Analysis 2-Uniform-Convergence of Fourier-Series, guaranteed and verified 100% Pass

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Uniform Convergence of Fourier Series



Up to now, we have talked mostly about 𝐿2 -convergence of Fourier series.
When does 𝑆𝑛 (𝑓) converge uniformly to 𝑓?



Before we answer this question we need another form of the Cauchy-Schwarz
inequality. To do this we first want to consider the set of all real sequences,
𝑥 = {𝑥𝑛 }, such that ∑∞ 𝑝
𝑛=1|𝑥𝑛 | < ∞ for 1 ≤ 𝑝 < ∞. We call this set 𝓵𝒑 .



In fact, ℓ𝑝 is a vector space under coordinatewise addition and we can define a
norm on this vector space by:


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‖𝑥 ‖𝑝 = (∑∞ 𝑝 𝑝
𝑛=1|𝑥𝑛 | ) where 𝑥 is a sequence in ℓ𝑝




To prove uniform convergence of a Fourier series to a function 𝑓(𝑥) under the
appropriate conditions we will need the following form of the Cauchy-Schwarz
inequality for ℓ2 .


Cauchy-Schwarz inequality: ∑∞
𝑖=1|𝑥𝑖 𝑦𝑖 | ≤ ‖𝑥 ‖2 ‖𝑦‖2



𝑥, 𝑦 ∈ ℓ2 (Note: ∑∞
𝑖=1 𝑥𝑖 𝑦𝑖 is a dot product for ℓ2 )

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Proof:

Let’s write < 𝑥, 𝑦 > = ∑∞
𝑖=1 𝑥𝑖 𝑦𝑖 .

Then < 𝑥, 𝑥 > = ∑∞ 2 2
𝑖=1 𝑥𝑖 = ‖𝑥 ‖2 .

If 𝑡 ∈ ℝ, then:

0 ≤ ‖𝑥 + 𝑡𝑦‖22 = < 𝑥 + 𝑡𝑦, 𝑥 + 𝑡𝑦 >
= < 𝑥, 𝑥 > +2𝑡 < 𝑥, 𝑦 > +𝑡 2 < 𝑦, 𝑦 >
= ‖𝑥 ‖22 + 2𝑡 < 𝑥, 𝑦 > +𝑡 2 ‖𝑦‖22 .

This is a quadratic in 𝑡 that’s nonnegative so:

𝐴𝑡 2 + 𝐵𝑡 + 𝐶 ≥ 0
𝐵2 − 4𝐴𝐶 ≤ 0.
Or in ths case,

(2 < 𝑥, 𝑦 >)2 − 4‖𝑥 ‖22 ‖𝑦‖22 ≤ 0
< 𝑥, 𝑦 >2 ≤ ‖𝑥 ‖22 ‖𝑦‖22
|< 𝑥, 𝑦 >| ≤ ‖𝑥 ‖2 ‖𝑦‖2
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|∑∞
𝑖=1 𝑥𝑖 𝑦𝑖 | ≤ (∑ ∞ 2 2 ∞ 2 2
𝑖=1 𝑥𝑖 ) (∑𝑖=1(𝑦𝑖 ) ) .

The same holds for:

𝑥 = (|𝑥1 |, |𝑥2 |, |𝑥3 |, … )
𝑦 = (|𝑦1 |, |𝑦2 |, |𝑦3 |, … ), so
1 1
∑∞
𝑖=1|𝑥𝑖 ||𝑦𝑖 | ≤ (∑ ∞ 2 2 ∞ 2 2
𝑖=1 𝑥𝑖 ) (∑𝑖=1 𝑦𝑖 )
1 1
∑∞
𝑖=1|𝑥𝑖 𝑦𝑖 | ≤ (∑ ∞ 2 2 ∞ 2 2
𝑖=1 𝑥𝑖 ) (∑𝑖=1 𝑦𝑖 ) .