Hey there, let's break this down step by step to clear up that confusion! First, let's ground ourselves in the definitions and how they connect to the proof of $|T| < \infty$ for $T \in B(X,Y)$.

First: What "Bounded Subset" Means in a Normed Space #

In any normed space (like $X$ or $Y$ here), a subset $S$ is bounded if and only if there exists some positive real number $M$ such that $|s| \leq M$ for every $s \in S$.

The phrase "a subset is bounded iff it's contained in some multiple of the unit ball" is just a rephrasing of this definition:

• The unit ball of a space is $B(0,1) = {z \mid |z| \leq 1}$.
• A "multiple of the unit ball" is $M \cdot B(0,1) = {M \cdot z \mid z \in B(0,1)}$, which is exactly the set ${y \mid |y| \leq M}$.
• So saying $S \subseteq M \cdot B(0,1)$ is the same as saying every element of $S$ has norm at most $M$—which is exactly the definition of $S$ being bounded.

Connecting This to the Operator Norm Proof #

Now, let's tie this to Rudin's proof for $|T| < \infty$:

1. We're looking at the set $S = { Tx \mid x \in X, |x| \leq 1 }$—this is the image of $X$'s unit ball under $T$, and it's a subset of $Y$.
2. By definition, $T \in B(X,Y)$ means $T$ is a bounded linear operator: there exists some $M \geq 0$ such that for every $x \in X$, $|Tx| \leq M|x|$.
3. Take any $Tx \in S$: since $|x| \leq 1$, we have $|Tx| \leq M|x| \leq M \cdot 1 = M$.
4. This means every element of $S$ has norm at most $M$, so $S \subseteq M \cdot B_Y(0,1)$ (the $M$-multiple of $Y$'s unit ball).
5. Using the boundedness criterion we just clarified, $S$ is a bounded subset of $Y$. The supremum of a bounded set of real numbers (in this case, the set ${|Tx| \mid Tx \in S}$) is always finite—so $|T| = \sup S < \infty$.

A Quick Note on Operator "Boundedness" vs. Set Boundedness #

One common point of confusion: the term "bounded" for linear operators doesn't mean the operator itself is bounded (which doesn't make sense, since we can scale $x$ to make $Tx$ as large as we want). Instead, it means the operator maps bounded subsets of $X$ to bounded subsets of $Y$—which is exactly what we used here: $T$ maps $X$'s bounded unit ball to $Y$'s bounded subset $S$.

内容的提问来源于stack exchange，提问作者theQman