A question about linear, bounded map in $L^p$ spaces and $C[0,1]$

Question

Let $l^{\infty}$ be the space of all complex bounded sequences with the supremum norm. Let $a_n\in l^{\infty}$ Define $T: l^1\to C[0,1]$ such that:

$\forall b_n\in l^1 , x\in [0,1]$, $$T(b_n)=\sum_{n=1}^{\infty} a_nb_nx^n.$$

Show that $T$ is well-defined, linear continuos and $||T||=||a_n||_{\infty}$ where $C[0,1]$ with the maximum norm.

Well defined: I think it means to show that for $b_n\in l^1$ the defined sum is in $C[0, 1].$ Given that $b_n\in l^1 , a_n\in l^{\infty}$ thus $a_nb_n\in l^1$ and $||a_nb_n||_1\leq ||a_n||_\infty ||b_n||_1$ and this is a finite value since $x\in [0, 1]$ so the sum converges and thus is continuos. I'm not sure about it.

T is linear: Let $b_n , c_n \in l^1$ and $d, e \in \mathbb{C}$ then \begin{align} T(cb_n+dc_n)(x)&= \sum_n a_n(cb_n+dc_n) x^n\\ &= \sum_n(ca_nb_n+da_nc_n)x^n\\ &= \sum_n ca_nb +\sum_n da_nc_n\\ &= c\sum_n a_nb_n+d\sum_n a_nc_n\\ &= cT(b_n)(x)+dT(c_n)(x) \end{align}

$T$ is continuous:

We can show that $T$ is bounded so since $T$ is linear, by theorem we get that $T$ is continuous.

According to the definition we have to show that there is $M>0$ such that $$||T(b_n)||\leq M ||b_n||.$$

To this end, \begin{align} ||T(b_n)(x)||_{C[0,1]}&=||\sum_n a_nb_nx^n||_{C[0,1]}\\ &=\max |\sum_n a_nb_nx^n|\\ &\leq \max \sum_n |a_nb_nx^n|. \end{align} I did not know how to use the information that $a_n\in l^\infty$ and $b_n\in l^1$ to finish this method.

The last part can be shown by the equivalent definitions $||T||=sup_{||_n||=1} ||T(b_n)||=sup_{b_n\neq0 \in l^1} \frac{||T(b_n)||}{||b_n||}$.

I had a problem with applying the definition that I know, in this question. I would be glad if you can help me in understanding this.

Answer 1

Well defined: we know that $\sum\limits_{n=1}^{\infty}a_nb_n$ converges absolutly. Since for every $x$ in $[0,1]$ we have $|a_nb_nx^n| \leq |a_nb_n|$, then Weierstrass M-test gives us uniformly convergance of series $\sum\limits_{n=1}^{\infty}a_nb_nx^n$ on $[0,1]$. Since $u_n(x)=a_nb_nx^n$ is continuous, then $\sum\limits_{n=1}^{\infty}a_nb_nx^n$ is continuous on $[0,1]$

Boundness: $|T(b_n)(x)|=|\sum\limits_{n=1}^{\infty}a_nb_nx^n|\leq \sum\limits_{n=1}^{\infty}|a_nb_nx^n| \leq \sum\limits_{n=1}^{\infty}|a_nb_n| \leq \|a_n\|_{\infty}\|b_n\|_1$. So, $T$ is bounded (and then continuos) and $\|T\|\leq \|a_n\|_{\infty}.$

Norm: let $b=(0,0,...,b_n=1,0,...) \in l^1$ with $\|b\|_1=1$. Then $\max\limits_{[0,1]}|T(b)(x)|=\max\limits_{[0,1]}|a_nx^n|=|a_n|\|b\|_1$, that is why $\|T\| \geq |a_n|$ for any $n$. Taking supremum over all positive integer $n$ we have $\|T\| \geq \|a_n\|_{\infty}$ compliting the proof.

Answer 2

You are almost there. Just a small point, you are right that to show that $T$ is well-defined, you need to show that $T(b_n)\in C[0, 1].$ The proof, however, you have given above only argues that $T(b_n)(x)<\infty$ for each $x\in [0, 1].$ This is because you have only argued that the sum converges for each $x.$ But it does not give you continuity. Recall that pointwise limits of continuous functions may fail to be continuous. To summarize, you proof is incomplete. It might be a helpful idea to try the Weiestrass-M test to obtain uniform convergence and hence conclude that the limiting function is indeed continuous.

Your proof of linearity is all right.

Finally to prove that $T$ is bounded. You have already obtained \begin{equation} ||T(b_n)||_{C[0, 1]}\le \max_{x\in [0, 1]}\sum_{n}| a_nb_nx^n|. \end{equation} To finish, just observe that for $|a_nb_nx^n|\le |a_nb_n|$ for all $x\in [0, 1].$ Therefore, $$||T(b_n)||_{C[0, 1]}\le \sum_{n}| a_nb_n|\le ||a_n||_{\infty}\sum_{n}|b_n|=||a_n||_{\infty}||b_n||_1.$$

Answer 3

As you have noted, the linearity of T is clear. Then, boundedness (continuity) is proved as you have suspected. Here are the details: \begin{align*} \|T(b)\|_{C[0,1]}&=\sup_{x\in[0,1]}\left|\sum_{n=1}^{\infty}a_{n}b_{n}x^{n}\right| \\ &\leq \sum_{n=1}^{\infty}|a_{n}||b_{n}|\leq\|a\|_{\infty}\|b\|_{1}\end{align*} where $a=(a_{n})_{n=1}^{\infty}\in\ell_{\infty}$ and $b=(b_{n})_{n=1}^{\infty}\in\ell_{1}$. This is because $|a_{n}|\leq\|a\|_{\infty}$ for all $n$ and because $b$ is absolutely summable by assumption. Well-definedness also follows as noted in the first (and second) answer.