality of supremums

Hi all,

If $f,g$ and $h$ are 3 continous functions on a set A,
that verify:

$\sup({f(t),t\in A})<\sup({h(t),t\in A})+\sup({g(t),t\in A})$

(beware the $<$ and not $\leq$)

can I say that there exists $x\in A$ such that:
$f(x)

thank you

Deubelte

2. Let $A=\mathbb{R}$ now define $f(x)=4,~g(x)=1,~\&~h(x)=2$.
Is that statement still true.
You must have left something out of the statement of the question.

3. Originally Posted by deubelte
If $f,g$ and $h$ are 3 continuous functions on a set A,
that verify:

$\sup({f(t),t\in A})<\sup({h(t),t\in A})+\sup({g(t),t\in A})$

(beware the $<$ and not $\leq$)

can I say that there exists $x\in A$ such that:
$f(x)
No. Take $A = \mathbb{R}$ and define $g(x) = \frac1{(x-10)^2+1}$, $h(x) = \frac1{(x+10)^2+1}$ and $f(x) = 1.5$ (constant).

Then $\sup\{g(x)\} = 1$ (attained when x=10) and $\sup\{h(x)\} = 1$ (attained when x=–10), so $1.5 = \sup\{f(x)\} < \sup\{g(x)\}+\sup\{h(x\} = 2$.

But $x\geqslant0\;\Rightarrow\; h(x)\leqslant0.01$, and $x\leqslant0\;\Rightarrow\; g(x)\leqslant0.01$. So $g(x)+h(x)$ is always less than 1.01, which is less than $f(x)$.

4. I have an other question. supposing we have two continuous functions
$f$ and $g$
such that:
$\sup(g(t),t\in A)<\sup(h(t),t\in A)$

can I say that:

there exists x in A such that:
1) $g(x)\leq h(x)$

2) $forall\epsilon$ there exists x such that:
$g(x)-\epsilon

thx

5. Originally Posted by deubelte
I have an other question. supposing we have two continuous functions
$f$ and $g$
such that:
$\sup(g(t),t\in A)<\sup(h(t),t\in A)$

can I say that:

there exists x in A such that:
1) $g(x)\leq h(x)$

2) $\forall\epsilon$ there exists x such that:
$g(x)-\epsilon
Choose k with $\sup\{g(t),t\in A\}. Then there exists x in A such that $h(x)>k$. But $g(x), and therefore $g(x).

6. Ok, so I guess that my second statement is also true ??

Merci l'Anglais.