Functional analysis,$\ell^p$, continuous and uncontinuous operator

Question

We observe $C[0,1]$ (the $\mathbb{C}$-vectorspace of all continous functions in $\mathbb{C}$ on $[0,1]$) as subspace of $L^p([0,1],\lambda)$ where $\lambda$ notes the Lebesgue-measure on $[0,1]$.

Observe $c_{00}$ as subspace of $\ell^p$ for every $1\leq p\leq\infty$. Let $T: (c_{00},\|\cdot\|_p)\to (c_{00},\|\cdot\|_1)$, $T(f)=f$. Show, that in this case for $p>1$ the operator $T$ is uncontinuous and $T^{-1}$ is continuous. Calculate $\|T^{-1}\|_{\operatorname{op}}$

I do not know if it is a common definition. It is $c_{00}:=\{f\in c_0|\exists N\in\mathbb{N}~~ \text{with}~~ f(n)=0~~\forall n\geq N\}$ and $c_0$ is the set of all sequences $f:\mathbb{N}\to\mathbb{C}$ which have the limit $0$.

To show, that $T$ is not continuous, I tried to show, that it is unbounded. And to show that $T^{-1}$ is continuous, I want to show, that it is bounded.

Can you give me a hint, on how to do this and how to calculate $\|T^{-1}\|_{\operatorname{op}}$?

Thanks in advance.

Edit1: To show, that $T^{-1}$ is bounded, I have to find $c>0$ such that $\|T^{-1}f\|_p\leq c\|T^{-1}f\|_1$.

It is $\|T^{-1}f\|_p=\|f\|_p=\left(\sum_{n=1}^\infty |a_i|^p\right)^{1/p}\leq \sum_{n=1}^\infty |a_i|=1\cdot\|f\|_1$

The sum exists (therefore is finite), since $f\in c_{00}$. Hence the sum converges. We see, that we can choose $c=1$ and $T^{-1}$ is bounded.

Edit2: Now I want to show, that $T$ is not bounded. Assume $T$ is bounded. Then there is a $c>0$ such that $\|f\|_1\leq c\|f\|_p$.

We take $f_n(k)=\begin{cases} \left(\frac1n\right)^{1/p}, ~\text{if}~ k\leq n\\ 0,~\text{else}\end{cases}$.

Hence $c\geq \frac{\|f_n\|_1}{\|f_n\|_p}$. Since $p>1$ it is $c^p\geq \frac{\|f_n\|_1^p}{\|f_n\|_p^p}$

Hence: $\frac{\left(\sum_{i=1}^n \frac{1}{n}\right)^p}{\sum_{i=1}^n \frac{1}{n^p}}\geq 1$. If we take the limit $\lim_{n\to\infty}\frac{\left(\sum_{i=1}^n \frac{1}{n}\right)^p}{\sum_{i=1}^n \frac{1}{n^p}}=\infty$.

Contradiction to $T$ is bounded.

Answer 1

First we prove $T^{-1}$ is continuous. Note that for any $a,b \geq 0$ you have that $(a^p+b^p)^{\frac{1}{p}} \leq a +b$. This follows from the fact that $a^p +b^P \leq (a+b)^p$. So take any $f \in c_{00}$ and choose $n \in \mathbb{N}$ such that $f(i) = 0$ for $i \geq n$. We then find that $$ \| T^{-1} f\|_p = \left( \sum_{k=1}^n \lvert f(i) \rvert^p \right)^{1/p} \leq \sum_{k=1}^n \lvert f(i) \rvert = \|f\|_1. $$ So $T^{-1}$ is bounded. Moreover we immediately find that $\|T^{-1}\| \leq 1$. To prove that $\|T^{-1}\| = 1$, note that if we define $f \in c_{00}$ by $f(k) = 1$ for $k=1$ and $f(k) = 0$ for $k \neq 1$. Then $\|T^{-1} f\|_p = 1 = \|f\|_p$, so we must have $\|T^{-1}\| \geq 1$ and ultimately also that $\|T^{-1}\| =1$.

To prove $T$ is unbounded, consider the sequence $(f_n)_n$ in $c_{00}$ defined by $f_n(k) = \left( \frac{1}{n}\right)^{\frac{1}{p}}$ if $k \leq n$ and $0$ if $k > n$. Now note that $\| f_n\|_p =1$ but $\|f_n\|_1 = n \frac{1}{n^{1/p}} = n^{1-\frac{1}{p}}$. For any $c > 0$, there is an $n \in \mathbb{N}$ such that $n^{1-\frac{1}{p}} > c$ or in other words $\|Tf_n\|_1 > c \|f_n\|_p$. This means $T$ is unbounded.