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# Thread: Intermediate value theorem 2

1. If:

f is a real value function continuous over [a,b] ,a<b and f(a)<f(b)

,then prove that for all η,
f(a)<η<f(b) ,there exists an ξ ,such that : f(ξ)=η

Proof.

1)Let S be a set of all xε[a,b] such that :f(x)<η.

Since a belongs to S and b is an upper bound then S has a Supremum

2)Let ξ= SupS

3) Will show that : a) $f(\xi)\geq\eta$ ,b) $f(\xi)\leq\eta$ and hence : $f(\xi)=\eta$

3a) By the definition of supremum, for every positive ε, there is a number x' of S with
$\xi-\epsilon\leq x'\leq\xi$ .
For this x', f(x')<η.
Since f is contiguous at ξ, $f(\xi)\leq\eta$ which is : 3a

3b) Any x greater than ξ is not in S and so $f(x)\geq \eta$.By continuity ,f(ξ) is the limit of f(x) as x tends to ξ through values greater than ξ ,and so :$f(\xi)\geq\eta$

I have no idea how he gets $f(\xi)\geq \eta$ in part 3b and $f(\xi)\leq\eta$ in part 3a

2.

3. Originally Posted by solakis
If:

f is a real value function continuous over [a,b] ,a<b and f(a)<f(b)

,then prove that for all η,
f(a)<η<f(b) ,there exists an ξ ,such that : f(ξ)=η

Proof.

1)Let S be a set of all xε[a,b] such that :f(x)<η.

Since a belongs to S and b is an upper bound then S has a Supremum

2)Let ξ= SupS

3) Will show that : a) $f(\xi)\geq\eta$ ,b) $f(\xi)\leq\eta$ and hence : $f(\xi)=\eta$

3a) By the definition of supremum, for every positive ε, there is a number x' of S with
$\xi-\epsilon\leq x'\leq\xi$ .
For this x', f(x')<η.
Since f is contiguous at ξ, $f(\xi)\leq\eta$ which is : 3a

3b) Any x greater than ξ is not in S and so $f(x)\geq \eta$.By continuity ,f(ξ) is the limit of f(x) as x tends to ξ through values greater than ξ ,and so :$f(\xi)\geq\eta$

I have no idea how he gets $f(\xi)\geq \eta$ in part 3b and $f(\xi)\leq\eta$ in part 3a
3a) $\displaystyle f(\xi) \leq \eta$
Pick $\displaystyle \epsilon = \eta - f(\xi)$ as you should know
f is continuous at ξ that means for any $\displaystyle \epsilon >0$ there exist $\displaystyle \delta >0$
such that whenever $\displaystyle \mid x - \xi \mid < \delta$ we have $\displaystyle \mid f(x) - f(\xi) \mid < \epsilon$
remember that epsilon>0 so if you choose a wrong epsilon the criteria for the continuity will be affected, back to our proof after we choose epsilon sub it
$\displaystyle \mid f(x) - f(\xi) \mid < \eta - f(\xi)$

$\displaystyle -\eta + f(\xi) < f(x) - f(\xi) < \eta - f(\xi)$

$\displaystyle f(x) - f(\xi) < \eta - f(\xi) \Rightarrow f(x) < \eta$ for

$\displaystyle \mid x - \xi \mid < \delta \Rightarrow \xi - \delta < x < \xi + \delta$

but $\displaystyle \xi$ is an upper bound for S, $\displaystyle \xi < \xi +\frac{\delta}{2}$
and
$\displaystyle \xi + \frac{\delta}{2} \in S$ since
$\displaystyle f\left(\xi + \frac{\delta}{2}\right) < \eta$ but $\displaystyle \xi$ is an upper bound for S contradiction!! so $\displaystyle \eta - f(\xi)$ is not positive so still two choices less than or equal zero

Lets see if $\displaystyle f(\xi) > \eta$
f is continuous at $\displaystyle \xi$ , now pick $\displaystyle \epsilon = f(\xi) - \eta$
there exist delta

$\displaystyle \mid f(x) - f(\xi) \mid < f(\xi) - \eta$ for $\displaystyle \mid x - \xi \mid < \delta$

$\displaystyle -f(\xi) + \eta < f(x) - f(\xi)$

$\displaystyle \eta < f(x)$ for $\displaystyle \xi - \delta < x < \xi + \delta$

$\displaystyle \xi - \delta$ is in S the definition of the upper bound which means $\displaystyle f \left(\xi - \delta \right)$ should be less than $\displaystyle \eta$ not larger contradiction so our epsilon is not positive

This idea from Wiki see

Originally Posted by Amer
3a) $\displaystyle f(\xi) \leq \eta$
Pick $\displaystyle \epsilon = \eta - f(\xi)$ as you should know
f is continuous at ξ that means for any $\displaystyle \epsilon >0$ there exist $\displaystyle \delta >0$
such that whenever $\displaystyle \mid x - \xi \mid < \delta$ we have $\displaystyle \mid f(x) - f(\xi) \mid < \epsilon$
remember that epsilon>0 so if you choose a wrong epsilon the criteria for the continuity will be affected, back to our proof after we choose epsilon sub it
$\displaystyle \mid f(x) - f(\xi) \mid < \eta - f(\xi)$

$\displaystyle -\eta + f(\xi) < f(x) - f(\xi) < \eta - f(\xi)$

$\displaystyle f(x) - f(\xi) < \eta - f(\xi) \Rightarrow f(x) < \eta$ for

$\displaystyle \mid x - \xi \mid < \delta \Rightarrow \xi - \delta < x < \xi + \delta$

but $\displaystyle \xi$ is an upper bound for S, $\displaystyle \xi < \xi +\frac{\delta}{2}$
and
$\displaystyle \xi + \frac{\delta}{2} \in S$ since
$\displaystyle f\left(\xi + \frac{\delta}{2}\right) < \eta$ but $\displaystyle \xi$ is an upper bound for S contradiction!! so $\displaystyle \eta - f(\xi)$ is not positive so still two choices less than or equal zero

Lets see if $\displaystyle f(\xi) > \eta$
f is continuous at $\displaystyle \xi$ , now pick $\displaystyle \epsilon = f(\xi) - \eta$
there exist delta

$\displaystyle \mid f(x) - f(\xi) \mid < f(\xi) - \eta$ for $\displaystyle \mid x - \xi \mid < \delta$

$\displaystyle -f(\xi) + \eta < f(x) - f(\xi)$

$\displaystyle \eta < f(x)$ for $\displaystyle \xi - \delta < x < \xi + \delta$

$\displaystyle \xi - \delta$ is in S the definition of the upper bound which means $\displaystyle f \left(\xi - \delta \right)$ should be less than $\displaystyle \eta$ not larger contradiction so our epsilon is not positive

This idea from Wiki see
The proof you proposed is completely different from the proof i wrote in my OP.

Sorry but i am not asking for a proof of the IVT but an explanation of the OP's proof

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