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Let $(a_k)_{k \in \mathbb{N}}$ be a sequence such that $$\sum_{k=1}^{\infty}{|a_k\xi_k|} < \infty$$ for all $x=(\xi_k)_{k \in \mathbb{N}} \in l_1$

Show that $T: l_1 \to l_1$, $T(x) = (a_k\xi_k)_{k \in \mathbb{N}}$ is linear and continuous.

My attempts

1)

We have $\|T(x)\| = \sum_{k=1}^{\infty}{|a_k\xi_k|} \leq \sum_{k=1}^{\infty}{|a_k|\|x\|} = \bigg(\sum_{k=1}^{\infty}{|a_k|}\bigg)\|x\| $.

If $\sum_{k=1}^{\infty}{|a_k|} < \infty$, it's done. However, I couldn't prove it.

2)

Since $l_1$ is a Banach space, we can apply the Closed Graph Theorem.

Thus, $G(T)$ closed $\Rightarrow$ $T$ is continuous. So I tried to prove that $G(T)$ is closed:

Let $(x,y) \in \overline{G(T)}$. Then there exists a sequence $\big((x_n,T(x_n))\big)_{n \in \mathbb{N}} \subset G(T)$ such that $\big((x_n,T(x_n))\big)_{n \in \mathbb{N}} \to (x,y)$, where $x_n = (\xi_k^{(n)})_{k \in \mathbb{N}}$

Then $x_n \to x$ and $T(x_n) \to y$. We must prove that $y = T(x)$.

By triangle inequality, we have:

$$\|T(x)-y\| = \|T(x)-T(x_n) + T(x_n) - y\| \leq \|T(x)-T(x_n) \| + \|T(x_n) - y\| $$

$\|T(x_n) - y\| $ is small for large $n$ since $T(x_n) \to y$.

I tried to prove that $\|T(x)-T(x_n) \| $ is also small for large $n$.

We have:

$$\|T(x_n) - T(x)\| = \sum_{k=1}^{\infty}{|a_k||\xi_k - \xi_k^{(n)}|} \leq \sum_{k=1}^{\infty}{|a_k|\|x - x_n\|} = \bigg(\sum_{k=1}^{\infty}{|a_k|}\bigg)\|x - x_n\|$$

Again, if $\sum_{k=1}^{\infty}{|a_k|} < \infty$, it's done.

I'm stuck here.

I appreciate if you could give some hints (not an entire solution).

Thanks.

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  • $\begingroup$ Summation by parts (along with $x \in l_1$) may be useful to convert the condition $\sum_i |a_i \xi_i| < \infty$ into a condition on the $a_i$ alone. $\endgroup$ – Jon Warneke Apr 21 '16 at 19:31
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Notation: $\xi=\{\xi_n\}\in l^1$.

For each $k>0$, let $T_k:l^1\to l^1$ given by $T_k(\xi)=\sum_{n=1}^ka_n\xi_n$. Let $F=\{T_k\}$. For each $\xi\in l^1$ we have $$\|T_k(\xi)\|=\sum_{n=1}^k|a_n\xi_n|\le\sum_{n=1}^\infty |a_n\xi_n|=\|T(\xi)\|<\infty.$$

Thus, $\sup_{k}\|T_k(\xi)\|<\infty$. By the Uniform Boundedness Principle, there is $M$ such that $\|T_k\|\le M$ for all $k$.

Now, let $\Xi_k=(0,0,...,0,1,0,...)\in l^1$. $\|\Xi_k\|=1$ and $M\ge\|T_k\|=\sup_{\|\xi\|=1}\|T_k\xi\|\ge\|T_k\Xi_k\|=|a_k|$. Conclusion: $\{a_n\}$ is bounded.

Now it is easy to finish:

$\|T\xi\|=\sum_{k=1}^\infty|a_n\xi_n|\le\sum_{k=1}^\infty M|\xi_n|=M\|\xi\|$, so $T$ is bounded.

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