Why is this "proof" of the Uniform Boundedness Principle via the contrapositive erroneous?

Question

In class, we stated the Uniform Boundedness Principle as follows:

Let $X$ be a Banach space, let $Y$ be a normed vector space, and let $A \subseteq L(X, Y)$ be a family of bounded linear operators. If for each $x \in X$, $\sup\{\Vert T(x) \Vert \; \vert \; T \in A \} \lt \infty$, then $\sup\{\Vert T \Vert \; \vert \; T \in A \} \lt \infty$.

We then proved the result directly, making use of the Baire Category Theorem along the way.

I had a notion that the proof of the UBP would have been much easier using the contrapositive. I immediately suspected that I was wrong, but I asked my professor about this because I couldn't discern why my "proof" would be wrong. He told me that I was negating the conclusion incorrectly, but I still don't really understand why. Can someone please explain to me why the following "proof" is erroneous?

Proof: We proceed via the contrapositive. Assume that $\sup\{\Vert T \Vert \; \vert \; T \in A \} = \infty$. Then there exists some $S \in A$ such that $S$ is unbounded. Hence, by the definition of the operator norm, it follows that there exists some $y \in X$ such that $\Vert S(y) \Vert = \infty$. QED.

Actually, I think I may have answered my own question in the process of formally typing it up. The fact that $\sup\{\Vert T \Vert \; \vert \; T \in A \} = \infty$ can still hold even when every operator in $A$ is actually bounded, right? Did I find my own mistake, or am I still missing something important?

Thank you.

Answer 1

You are correct about your mistake. For instance, $A$ might have elements $T_n$ such that $\|T_n\|=n$ for each $n\in\mathbb{N}$, and then $\sup\{\Vert T \Vert \; \vert \; T \in A \} = \infty$ without any individual element of $A$ being unbounded.

Note that in fact it is impossible for $A$ to have an unbounded element, since by assumption $A$ is "a family of bounded linear operators".

(Moreover, even if $S$ is unbounded, that does not mean there exists $y\in X$ such that $\|S(y)\|=\infty$, for the same reason: it just means there are unit vectors which $S$ takes to vectors of arbitrarily large (but finite) norm. Indeed, it doesn't make sense to write $\|S(y)\|=\infty$ since no element of $Y$ can have infinite norm (a norm is defined as a function $Y\to [0,\infty)$).)