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I was reading the following theorem and its proof:

If $(f_n)$ and $(g_n)$ are uniformly convergent sequences of functions, then $(f_n+g_n)$ is a uniformly convergent sequence of functions.

Proof. Let $\epsilon>0$. Then there exists $K\in\mathbb{N}$ such that $k\geq K$ implies $\left \| f_k-f \right \|_A<\epsilon/2$ since $(f_n)$ is uniformly convergent to $f$ on $A$. Similarly, there exists $M\in\mathbb{N}$ such that $m\geq M$ implies $\left \| g_m-g \right \|_A<\epsilon/2$. Set $N=\max\{K,M\}$. It follows that for $n\geq N$, we have $$\left \| (f_n+g_n)-(f+g) \right \|_A=\left \| (f_n-f)+(g_n-g) \right \|_A\leq\left \| f_n-f \right \|_A+\left \| g_n-g \right \|_A<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.$$ Therefore, $(f_n+g_n)$ is uniformly convergent to $(f+g)$ on $A$. $\square$

My quesiton is: What is the meaning of those double absolute value bars, and what do their subscripts denote? Thanks!

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Is $A$ a normed vector space? If so, $\|\cdot\|_A$ likely denotes the norm of an element of $A$. –  Ilmari Karonen Mar 8 '12 at 19:58

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up vote 2 down vote accepted

It looks like $$ \|f\|_A=\sup\{|f(t)|:\ t\in A\}. $$

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Impossible to tell without context. Presumably $A$ is some set, the function are defined on $A$, and the double bars indicate some metric defined on functions on $A$. But no one would write a paragraph like that without having previously explained all the notation, so just go back a few pages to see what it all means.

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