# Baby Rudin - Notes on example 7.6

## Contents

I am currently going through the "Principles of Mathematical Analysis" by Walter Ruding and just below theorem 7.12 he mentions that an interesting property can be proved about one of the examples. He does not explain how and I was not able to find any solution on the internet. Maybe the proof is trivial to anybody but me, but I'll share it here anyway.

First off, what are we talking about anyway? The example is in the context of sequences of functions. He says that there are sequences of continuous functions that converge to a continuous function, but do not converge uniformly to that function. He gives a sequence from an earlier example:

\[f_n(x) = n^2x(1-x^2)^n \qquad (\text{for } 0 \le x \le 1, n \in \mathbb{N}^+).\]

It's easy to show (and Rudin does it), that for every \(x\) in the domain:

\[f(x) = \lim_{n \to \infty} f_n(x) = 0.\]

Therefore \(f\) is continuous. \(f_n\) is continuous for every \(n\) since it is defined as a combination of elementary functions.

As Rudin suggests we might want to apply theorem 7.9, which I'll state for easy reference:

Define \[M_n = \sup |f_n(x) - f(x)|\] then \(\{f_n\}\) converges uniformly to \(f\) iff \(M_n\) converges to \(0\).

So, how do we apply that to our example?

By integration we get \[\int_0^1 f_n(x) \mathrm{d}x = \frac{n^2}{2n + 2}.\]

Since for all $x$ we have $f(x)=0$, we get \(M_n = \sup f_n(x)\).

By theorem 6.12 we can put a lower bound on the supremum using the integral:

\[\int_0^1 f_n(x) \le M_n \cdot (1 - 0).\]

By substitution we get

\[M_n \ge \frac{n^2}{2n + 2}\].

We can conclude using the comparison test that \(M_n\) diverges and therefore \(\{f_n\}\) does not converge uniformly by theorem 7.9.

Author Ben Justus Bals

LastMod 2020-04-11