Real-Analysis-Approximation-of-Lp-Functions, guaranteed and verified 100% Pass

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Approximation of 𝐿𝑝 Functions


We have already seen that if 𝑓 ∈ 𝐿1 (ℝ) then given any 𝜖 > 0 there is a simple
function 𝜂, a step function 𝑠, and a continuous function 𝑔, all with finite support
such that:

∫ℝ |𝑓 − 𝜂 | < 𝜖 , ∫ℝ |𝑓 − 𝑠| < 𝜖 , and ∫ℝ |𝑓 − 𝑔| < 𝜖 .
We will now extend these results to 𝑓𝜖𝐿𝑝 (𝐸).



Def. Let 𝑋 be a normed linear space. Given two subsets 𝐹 and 𝐺 of 𝑋 with
𝐹 ⊆ 𝐺, we say 𝐹 is dense in 𝐺 if for each 𝑔 ∈ 𝐺 and 𝜖 > 0 there is an element
𝑓 ∈ 𝐹 where ‖𝑓 − 𝑔‖ < 𝜖.


Notice that 𝐹 is dense in 𝐺 if and only if for each 𝑔 ∈ 𝐺 there is a sequence {𝑓𝑛 }
in 𝐹 such that lim 𝑓𝑛 = 𝑔 in 𝑋.
𝑛→∞

In addition for 𝐹 ⊆ 𝐺 ⊆ 𝐻 ⊆ 𝑋 if 𝐹 is dense in 𝐺 and 𝐺 is dense in 𝐻 then 𝐹 is
dense in 𝐻.



Ex. The rational numbers are dense in ℝ.



Ex. Simple functions, step functions, and continuous functions, all with finite
support are dense in 𝐿1 (ℝ) (and 𝐿1 (𝐸)).

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Prop. Let 𝐸 be a measurable set and 1 ≤ 𝑝 ≤ ∞. Then the simple functions in
𝐿𝑝 (𝐸) are dense in 𝐿𝑝 (𝐸).


Proof: Let 𝑓 ∈ 𝐿𝑝 (𝐸).

First let 𝑝 = ∞.

By definition, there is a subset of measure 0, 𝐸0 , such that 𝑓 is bounded on
𝐸~𝐸0 .

From the simple approximation lemma, there is a sequence of simple functions
{𝜑𝑛 } on 𝐸~𝐸0 such that |𝜑𝑛 | ≤ |𝑓| for all 𝑛 on 𝐸~𝐸0 with 𝜑𝑛 → 𝑓
uniformly on 𝐸~𝐸0 .


Since 𝑓 ∈ 𝐿∞ (𝐸), it is bounded a.e and since 𝜑𝑛 → 𝑓 uniformly on 𝐸~𝐸0 , it
converges in 𝐿∞ (𝐸).

Thus simple functions are dense in 𝐿∞ (𝐸).



Now assume 1 ≤ 𝑝 < ∞.

By the simple approximation theorem, there is a sequence of simple
functions 𝜑𝑛 → 𝑓 pointwise on 𝐸 with |𝜑𝑛 | ≤ |𝑓| for all 𝑛.


By the integral comparison test 𝜑𝑛 ∈ 𝐿𝑝 (𝐸) for all 𝑛 since

|𝜑𝑛 |𝑝 ≤ |𝑓|𝑝 thus ∫𝐸 |𝜑𝑛 |𝑝 ≤ ∫𝐸 |𝑓|𝑝 < ∞.

Now notice that:

|𝜑𝑛 − 𝑓|𝑝 ≤ 2𝑝 (|𝜑𝑛 |𝑝 + |𝑓|𝑝 ) ≤ 2𝑝+1 |𝑓|𝑝 on 𝐸.