Real-Analysis-The-Simple-Approximation-Theorem, guaranteed and verified 100% Pass

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The Simple Approximation Theorem


Def. Given a sequence of functions {𝑓𝑛 } with common domain 𝐸, a function 𝑓
on 𝐸 and a subset 𝐺 ⊆ 𝐸, we say:

1) The sequence {𝒇𝒏 } converges to 𝒇 pointwise on 𝑮 if
lim 𝑓𝑛 (𝑥 ) = 𝑓(𝑥) for all 𝑥 ∈ 𝐺.
𝑛→∞


2) The sequence {𝒇𝒏 } converges to 𝒇 pointwise a.e. on 𝑮 if it converges to 𝑓
pointwise on 𝐺~𝐻, where 𝑚(𝐻 ) = 0.

3) The sequence {𝒇𝒏 } converges to 𝒇 uniformly on 𝑮 if for each 𝜖 > 0,
there is an 𝑁 such that if 𝑛 ≥ 𝑁 then:
|𝑓(𝑥 ) − 𝑓𝑛 (𝑥 )| < 𝜖 on 𝐺

The pointwise limit of continuous functions need not be continuous.

Ex. In the example below prove that 𝑓𝑛 → 𝑓 pointwise on 0 ≤ 𝑥 ≤ 1 but not
uniformly.

𝑓𝑛 (𝑥 ) = 𝑥 𝑛 ; 0 ≤ 𝑥 ≤ 1.
𝑓(𝑥 ) = 0 if 0 ≤ 𝑥 < 1
=1 if 𝑥 = 1


To prove pointwise convergence we must show that given any 𝜖 > 0 there exists
an 𝑁 such that if 𝑛 ≥ 𝑁 then |𝑓 (𝑥 ) − 𝑓𝑛 (𝑥 )| < 𝜖.

Note: for pointwise convergence 𝑁 can depend on the point 𝑥.

At 𝑥 = 0, 𝑓𝑛 (0) = 0 = 𝑓(0) for all 𝑛, thus 𝑓𝑛 (0) → 𝑓(0).

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If 0 < 𝑥 < 1, then 𝑓 (𝑥 ) = 0.

|𝑓(𝑥 ) − 𝑓𝑛 (𝑥 )| = |0 − 𝑥 𝑛 | = 𝑥 𝑛 ; since 0 < 𝑥 < 1.
So solve 𝑥 𝑛 < 𝜖 for 𝑛.

𝑛(𝑙𝑛𝑥) < 𝑙𝑛𝜖
𝑙𝑛𝜖
𝑛> .
ln(𝑥)
𝑙𝑛𝜖
So choose 𝑁 > .
ln(𝑥)

Notice our formula for 𝑁 depends on the point 𝑥.
𝑙𝑛𝜖 𝑙𝑛𝜖
So if 𝑁 > and 𝑛 ≥𝑁>
ln(𝑥) ln(𝑥)
𝑙𝑛𝜖
Then 𝑛>
ln(𝑥)

𝑛(𝑙𝑛𝑥) < 𝑙𝑛𝜖
𝑥𝑛 < 𝜖
|0 − 𝑥 𝑛 | < 𝜖.

So 𝑓𝑛 → 𝑓 pointwise on 0 ≤ 𝑥 < 1.

At 𝑥 = 1 , 𝑓𝑛 (1) = 1 = 𝑓(1) for all 𝑛 so 𝑓𝑛 → 𝑓 pointwise on 0 ≤ 𝑥 ≤ 1.


Notice that 𝑓𝑛 (𝑥) does not converge uniformly to 𝑓(𝑥) on [0,1] because:
𝑙𝑛𝜖 𝑙𝑛𝜖
|0 − 𝑥 𝑛 | < 𝜖 is equivalent to 𝑛 > , and as 𝑥 → 1, is unbounded
ln(𝑥) ln(𝑥)
above for any fixed 𝜖.

Thus there isn’t any fixed 𝑁 such that 𝑛 ≥ 𝑁 ⇒ |0 − 𝑥 𝑛 | < 𝜖 for all 0 ≤ 𝑥 ≤ 1.
Thus 𝑓𝑛 (𝑥) doesn’t converges uniformly to 𝑓(𝑥) for 0 ≤ 𝑥 ≤ 1.