Two Continuity Questions (1 Viewer)

mreditor16 · Apr 19, 2015

For Q74, I have done a) and b), but need help with c).

And for Q77, I have done a), but need help with b).

Thanks heaps

InteGrand · Apr 19, 2015

mreditor16 said:
For Q74, I have done a) and b), but need help with c).

And for Q77, I have done a), but need help with b).

Thanks heaps

Proof outline for 74 (c):

InteGrand said:
PROOF OUTLINE:

Let $f(\xi)=\Xi >0$ .

By definition of limits to infinity, there exist $M>N$ such that $|f(x)-0|<\Xi$ whenever $x>M$ and whenever $x<N$ . So $\xi \in [N,M]$ , and:

(1) $f(x)<\Xi \text{ }\forall x \in (-\infty, N)$

(2) $f(x)<\Xi \text{ }\forall x\in(M,\infty)$ .

By the extreme value theorem (since f is continuous), $f$ attains a maximum value $K$ on $[N,M]$ .

Since $f(x)=\Xi >0$ when $x= \xi \in [N,M]$ , this maximum $K$ satisfies $K\geq \Xi$ . (3)

(1), (2) and (3) imply that f attains a maximum value on $\mathbb{R}$ , namely $f_\text{max.}=K$ .

mreditor16 · Apr 19, 2015

Oops, Integrand, yes I have asked that Q before :/ Anyone willing to help with 77 b) though?

VBN2470 · Apr 19, 2015

I hope this helps

Sy123 said:
I did not use the hint, I did this instead:

$\\ $a) Take some$ \ k \ $in$ \ [0,1] \ $since$ \ f \ $is continuous on$ \ [0,1] \ $then is continuous on$ \ [0,k] \ $and$ \ [k, 1] \\ $By the intermediate value theorem then$ \ f \ $takes every value in$ \ [f(0) , f(k)] \ $by the first interval and takes every value in$ \ [f(k) , f(1)] \ $by the second interval$ \ \\ $Since$ \ f(0) = f(1) \ $then it takes every value from$ \ [0,1] \ $twice$ \\ $Hence for some arbitrary$ \ m \ $in$ \ [0,1] \ $there is$ \ n \ $in$ \ [0,1] \ $so that$ \ \ f(m) = f(j) \\ $Since this holds for any$ \ m \ $in$ \ [0,1] \ $then pick$ \ a \ $such that$ \ j = a + \frac{1}{2} \\ $hence$ \ f(a) = f\left( a + \frac{1}{2} \right) \ $and since$ \ \frac{1}{2} \leq a + \frac{1}{2} \leq 1 \ $then$ \ a \ $must be in$ \ [0 , \frac{1}{2}]$

$\\ $b) Using the same method we pick some$ \ a \ $such that$ \ j = a + \frac{1}{n} \\ $and since$ \ j \leq 1 \ $then$ \ a \ $is in$ \ [0, \frac{n-1}{n} ]$

Sy123 said:
But using the hint:

$g(x) = f\left( x + \frac{1}{2} \right) - f(x)$

$\\g\left(\frac{1}{2} \right) + g(0) = f\left(1\right) - f \left(\frac{1}{2}\right) + f\left(\frac{1}{2} \right) - f(0) = 0 \\ $Hence$ \ g\left(\frac{1}{2} \right) = - g(0)$

$\\ g \ $is continuous over$ \ [0, 1/2] \ $and since$ \ g(0) < 0 < g(1/2) \ $or$ \ g(1/2) < 0 < g(0) \ $there must be some$ \ x \ $so that$ \ g(x) = 0 \\ $Hence, some$ \ x= a \ g(a) = 0 \Rightarrow \ f(a+1/2) = f(a)$

.

mreditor16 · Apr 19, 2015

You asked the same Q, last year. Thanks VBN

InteGrand · Apr 19, 2015

VBN2470 said:
I hope this helps

.

How did we conclude that there exists a such that $f(a) = f\left(a + \frac{1}{n}\right)$ ? We said there are at least two x-values that map to the same y-value, but how did we conclude that we can make these x-values differ by $\frac{1}{n}$ ?

VBN2470 · Apr 19, 2015

Sy123 concluded it for me

Sy123 said:
Similarly:

$g(x) = f(x + 1/n) - f(x)$

$g(0) = f(1/n) - f(0)$

$g(1/n) = f(2/n) - f(1/n)$

$g(2/n) = f(3/n) - f(2/n)$

.
.
.

$g((n-1)/n) = f(1) - f(0)$

Add them side by side:

$(1): \ \ \ \sum_{k=0}^{n-1} g\left(\frac{k}{n} \right) = 0 \ \ \ $(telescoping and cancellation of$ \ f(1) \ $and$ \ f(0) $)$$

$\\ $Assume that$ \ g(x) \neq 0 \ $then$ \ g\left(\frac{k}{n} \right) > 0 \ $or$ \ g\left(\frac{k}{n} \right) < 0 \ \ $for all$ \k \\ $however this contradicts$ \ (1) \ $Therefore a contradiction arises meaning there is at least one$ \ k \ $so that$ \ g\left(\frac{k}{n} \right) < 0 \\ $Since$ \ g \ $is continuous, by IVT it follows that there must be a root to$ \ g \ $(similar logic to first part$ \\ $Therefore there is at least some$ \ a \ $so that$ \ g(a) = 0 \Rightarrow \ f(a + 1 /n) = f(a)$

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Two Continuity Questions (1 Viewer)

mreditor16

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