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Equip the space $C([0,1])$ with the usual supremum norm. Choose and fix a sequence ${\{t_n\}_{n=1}^{\infty}}$ of different points in $[0,1]$. For $\alpha =(\alpha_1,\alpha_2,...)$ in $l^1$, define $\beta_\alpha:C([0,1]) \to \mathbb{C}$ by

$$\beta_\alpha(f) = \sum_{n=1}^{\infty}\alpha_nf(t_n)$$ for $f \in C([0,1]).$

a) Show that $\beta_\alpha$ is a continuous linear functional on $C([0,1])$

I tried that with the definition of continuous transformation: $$ |\beta_\alpha(f)| \leq \sum_{n=1}^{\infty} |\alpha_n||f(t_n)| \leq \sum_{n=1}^{\infty} |\alpha_n|\sum_{n=1}^{\infty} |f(t_n)|$$

Since $\alpha$ is in $l^1$ the first factor is bounded but I have no idea how to get that to $\leq c ||f||_\infty$. Or do I have that completely different?

b) Show that the map $j: l^1 \to C([0,1])$, defined by $j(\alpha)=\beta_\alpha$ for $\alpha \in l^1$, is an isometric linear map.

Here I need to show, that this map is bijectiv is that right? How do I do that?

c) Show that $j(l^1)$ is a closed subspace of C([0,1])'.

It is closed if it contains all the limit points.. but I dont know how to prove that.

d) Show that C([0,1]) is not refelxiv.

I just really have no idea how to prove any of those statements. Can anyone help?

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  • $\begingroup$ Start with the definition(s). Post what you have tried. $\endgroup$ – user251257 Dec 6 '16 at 19:22
  • $\begingroup$ I edited my question, but I would really need some help! $\endgroup$ – Yuhe Dec 7 '16 at 11:29
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$(l^1)^*$ is not separable. (It does not have a countable dense subset.) Since $l_1$ is homeomorphic to a subspace of $C[0,1],$ therefore $C[0,1]^*$ is not separable. Therefore $C[0,1]^{**}$ is not separable. But $C[0,1]$ is separable.

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(1). $\sum_{n=1}^{\infty}|f(t_n)|$ might very well be infinite, for example if $f(x)=1$ for all $x.$

Let $A=\sum_{n=1}|\alpha_n|.$ By hypothesis, $A<\infty .$

Since $\|f\|=\max \{|f(x)|:x\in [0,1]\}<\infty$ we have $$|\beta_{\alpha}(f)|\leq \sum_{n=1}^{\infty}|\alpha_n|\cdot |f(t_n)|\leq \sum_{n=1}^{\infty}(|\alpha_n|\cdot \|f\|)=$$ $$=\|f\|\sum_{n=1}^{\infty}|\alpha_n|=\|f\|A.$$ Therefore $\beta_{\alpha}$ is bounded, hence continuous ( because it is linear) with $\|\beta_{\alpha}\|\leq A.$

(2). Regardless of whether we are speaking of the space $C[0,1]$ of continuous real or continuous complex-valued functions: Let $\epsilon >0.$ Take $n\in \mathbb N$ such that $\sum_{n'>n}|\alpha_{n'}|<\epsilon.$

There exists $f\in C[0,1]$ with $\|f\|=1$ such that for all $m\leq n$ we have $\alpha_mf(t_m)=|\alpha_m|.$ Then $$\|\beta_{\alpha}\|\geq |\beta_{\alpha}(f)|\geq |\sum_{m=1}^n \alpha_mf(t_m)|-\sum_{n'>n}|\alpha_{n'}|\cdot |f(t_{n'}|= $$ $$=\sum_{j=1}^n |\alpha_j|-\sum_{n'>n}|\alpha_{n'}|\cdot |f(t_{n'}|>$$ $$>(A-\epsilon)-\sum_{n'>n} |\alpha_{n'}|\cdot \|f\|=$$ $$=(A-\epsilon)-\sum_{n'>n}|\alpha_{n'}|>A-2\epsilon.$$ Therefore $\|\beta_{\alpha}\|\geq A.$ Since we already have $\|\beta_{\alpha}\|\leq A,$ we conclude that its norm is $A.$

(3). For Q (c): From the above line, for $a,a'\in l_1$ the norm of $j(a-a')$ as a functional on $C[0,1]$ is the $l_1$ norm of $(a-a').$ And (obviously) $j(a-a')=j(a)-j(a').$

Let $(j(a_n))_n$ be a Cauchy sequence in $C[0,1]^*$. Then $(a_n)_n$ is a Cauchy sequence in $l_1$. Since $l_1$ is complete, there exists $a\in l_1$ such that $\lim_{n\to \infty}\|a_n-a\|_{l_1}=0.$ Using the subscript "$*$" on the norm to denote the norm of a functional on $C[0,1],$ we have $$ 0=\lim_{n\to \infty} \|a_n-a\|_{l_1}=\lim_{n\to \infty}\|j(a_n)-j(a)\|_{*}.$$

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  • $\begingroup$ Do you really mean Q(b) in 3.? $\endgroup$ – Yuhe Dec 7 '16 at 12:56
  • $\begingroup$ thanks for your help!! Could you help me with the rest too? $\endgroup$ – Yuhe Dec 9 '16 at 9:47
  • $\begingroup$ (3) is for (c). I fixed that. (2) takes care of (b). The linearity of $j$ should be obvious. As for (d) i forget what reflexive means in this context but I might be able to answer if I knew $\endgroup$ – DanielWainfleet Dec 9 '16 at 12:51
  • $\begingroup$ X is reflexive if J(X) = X'', do you know what I mean? $\endgroup$ – Yuhe Dec 9 '16 at 12:57
  • $\begingroup$ And for c, you are showing that it is complete and so closed? is that right? why is j(a_n) not in j(l^1)? Aren't you showing that C[0,1] is complete? $\endgroup$ – Yuhe Dec 9 '16 at 12:59

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