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Intermediate value theorem 2

solakis

Active member
Dec 9, 2012
322
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
 

Amer

Active member
Mar 1, 2012
275
Re: intermediate value theorem 2

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) [tex] f(\xi) \leq \eta [/tex]
Pick [tex] \epsilon = \eta - f(\xi) [/tex] as you should know
f is continuous at ξ that means for any [tex] \epsilon >0 [/tex] there exist [tex] \delta >0 [/tex]
such that whenever [tex] \mid x - \xi \mid < \delta [/tex] we have [tex] \mid f(x) - f(\xi) \mid < \epsilon [/tex]
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
[tex] \mid f(x) - f(\xi) \mid < \eta - f(\xi) [/tex]

[tex] -\eta + f(\xi) < f(x) - f(\xi) < \eta - f(\xi) [/tex]

[tex] f(x) - f(\xi) < \eta - f(\xi) \Rightarrow f(x) < \eta [/tex] for

[tex] \mid x - \xi \mid < \delta \Rightarrow \xi - \delta < x < \xi + \delta [/tex]

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

Lets see if [tex] f(\xi) > \eta [/tex]
f is continuous at [tex] \xi [/tex] , now pick [tex] \epsilon = f(\xi) - \eta [/tex]
there exist delta

[tex] \mid f(x) - f(\xi) \mid < f(\xi) - \eta [/tex] for [tex] \mid x - \xi \mid < \delta [/tex]

[tex] -f(\xi) + \eta < f(x) - f(\xi) [/tex]

[tex] \eta < f(x) [/tex] for [tex] \xi - \delta < x < \xi + \delta [/tex]

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

This idea from Wiki see this
 
Last edited:

solakis

Active member
Dec 9, 2012
322
Re: intermediate value theorem 2

3a) [tex] f(\xi) \leq \eta [/tex]
Pick [tex] \epsilon = \eta - f(\xi) [/tex] as you should know
f is continuous at ξ that means for any [tex] \epsilon >0 [/tex] there exist [tex] \delta >0 [/tex]
such that whenever [tex] \mid x - \xi \mid < \delta [/tex] we have [tex] \mid f(x) - f(\xi) \mid < \epsilon [/tex]
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
[tex] \mid f(x) - f(\xi) \mid < \eta - f(\xi) [/tex]

[tex] -\eta + f(\xi) < f(x) - f(\xi) < \eta - f(\xi) [/tex]

[tex] f(x) - f(\xi) < \eta - f(\xi) \Rightarrow f(x) < \eta [/tex] for

[tex] \mid x - \xi \mid < \delta \Rightarrow \xi - \delta < x < \xi + \delta [/tex]

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

Lets see if [tex] f(\xi) > \eta [/tex]
f is continuous at [tex] \xi [/tex] , now pick [tex] \epsilon = f(\xi) - \eta [/tex]
there exist delta

[tex] \mid f(x) - f(\xi) \mid < f(\xi) - \eta [/tex] for [tex] \mid x - \xi \mid < \delta [/tex]

[tex] -f(\xi) + \eta < f(x) - f(\xi) [/tex]

[tex] \eta < f(x) [/tex] for [tex] \xi - \delta < x < \xi + \delta [/tex]

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

This idea from Wiki see this
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