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.

  • $\begingroup$ 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! $\endgroup$ – Steven Wagter Mar 10 at 18:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.