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I'm trying some practice questions and came upon the following operator $T$ defined on $l^{\infty}$. For a sequence $x\in l^{\infty}$, we define $T(x)_n=\frac{1}{n}\displaystyle\sum_{i=1}^{n}x_i$, i.e., the sequence $T(x)$ is the sequence of Cesaro sums of the sequence $x$. The exercise asks to investigate the boundedness and compactness of this operator for the space $l^{\infty}$ and for its restriction to $l^1$. I managed to prove that it is in fact bounded on $l^{\infty}$ and has norm 1. However, I can't quite prove (or disprove) boundedness for when it's restricted to $l^1$. Also, I have no idea how to prove compactness. I tried finding some sequence with no convergent subsequence in the sequences $T(x)$, but it seems that it involves some trick to find an example. I'd very much appreciate some help on this. thanks in advance.

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    If $x=(1,0,0,0,...)$ then $Tx=(1,1/2,1/3,1/4,...)$ which is not in $\ell^1$. So I don't think $T$ is even a well defined operator on $\ell^1$. Maybe I'm missing something – 2017-02-09
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    You're right, I think I was muddled in how I phrased my question. What I meant was that the operator restricted to $l^1$. I've edited it so it makes more sense. – 2017-02-09
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    $ l^1\subset l^{\infty}$. . We have $\|T(x)\|_{\infty}\leq \|x\|_{\infty}$ for all $x\in l^{\infty}$, this will hold for $x\in l^1$. But $T$ does not map $l^1$ into $l^1$ as noted above by Shalop. – 2017-02-09

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Consider $T: \ell^{\infty} \to \ell^{\infty}$. Define a sequence $(x^n)_n \subset \ell^{\infty}$ by $x^n(m) = 1_{[m \geq n]}$. Then $\|x^n\|_{\infty} = 1$ for all $n$. And $(Tx^n)(m) \to 0$ for all $m$, but $\|Tx^n\|_{\infty}=1$ since Cesaro means converge to the same limit as the original sequence. Thus we found a uniformly bounded sequence $(x^n)$ such that $(Tx^n)$ has no uniformly convergent subsequence. So $T: \ell^{\infty} \to \ell^{\infty}$ is not compact.

Now consider $T$ as an operator from $ \ell^1 \to \ell^{\infty}$. It is easily verified that this has operator norm $1$ also. To show it is compact, note that if $(x^n)$ is a sequence in $\ell^1$ such that $\|x^n\|_1 \leq 1$ for all $n$, then (by compactness of $[0,1]^{\Bbb N}$ in the product topology), there exists some $u \in \ell^{\infty}$ and a subsequence $x^{n_k}$ such that $x^{n_k}(m) \to u(m)$ for all $m \in \Bbb N$. Then we also have that $Tx^{n_k}(m) \to Tu(m)$ for all $m$. Fixing $\epsilon>0$ choose some $N$ such that $2/N< \epsilon$, and then choose $K>0$ such $|Tx^{n_k}(m)-Tx^{n_j}(m)|<\epsilon$ whenever $m \leq N$ and $j,k \geq K$. By construction, for $m \geq N$ and any $j,k$ we have that $$|Tx^{n_k}(m)-Tx^{n_j}(m)| \leq \frac{1}{m} \sum_{i=1}^m |x^{n_k}(i) - x^{n_j}(i)| \leq \frac{1}{m} (\|x^{n_k}\|_1+ \|x^{n_j}\|_1) \leq \frac{2}{m}<\epsilon$$ So, $\|Tx^{n_j}-Tx^{n_k}\|_{\infty} < \epsilon$ for $j,k \geq K$. So that $(Tx^n)$ has a Cauchy subsequence, and thus $T$ is compact from $\ell^1 \to \ell^{\infty}$.