Answer: Prove Oscillation of Subintervals in [c,d] with η < ω_f (x)

[GridironCPJ]

Q: Suppose that the oscillation ω_f (x) of a function f is smaller than η at each point x of an interval [c,d]. Show that there must be a partition π of [c,d] s.t. the oscillation
ωf([x_(k-1),x_k ])<η
on each member of the partition.

My solution (Rough sketch):

This condition on x is local, so it must be true for a δ-neightborhood of x s.t. ωf(δ(x))<η. Now take a partition s.t. each subinterval [x_(k-1),x_k ]<δ. Thus, each subinterval is less than the δ from the δ-neightborhood of x, so then
ωf([x_(k-1),x_k ])$\leq$ωf(δ(x))<η. QED

Is this logic too sloppy? If so, does anyone have any suggestions as to a more proper way to prove this?

 

[DonAntonio]

Q: Suppose that the oscillation ω_f (x) of a function f is smaller than η at each point x of an interval [c,d]. Show that there must be a partition π of [c,d] s.t. the oscillation
ωf([x_(k-1),x_k ])<η
on each member of the partition.

My solution (Rough sketch):

This condition on x is local, so it must be true for a δ-neightborhood of x s.t. ωf(δ(x))<η. Now take a partition s.t. each subinterval [x_(k-1),x_k ]<δ. Thus, each subinterval is less than the δ from the δ-neightborhood of x, so then
ωf([x_(k-1),x_k ])$\leq$ωf(δ(x))<η. QED

Is this logic too sloppy? If so, does anyone have any suggestions as to a more proper way to prove this?

"The oscillation ω_f (x) of a function f is smaller than η at each point x of an interval [c,d]" means $\lim_{t \to 0^+} \omega_{[x-t,x+t]}f(x)=h<\eta\Longrightarrow \forall \epsilon>0\,\,\exists \delta>0\,\,s.t.$

$0<t<\delta\Longrightarrow \left|\omega_{[x-t, x+t]}f(x)-h\right|<\epsilon$ .

Can you take it from here?

DonAntonio

 

[GridironCPJ]

"The oscillation ω_f (x) of a function f is smaller than η at each point x of an interval [c,d]" means $\lim_{t \to 0^+} \omega_{[x-t,x+t]}f(x)=h<\eta\Longrightarrow \forall \epsilon>0\,\,\exists \delta>0\,\,s.t.$

$0<t<\delta\Longrightarrow \left|\omega_{[x-t, x+t]}f(x)-h\right|<\epsilon$ .

Can you take it from here?

DonAntonio

That's a much nicer definition of what's going on. Now, I can simply just take a partition of [c, d] s.t. each subinterval of the partition is of equal length, specifically [x-t, x+t], which satisfies the definition of being less than η. Correct?

 

[DonAntonio]

That's a much nicer definition of what's going on. Now, I can simply just take a partition of [c, d] s.t. each subinterval of the partition is of equal length, specifically [x-t, x+t], which satisfies the definition of being less than η. Correct?

Well, no, since "t" depends on the particular $x\in [c,d]$ are we working with -- and thus it' would have been wiser

to denote it by $t_x$ --, but then you can argue as follows:

Since clearly our original interval $[c,d]\subset \cup_{x\in [c,d]}(x-t_x,x+t_x)$ and it is a compact set in the

real line, there exists a finite number of points...etc.

DonAntonio

 

[GridironCPJ]

Well, no, since "t" depends on the particular $x\in [c,d]$ are we working with -- and thus it' would have been wiser

to denote it by $t_x$ --, but then you can argue as follows:

Since clearly our original interval $[c,d]\subset \cup_{x\in [c,d]}(x-t_x,x+t_x)$ and it is a compact set in the

real line, there exists a finite number of points...etc.

DonAntonio

You lost me on the finite number of points part. I appologize, as my compactness knowledge is quite scarse.

 

[DonAntonio]

You lost me on the finite number of points part. I appologize, as my compactness knowledge is quite scarse.

Well, if you haven't yet studied compact sets I, for one, cannot help you. The continuation of my idea is:

By compactness of [c,d] there exist a finite number of points $x_1,...,x_n\,\,s.t.\,\,[c,d]\subset \cup_{i=1}^n (x_i-t_{x_i},x_i+t_{x_i})$ , so now we can choose

$t:=\max_i\{t_{x_i}\}$ and now yes: end the proof as you wanted before.

DonAntonio

 

[GridironCPJ]

Well, if you haven't yet studied compact sets I, for one, cannot help you. The continuation of my idea is:

By compactness of [c,d] there exist a finite number of points $x_1,...,x_n\,\,s.t.\,\,[c,d]\subset \cup_{i=1}^n (x_i-t_{x_i},x_i+t_{x_i})$ , so now we can choose

$t:=\max_i\{t_{x_i}\}$ and now yes: end the proof as you wanted before.

DonAntonio

I've been briefly introduced to a few compactness arguments. I see what you mean now, thank you for your help.