| Thread Closed |
pointwise => uniform convergence when? |
Share Thread | Thread Tools |
| Oct4-08, 02:21 PM | #1 |
|
|
pointwise => uniform convergence when?
If we have that:
[tex]f_T \left( x \right) \to f\left( x \right)[/tex] for each x (pointwise convergence) and also that: [tex]f_T[/tex] and [tex]f[/tex] are bounded or/and uniformly continuous functions then can we show that there is also uniform convergence? If no why not? Can you show it with an example? In general, are there any cases where pointwise convergence implies uniform convergence? I can't find any proof on that. Can you guide me on this? Thanks in advance |
| Oct4-08, 02:49 PM | #2 |
|
|
Having functions bounded doesn't help with uniform convergence at least.
It is not clear to me what you mean by uniform continuity. Do you mean that each [itex]f_T[/itex], with fixed T, is uniformly continuous so that the choice of [itex]\delta[/itex], with given [itex]\epsilon[/itex], doesn't depend on the point [itex]x[/itex]? If so, it doesn't help with the uniform convergence. If you instead meant that the collection of all [itex]f_T[/itex] is uniformly continuous so that the choice of [itex]\delta[/itex] doesn't depend on T, then I'm not sure right now. |
| Oct4-08, 02:55 PM | #3 |
|
|
I actually meant the first thing you said.
So, there is no known way where pointwise implies uniform convergence? |
| Oct4-08, 03:10 PM | #4 |
|
|
pointwise => uniform convergence when?
First consider this example:
[tex] f_N:[0,1]\to\mathbb{R}, \quad f_N(x) = \left\{\begin{array}{ll} 1,\quad &\frac{1}{N^2} \leq x \leq \frac{1}{N}\\ 0,\quad &0\leq x < \frac{1}{N^2}\;\textrm{or}\;\frac{1}{N} < x \leq 1\\ \end{array}\right. [/tex] Now [tex]\lim_{N\to \infty}f_N(x)\to 0[/tex] for all [tex]x\in [0,1][/tex], but [tex]f_N[/tex] do not converge to zero uniformly. This was not an example with continuous functions, but it is easy to see that you can smoothen these functions a little bit so that each of them alone becomes uniformly continuous, and still the main effect remains the same. This way you get a sequence of continuous, and uniformly bounded functions, which converge towards a continuous function, but still the convergence is not uniform. |
| Oct4-08, 03:31 PM | #5 |
|
|
It could be that this kind of claim is right:
Let [tex]f_N:X\to Y[/tex] be a sequence of continuous functions between two metric spaces, of which [tex]X[/tex] is compact. If [tex]f_N\to f[/tex] pointwisely, and if the sequence is uniformly continuous so that for all [tex]\epsilon > 0[/tex] there exists a [tex]\delta > 0 [/tex] such that [tex] f_N(B(x,\delta))\subset B(f_N(x),\epsilon)\quad\forall x\in X,\; \forall N\in\mathbb{N}, [/tex] then also [tex]f[/tex] is continuous, and [tex]f_N\to f[/tex] uniformly. (If that's not right, then it could be that something similar looking is still right) |
| Oct5-08, 04:24 PM | #6 |
|
|
Thanks again |
| Thread Closed |
| Thread Tools | |
Similar Threads for: pointwise => uniform convergence when?
|
||||
| Thread | Forum | Replies | ||
| Pointwise vs. Uniform Convergence. | Calculus & Beyond Homework | 4 | ||
| Pointwise vs Uniform Convergence | Calculus | 2 | ||
| Uniform Convergence | Calculus & Beyond Homework | 4 | ||
| Pointwise convergence | Calculus | 5 | ||
| Pointwise convergence of fourier expansion | Quantum Physics | 4 | ||