MHB Show that there is a z such that f(z)=0

[evinda]

Hey! :)
I am given the following exercise:
$f:[a,b] \to \mathbb{R}$ continuous and $\forall x$ there is a $y$ such that $|f(y)| < \frac{|f(x)|}{2}$ .Show that there is a $z$ such that $f(z)=0$.
That's what I have tried:
Suppose that there is not a $z$ such that $f(z)=0$.Then $f(x)>0 , \forall x$ or $f(x)<0 , \forall x$.
If $f(x)>0 , \forall x$ then from the relation $|f(y)| <\frac{|f(x)|}{2}$ for $x=y$ we find $\frac{-f(x)}{2}>0$,that can't be true,as we have supposed that $f$ is positive $\forall$ x.
Now,suppose that $f(x)<0 , \forall x$,from the relation $|f(y)| < \frac{|f(x)|}{2}$ for $x=y$ we find $|f(x)|<0$,that also can't be true.

So,we conclude that it can't be true that $f$ doesn't change sign,so there has to be a $z$ such that $f(z)=0$.

Could you tell me if it is right?

 

[Opalg]

Hey! :)
I am given the following exercise:
$f:[a,b] \to \mathbb{R}$ continuous and $\forall x$ there is a $y$ such that $|f(y)| < \frac{|f(x)|}{2}$ .Show that there is a $z$ such that $f(z)=0$.
That's what I have tried:
Suppose that there is not a $z$ such that $f(z)=0$.Then $f(x)>0 , \forall x$ or $f(x)<0 , \forall x$.
If $f(x)>0 , \forall x$ then from the relation $|f(y)| <\frac{|f(x)|}{2}$ for $x=y$ we find $\frac{-f(x)}{2}>0$,that can't be true,as we have supposed that $f$ is positive $\forall$ x.
Now,suppose that $f(x)<0 , \forall x$,from the relation $|f(y)| < \frac{|f(x)|}{2}$ for $x=y$ we find $|f(x)|<0$,that also can't be true.

So,we conclude that it can't be true that $f$ doesn't change sign,so there has to be a $z$ such that $f(z)=0$.

Could you tell me if it is right?

No, that argument does not work: you are not permitted to take $y=x$. You are given that for each $x$ in $[a,b]$ there exists a $y$ in $[a,b]$ such that $|f(y)| < |f(x)|/2$, but there is no reason to think that $y=x$.

Start with an arbitrary point $x_1$ in $[a,b]$, and use the given condition to construct inductively a sequence $(x_n)$ such that $|f(x_{n+1})| < |f(x_n)|/2$ for each $n$. Then apply a theorem which says that this sequence must have a convergent subsequence.

 

[evinda]

No, that argument does not work: you are not permitted to take $y=x$. You are given that for each $x$ in $[a,b]$ there exists a $y$ in $[a,b]$ such that $|f(y)| < |f(x)|/2$, but there is no reason to think that $y=x$.

Start with an arbitrary point $x_1$ in $[a,b]$, and use the given condition to construct inductively a sequence $(x_n)$ such that $|f(x_{n+1})| < |f(x_n)|/2$ for each $n$. Then apply a theorem which says that this sequence must have a convergent subsequence.

Which theorem could I use for example?

 

[Opalg]

Which theorem could I use for example?

Bolzano–Weierstrass.

 

[ThePerfectHacker]

Here is another idea.

Consider, $|f|:[a,b]\to \mathbb{R}$ this function is continous.
By EVT it has a minimal value $m$.

If $m=0$ the proof of your claim is complete.

If $m>0$ choose $x$ such that $|f(x)| = m$. By hypothesis of your problem there is a $y$ such that $|f(y)| < \tfrac{1}{2}|f(x)|$. This leads to a contradiction ...

 

[evinda]

Bolzano–Weierstrass.

Ok!Thanks a lot! :)

- - - Updated - - -

Here is another idea.

Consider, $|f|:[a,b]\to \mathbb{R}$ this function is continous.
By EVT it has a minimal value $m$.

If $m=0$ the proof of your claim is complete.

If $m>0$ choose $x$ such that $|f(x)| = m$. By hypothesis of your problem there is a $y$ such that $|f(y)| < \tfrac{1}{2}|f(x)|$. This leads to a contradiction ...

Ok..Thank you!