Prove that the formula
$T(b_1,b_2,b_3,…,b_n,…) = (b_1,
b_2/2
,…,
b_n/n
,…)$
defines a bounded linear operator $T : (ℓ^∞,∥·∥_∞)→(ℓ^∞,∥·∥_∞)$.
Proving that it is linear is easy. Need help with the bounded part. But it seems obviously true:
If it is bounded then $\exists C \ge 0$ such that $\|Tx\|_{\infty} \le C \|x\|_{\infty}$ for all $x \in X$. We can see that $\|Tx\|=\|b_n/n\|=\sup_{n \in \mathbb N}|b_n/n| \leq \|b_n\|=\sup_{n \in \mathbb N} |b_n|$ for all $n \ge 1$ so we can let $C=1$.
Is this how we are meant to generally show boundedness of linear operators? Although I have heard of a theorem that says continuous $\iff$ bounded. But showing continuity would be long…
Best Answer
If $T$ is linear, continuity is equivalent to continuity at $0$.
If $T$ is bounded, then $T(B(0,1)) \subset B(0, \|T\|)$ and hence $B(0,1) \subset T^{-1} (B(0, \|T\|))$ and so for any $\epsilon>0$ we have $B(0,\epsilon) \subset T^{-1} (B(0, \epsilon\|T\|))$. It follows that $T$ is continuous at $x=0$.
Conversely, if $T$ is continuous at $x=0$, then $T^{-1}(B(0,1))$ contains some open set $U$ that has $0 \in U$. Hence $B(0,{1 \over K}) \subset T^{-1}(B(0,1))$ for sufficiently large $K$. In other words, $B(0,1) \subset T^{-1}(B(0,K)$ and so $T(B(0,1)) \subset B(0, K)$, hence $T$ is bounded (and $\|T\| \le K$).
Hence a demonstration of boundedness is directly equivalent to demonstrating continuity and vice versa.
In the questions, it is clear that $|[Tx]_k| \le |x_k|$ for each $k$, hence $\|Tx\|_\infty \le \|x\|_\infty$ and so $\|T\| \le 1$. Since $Te_1 = e_1$ we have equality.