Boundedness requirement Theorem 7.29 Baby Rudin

7.29 Theorem - Let $$\mathscr B$$ be the uniform closure of an algebra $$\mathscr A$$ of bounded functions. Then $$\mathscr B$$ is a uniformly closed algebra.

An algebra $$\mathscr A$$ is a family of complex functions defined on a set $$E$$ which is closed under addition of two functions, multiplication of two functions, and multiplication of a function by a constant.

Definition of uniform closure:

Let $$\mathscr B$$ be the set of all functions which are limits of uniformly convergent sequences of members of an algebra $$\mathscr A$$. Then $$\mathscr B$$ is called the uniform closure of $$\mathscr A$$.

Definition of uniform closed:

If $$\mathscr A$$ has the property that $$f \in \mathscr A$$ whenever $$f_n \in \mathscr A (n = 1,2,3,...)$$ and $$f_n \to f$$ uniformly on $$E$$, then $$\mathscr A$$ is said to be uniformly closed.

My question is: why do we require the functions to be bounded in theorem 7.29? For example, in proving $$f_n +g_n \to f + n$$ whenever $$f_n \to f, g_n \to g$$ uniformly, we can just appeal to theorem 3.3a (which states $$\lim_{n \to \infty} (s_n + t_n) = s + t$$ whenever $$\lim_{n \to \infty} s_n = s$$, $$\lim_{n \to \infty} t_n = t$$) to see that $$f_n +g_n \to f + n$$ uniformly. I don't see how boundedness is required here, nor anywhere else in proving the theorem.

So are the functions in $$\mathscr A$$ required to be bounded?

If we did not require boundedness, then $$\mathscr{B}$$ may not be closed under multiplication. In other words, it is false that if $$f_n\rightarrow f$$ uniformly and $$g_n\rightarrow g$$ uniformly, then $$f_ng_n\rightarrow fg$$ uniformly, unless we add an extra assumption like boundedness of at least one of the functions.
For example, on $$E=\mathbb{R}$$, consider $$f_n(x)=x+1/n=g_n(x)$$. Then, $$f_n,g_n\rightarrow x$$ uniformly, but the convergence of $$f_ng_n$$ to $$x^2$$ is not uniform.
• Exactly the answer I was looking for. My reasoning was that Theorem 3.3 yields an $N$ for each $\varepsilon > 0$ for all $x \in E$, for each of the three operations, which would work for all $x$ due to uniform convergence, but when you look at the proof of multiplication (3.3b) there is a little more to the story. The $N$ is not guaranteed to work for all $x$. Thanks again, +1! – Steven Wagter Mar 10 at 18:56