Essential Supremum Problem: Measurable Positive Functions

  • Thread starter Thread starter johnson123
  • Start date Start date
  • Tags Tags
    Supremum
johnson123
Messages
16
Reaction score
0
Problem: Show an example of a sequence of measurable positive functions on (0,1) so that
\left\|\underline{lim} f_{n}\right\| < \underline{lim}\left\|f_{n}\right\| for n\rightarrow\infty

My work: I think its just the indicator function I_{[n,n+1]}

Since \left\|\underline{lim} I_{[n,n+1]}\right\|= 0 < \underline{lim}\left\|I_{[n,n+1]}\right\| =1

For some reason I do not feel to confident in my answer, so any comments are welcome.
 
Physics news on Phys.org
correction

Problem: Show an example of a sequence of measurable positive functions on (0,1) so that
\left\|\underline{lim} f_{n}\right\|_{\infty} < \underline{lim}\left\|f_{n}\right\|_{\infty} for n\rightarrow\infty

My work: I think its just the indicator function I_{[n,n+1]}

Since \left\|\underline{lim} I_{[n,n+1]}\right\|_{\infty}= 0 < \underline{lim}\left\|I_{[n,n+1]}\right\|_{\infty} =1

For some reason I do not feel to confident in my answer, so any comments are welcome.
 
That's pretty good. But the domain of the indicator functions isn't (0,1). Can you build a very similar example using functions defined only on (0,1)?
 
Thanks for the response Dick.

If f_{n}=I_{(\frac{n-1}{n},1)}, then \left\|\underline{lim} I_{(\frac{n-1}{n},1)}\right\|_{\infty}= 0 < \underline{lim}\left\|I_{(\frac{n-1}{n},1)}\right\|_{\infty} =1

Please correct me if I am wrong.
 
Sure. That's fine. I was thinking of I_(0,1/n), but you can put stuff on the other side of the interval as well.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top