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I don't know how solve this problem. Please I need help.

Let $X =\mathcal{C}[0,1]$ with the uniform norm and let $\{p_j\}_{j\in\mathbb{N}}$, $\{q_j\}_{j\in\mathbb{N}}\subseteq X$ such that the series $\sum\limits_{j=1}^{\infty}p_j(s)q_j(t)$ uniformly converge to a continuous function $K:[0,1]\times [0,1]\rightarrow > \mathbb{R}$, i.e., $$\lim_{N\rightarrow +\infty}\left\{\max_{(s,t)\in [0,1]\times [0,1]}\left|K(s,t) - \sum_{j=1}^Np_j(s)q_j(t)\right|\right\}\ =\ 0.$$ Besides, let $F_j\in X^{\prime}$ defined by $F_j(u) := \int_0^1q_j(t)u(t)dt$, $\forall\ u\in X$. Show that, for all $G\in X^{\prime}$, the series $\sum\limits_{j=1}^{\infty}G(p_j)F_j$ is convergent in $X^{\prime}$. Identify the limit value, in terms of the adjoint of a convenient operator.

Thanks in advance.

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1  
Is $\,X'\,$ the dual space of $\,X\,$, what we mortals usually denote as $\,X^*\,$ ? –  DonAntonio Jun 3 '13 at 21:44
2  
Maybe I am not a mortal, because I denote $X^{\prime}$ as the dual of $X$ too. –  FASCH Jun 3 '13 at 22:06
1  
Yes, $X^{\prime}$ is the dual of $X$. –  user70195 Jun 3 '13 at 22:06

1 Answer 1

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For the first question, notice that for a fixed $u\in X$, we have $$\sum_{j=1}^NG(p_j)F_j(u)=\sum_{j=1}^NG((F_j(u))\cdot p_j)=G\left(\sum_{j=1}^NF_j(u)p_j\right),$$ hence given integers $M$ and $N$, we get $$\left|\sum_{j=M}^{M+N}G(p_j)F_j(u)\right|\leqslant \lVert G\rVert\cdot\max_x\left\lvert\sum_{j=M}^{M+N}p_j(x)\int_0^1q_j(t)dt\rvert\right|.$$ Recall that a dual space is complete.

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