Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to find a cauchy sequence in $C[0,1]$ that converges under $\|\cdot\|_2$ to a limit which isn't continuous.

Any ideas?

share|improve this question
    
Instead of "converges under $\|\cdot\|_2$ to a limit which isn't continuous," you probably want "does not converge under $\|\cdot\|_2$ to a limit which is continuous." –  Jonas Meyer May 11 '12 at 19:19

3 Answers 3

up vote 5 down vote accepted

Let $$f_n(x) = \left\{ \begin{array}{rl} 0 & \text{if } x \leq 1/2,\\ 1 & \text{if } x \geq 1/2+1/n,\\ n(x-1/2) & \text{if } 1/2\leq x\leq 1/2+1/n. \end{array} \right.$$

share|improve this answer
    
Does this sequence work for showing C[0,1] is not Banach for any $p<\infty$? I can see it works for $p=1$ & $p=2$. –  rk101 May 11 '12 at 13:23
    
Yes, I think so. –  Michael Greinecker May 11 '12 at 13:25
    
Yes it does actually, I have just checked it. –  rk101 May 11 '12 at 13:38
    
A quick question: Why $\|\cdot\|_1 \leq \|\cdot\|_2 \leq \|\cdot\|_{\infty}$ ? –  rk101 May 11 '12 at 13:56
    
I'm not completely ure. I think for simple functions you can use the corresponding reult for finite dimensioal spaces and then take limits. –  Michael Greinecker May 11 '12 at 14:23

Another one. Think of a discontinuous (bounded measurable) function. Say: $f(x) = 0$ on $[0,1/2]$ and $f(x) = 1$ on $(1/2,1]$. Write down its Fourier series. The partial sums are continuous. They converge in $L_2$ norm (to $f$) but do not converge to any element of $C[0,1]$.

share|improve this answer
1  
Note that $f$ should be not only discontinuous, but not a.e. equal to any continuous function. E.g. $f(1/2) = 1$, $f(x) = 0$ otherwise is discontinuous, but its Fourier series converges uniformly to the continuous function 0. –  Nate Eldredge May 11 '12 at 13:10

Think of the function that is $0$ on the interval $\left[0,1-\frac{1}{n}\right]$ and then is $y=n(x-1)+1$ on the remainder of the interval. As $n$ increases the "spike" gets sharper. The limit function is not continuous at $x=1$.

share|improve this answer
    
However, the limit function in the $L^2$ norm is $0$, which is in $C[0,1]$. These aren't the pointwise limits you're looking for... –  robjohn May 11 '12 at 15:33

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.