A continuous linear functional on a subspace extends to the entire normed space with the same norm. The example is deliberately concrete; it is a test of the statement, not a substitute for it.
Start locally
A Banach space \(X\) is complete in its norm, and a bounded linear operator \(T:X\to Y\) has norm \(\|T\|=\sup_{\|x\|\le1}\|Tx\|\). Completeness powers the major structural theorems.
The definition determines which expressions are legal; only then does the identity become meaningful. An equality in \(\mathcal A\) may change ambient meaning, so I keep \(\mathsf D\) separate from \(\mathsf C\).
Compute before generalising
Here is a concrete symbolic test. Reading it from left to right reveals which transformation is reversible and which is only an implication.
The global view
The two-row display is also a debugging tool: if the conclusion changes when only notation changes, some hidden choice has entered the argument.
Edge conditions
Finite-dimensional intuition can fail badly in infinite dimensions. Closed, bounded sets need not be compact, and linear maps need not be bounded automatically.
The important habit is to remember what was fixed before the calculation began and what was proved only afterward. The final display preserves that order.