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I have to prove that (1) if a function sequence $(f_n) \subset C[0,1]$ is weakly convergent to $f\in C[0,1]$ then $f_n(x)\rightarrow f(x)$ for any $x \in [0,1]$. I have also to (2) show, that pointwise convergence of $(f_n) \subset C[0,1]$ does not imply weak convergence.

My attempt:

(1) Let $(f_n) \subset C[0,1]$ be a sequence that is weakly convergence to $f\in C[0,1]$.We have: $$\forall \phi \in B(C[0,1],\mathbb{K}): |\phi f_n(x)-\phi f(x)|\rightarrow 0 \:\:\:\: \forall x\in[0,1]$$ Now take $\theta : f \rightarrow f(x)$ Obviously: $$\theta \in B(C[0,1],\mathbb{K})$$

we have:

$$|f_n(x)- f(x)| \rightarrow 0$$

(2) I consider the functional: $$\alpha(f)=\int_0^1f(t)dt$$ but I can't think of any sequence $(f_n)$ which is pointwise convergent to some $f$ and $| \alpha(f_n-f)|$ does not go to $0$. Any help?

luka5z
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    What is $Y$ here? – Daniel Fischer May 26 '14 at 19:06
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    (2) Consider the sequence $$f_n (t) =\begin{cases} 0 \mbox{ if } 0\leq t<\frac{1}{n+1} \ n^{100} \left(t-\frac{1}{n}\right)\left(t-\frac{1}{n+1}\right) \mbox{ if } \frac{1}{n+1} <t<\frac{1}{n} \ 0 \mbox{ if } \frac{1}{n} \leq t\leq 1\end{cases} $$ –  May 26 '14 at 19:18

1 Answers1

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(1) You only know the convergence $\phi(f_n) \rightarrow \phi(f)$ for bounded linear functionals $\phi$, not for arbitrary linear maps $\phi \in B(C([0,1]), Y)$ (and, as Daniel Fischer points out: What is $Y$?).

Convince yourself that the linear(!), bounded(!) functional $\phi_x(f) := f(x)$ does the job.

(2) Try something like $f_n = n \cdot \chi_{(0, 1/n)}$. Of course, you have to modify the idea to make it a continuous function.

PhoemueX
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  • $Y=\mathbb{R}$, and now what is wrong with part (1)? I will try (2) as you suggested – luka5z May 26 '14 at 19:13
  • The problem with part (1) is that you never defined $\phi(f)$ (i.e. you do not state your choice of $\phi$). At least it is not clear to me what you $\phi$ is. – PhoemueX May 26 '14 at 19:23
  • $\phi(f)$ is just a linear and continuous functional on $C[0,1]$ so $\phi \in B(C[0,1],\mathbb{K})$ – luka5z May 26 '14 at 19:25
  • That is exactly the problem. You have to choose $\phi : C([0,1]) \rightarrow \mathbb{K}, f \mapsto f(x)$. (Why is that continuous)? Consider for example the space $L^1([0,1])$. Here, weak convergence does not(!) imply pointwise convergence. Why is that? – PhoemueX May 26 '14 at 19:26
  • Ok I see now. From definition of weak convergence I have $\forall \phi \in B(C[0,1],\mathbb{K}): |\phi f_n(x)-\phi f(x)|\rightarrow 0 :::: \forall x\in[0,1]$ but I have to consider specific $\phi$s as you mentioned. $f\rightarrow f(x)$ is continuous because it can be bounded by supremum with functional norm 1 – luka5z May 26 '14 at 19:29
  • I edited (1), Now going for (2) – luka5z May 26 '14 at 19:34
  • I have no idea about (2) ;/ – luka5z May 26 '14 at 19:49
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    Just make the "rectangle" with height $n$ into a "triangle" so that the triangle has height $\sim 2n$ and is located between $0$ and $2/n$ (i.e. rise with slope $\sim n^2$ and fall with slope $\sim n^2$), or something like that. E.g. $f(x) = n^2x$ for $x \in [0,1/n]$, $f(x) = n^2 (2/n -x)$ for $x \in [1/n, 2/n]$ – PhoemueX May 26 '14 at 20:11
  • but to which function it is pointwise covnergent? to f=0? – luka5z May 27 '14 at 16:07
  • Yes, because $f_n(0)=0$ and for $x \in (0,1]$ we have $2/n < x$ for $n \geq n_0(x)$ and hence also $f_n(x) =0$ for $n \geq n_0(x)$. – PhoemueX May 27 '14 at 16:13