244 CHAPTER 13. SOME FUNDAMENTALS∗

To illustrate the difference in the two types of convergence, here is a standard example.

Example 13.6.2 Let

f (x)≡

{0 if x ∈ [0,1)1 if x = 1

Also let fn (x) ≡ xn for x ∈ [0,1] . Then fn converges pointwise to f on [0,1] but does notconverge uniformly to f on [0,1].

Note how the target function is not continuous although each function in the sequenceis. The next theorem shows that this kind of loss of continuity never occurs when you haveuniform convergence. The theorem holds generally when S⊆ X a normed linear space andf,fn have values in Y another normed linear space. You should fill in the details to be sureyou understand this. You simply replace |·| with ∥·∥ for an appropriate norm.

Theorem 13.6.3 Let fn : S→ Cq be continuous and let fn converge uniformly to f on S.Then if fn is continuous at x ∈ S, it follows that f is also continuous at x.

Proof: Let ε > 0 be given. Let N be such that if n≥ N, then

supy∈S|fn (y)−f (y)| ≡ ||fn−f ||

∞<

ε

3

Pick such an n. Then by continuity of fn at x, there exists δ > 0 such that if |y−x|< δ ,then |fn (y)−fn (x)|< ε

3 . Then if |y−x|< δ ,y ∈ S, then

|f (x)−f (y)| ≤ |f (x)−fn (x)|+ |fn (x)−fn (y)|+ |fn (y)−f (y)|

<ε

3+

ε

3+

ε

3= ε

Thus f is continuous at x as claimed. ■

13.7 Root TestThe root test has to do with when a series of real or complex numbers converges. I amassuming the reader has been exposed to infinite series. However, this that I am about toexplain is a little more general than what is usually seen in calculus. If you have a sequenceof real numbers {ak}∞

k=1 , ifAn ≡ sup

k≥nak

then the sequence {An} is decreasing. In the above, supk≥n ak means the least upper boundof all ak for k ≥ n or if there is no upper bound, An is simply said to equal ∞. This is justa formality to make it easy to give an easy discussion. Then, since {An} is a decreasingsequence, there are two cases. One is that it is bounded below and the other case is that itisn’t. In the first case, the sequence must converge to the greatest lower bound of the Anand in the second case, we say that the sequence converges to −∞. Then

lim supn→∞

an ≡ limn→∞

(supk≥n

ak

)Thus, if limsupn→∞ < r, it follows that for all n large enough every ak < r. If limsupn→∞ ak >r, it means there are infinitely many k such that ak > r.