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Convergence of Vectors

Module by: Michael Haag. E-mail the author

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Summary: This modules presents two common types of convergence, pointwise and norm, and discusses their properties, differences, and relationships with one another.

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Convergence of Vectors

We now discuss pointwise and norm convergence of vectors. Other types of convergence also exist, and one in particular, uniform convergence, can also be studied. For this discussion , we will assume that the vectors belong to a normed vector space.

Pointwise Convergence

A sequence gn |n=1∞ n 1 gn converges pointwise to the limit g g if each element of gn gn converges to the corresponding element in g g. Below are few examples to try and help illustrate this idea.

Example 1

gn= gn1gn2 = 1+1n2−1n gn gn 1 gn 2 1 1 n 2 1 n First we find the following limits for our two gn gn's: limn→∞gn1=1 n gn 1 1 limn→∞gn2=2 n gn 2 2 Therefore we have the following, limn→∞gn=g n gn g pointwise, where g=12 g 1 2 .

Example 2

∀t,t∈ℝ:gnt=tn t t gn t t n As done above, we first want to examine the limit limn→∞gnt0=limn→∞t0n=0 n gn t0 n t0 n 0 where t0∈ℝ t0 . Thus limn→∞gn=g n gn g pointwise where gt=0 g t 0 for all t∈ℝ t .

Norm Convergence

The sequence gn |n=1∞ n 1 gn converges to gg in norm if limn→∞∥gn−g∥=0 n gn g 0 . Here ∥˙∥ ˙ is the norm of the corresponding vector space of gn g n 's. Intuitively this means the distance between vectors gn g n and g g decreases to 00.

Example 3

gn=1+1n2−1n g n 1 1 n 2 1 n Let g=12 g 1 2

∥gn−g∥=1+1n−12+2−1n2=1n2+1n2=2n g n g 1 1 n 1 2 2 1 n 1 2 1 n 2 1 n 2 2 n (1)
Thus limn→∞∥gn−g∥=0 n g n g 0 Therefore, gn→g g n g in norm.

Example 4

g n t=tnif0≤t≤10otherwise g n t t n 0 t 1 0 Let gt=0 g t 0 for all tt.

∥ g n t−gt∥=∫01t2n2dt=t33n2|n=01=13n2 g n t g t t 1 0 t 2 n 2 n 0 1 t 3 3 n 2 1 3 n 2 (2)
Thus limn→∞∥ g n t−gt∥=0 n g n t g t 0 Therefore, g n t→gt g n t g t in norm.

Pointwise vs. Norm Convergence

Theorem 1

For ℝm m , pointwise and norm convergence are equivalent.

Proof: Pointwise ⇒ Norm

g n i→gi g n i g i Assuming the above, then ∥gn−g∥2=∑i=1m g n i−gi2 g n g 2 i m 1 g n i g i 2 Thus,

limn→∞∥gn−g∥2=limn→∞∑i=1m2=∑i=1mlimn→∞2=0 n g n g 2 n i m 1 g n i g i 2 i m 1 n g n i g i 2 0 (3)

Proof: Norm ⇒ Pointwise

∥gn−g∥→0 g n g 0

limn→∞∑i=1m2=∑i=1mlimn→∞2=0 n i m 1 g n i g i 2 i m 1 n g n i g i 2 0 (4)
Since each term is greater than or equal zero, all 'mm' terms must be zero. Thus, limn→∞2=0 n g n i g i 2 0 forall ii. Therefore, gn→g pointwise g n g pointwise

Note:

In infinite dimensional spaces the above theorem is no longer true. We prove this with counter examples shown below.

Counter Examples

Example 5: Pointwise ⇏ Norm

We are given the following function: g n t=nif0<t<1n0otherwise g n t n 0 t 1 n 0 Then limn→∞ g n t=0 n g n t 0 This means that, g n t→gt g n t g t pointwise where for all tt gt=0 g t 0 .

Now,

∥ g n ∥2=∫-∞∞| g n t|2dt=∫01nn2dt=n→∞ g n 2 t g n t 2 t 1 n 0 n 2 n (5)
Since the function norms blow up, they cannot converge to any function with finite norm.

Example 6: Norm ⇏ Pointwise

We are given the following function: g n t=1if0<t<1n0otherwise if n is even g n t 1 0 t 1 n 0 if n is even g n t=-1if0<t<1n0otherwise if n is odd g n t -1 0 t 1 n 0 if n is odd Then, ∥ g n −g∥=∫01n 1 dt=1n→0 g n g t 1 n 0 1 1 n 0 where gt=0 g t 0 for all tt. Therefore, g n →g in norm g n g in norm However, at t=0 t 0 , g n t g n t oscillates between -1 and 1, and so it does not converge. Thus, g n t g n t does not converge pointwise.

Problems

Prove if the following sequences are pointwise convergent, norm convergent, or both and then state their limits.

  1. g n t=1ntif0<t 0 ift≤0 g n t 1 n t 0 t 0 t 0
  2. g n t=ⅇ-ntift≥0 0 ift<0 g n t n t t 0 0 t 0

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