I was given a problem in my real analysis class and I'm having some trouble understanding a part of the question. The question reads as follows:
Let $(X, \lVert \cdot\rVert)$ be a normed vector space.
Prove that $X^* = \{f: X \to \mathbb{R}, f $ linear and continuous $\}$ with the "operator norm" $$ \lVert f \lVert = \sup_{\lVert x \lVert = 1} \lVert f(x) \lVert$$ is a Banach space.
My plan on how to approach the question was to construct a cauchy sequence in $(X^*, \lVert \cdot\lVert$) and show it converges in $(X^*, \lVert \cdot\lVert)$. But I'm confused as to how the "operator norm" works (I wrote it down as it appeared on the page). There are a few norms in the definition for the operator norm and I'm confused as to what each one is. My interpretation was that $\lVert f \lVert$ and $\lVert f(x) \lVert$ are the "operator norm" and $\lVert x \lVert$ was the norm on $(X, \lVert\cdot \lVert)$. Is that the correct way to read it?
Also, what is the significance of having $\lVert x \lVert = 1$ for the operator norm? Is there a reason that $\lVert x \lVert = 1$? Thanks for any help
Good question, often this issue is made more clear with notation I will show presently. The norm on the $x$ in your equation is the norm $X$ came with, and the norm on $f$ is the one on $\mathbb{R}$, where $f$ sends things, i.e. the absolute value. This is sometimes denoted $$ ||f||=\sup_{||x||_X} ||f||_\mathbb{R} $$
As for your idea, sounds great. It may be helpful to note that the supremum norm is often called "the norm of uniform convergence."