On a density principle found in Gröchenig's book.

Question

At the beginning of the appendix of the renown book by K.Gröchenig Introduction to Time-Frequency Analysis, there is a statement called density principle which I believe to be incorrect. I hope I will get some clarifications. This is particularly important because this principle is applied to prove other theorems in the book.

A.1 Density principle. Let $X$ be a dense subspace of $B_1$. Assume that $A:X\to B_2$ is a linear operator that satisfies the inequality $$\|Af\|_{B_2}\leq C\|f\|_{B_1},\qquad \text{ for all }f\in X\qquad (A.1) $$ then (A.1) holds for all $f\in B_1$.

Here, $B_1$ and $B_2$ are Banach spaces. Notice that $A$ is not assumed to be bounded on $B_1$, and this principle is applied elsewhere in the same book without this assumption, so it is not just a typo.

Why I believe the statement is incorrect. If $X$ is infinite dimensional, then it is possible to construct a linear operator $A:B_1\to B_2$ such that $A$ is bounded on $X$, yet unbounded on $B_1$. See for instance here and here. In this case, the inequality (A.1) cannot hold for all $f\in B_1$, yet it does hold for all $f\in X$ for a proper choice of the constant $C$.

The proof proposed in the book.

For every $x\in B_1$ there exists a sequence $f_n \in X$ such that $\lim_{n\to \infty}\|f_n-f\|_{B_1}=0$. Since $\|Af_n-Af_m\|_{B_2}=\|A(f_n-f_m)\|_{B_2}\leq \|A\|_{\text{op}}\|f_m-f_n\|_{B_1}$, $(Af)_{n\in \mathbb{N}}$ is a Cauchy sequence in $B_2$. Therefore $Af_n$ converges in $B_2$ and we may define $Af$ by $Af=\lim_{n\to \infty}Af_n$.

What I believe is wrong with the proof. This argument does not seem to prove the claim. We are starting with an operator that is already defined on $X$, so why would the proof conclude by redefining the operator $A$ on the whole space? This would make sense if we knew that the extension is unique. But this is only guaranteed if we assume that $A$ is bounded on $B_1$, which we have not.

What do you think, MSE?

EDIT: Actually the theorem is correct, although the sentence $(A.1)$ holds for all $f\in B_1$ is rather ambiguous. The issue is that this princple is recalled when proving another theorem (see here) where the operator $A$ was already defined on the whole $B_1$ (not just on $X$), so we cannot apply this principle since it requires to redefine the operator on $B_1\setminus X$.

Answer 1

The statement should be:

Then $A$ has a unique extension to $B_1$ such that (A.1) holds for all $f \in B_1$

with the proof as given.

Answer 2

You are correct: In this form, the theorem can’t be true, by the arguments you described. I do not have the book and therefore can’t check, but my guess is that the author wants to prove and then use the fact that you can extend the operator $A$ to a bounded one on $B_1$ (in fact, the operator norm stays the same) – at least that’s what the proof concludes. When he applies the theorem, is this result sufficient or does he really need that some given operator is bounded if it is on a dense subset?