Analysis 2-The Space Ralphaab, guaranteed and verified 100% Pass

1


The Space 𝑅𝛼 [𝑎, 𝑏]


Now we want to understand what kind of structure 𝑅𝛼 [𝑎, 𝑏] has. Is it a vector
space? If so, is there a norm on that vector space and is there a norm that will
make it complete?



Theorem: Let 𝑓, 𝑔 ∈ 𝑅𝛼 [𝑎, 𝑏] and let 𝑐 ∈ ℝ. Then
𝑏 𝑏
i. 𝑐𝑓 ∈ 𝑅𝛼 [𝑎, 𝑏] and ∫𝑎 𝑐𝑓𝑑𝛼 = 𝑐 ∫𝑎 𝑓𝑑𝛼 .
𝑏 𝑏 𝑏
ii. 𝑓 + 𝑔 ∈ 𝑅𝛼 [𝑎, 𝑏] and ∫𝑎 (𝑓 + 𝑔)𝑑𝛼 = ∫𝑎 𝑓𝑑𝛼 + ∫𝑎 𝑔𝑑𝛼 .
𝑏 𝑏
iii. ∫𝑎 𝑓𝑑𝛼 ≤ ∫𝑎 𝑔𝑑𝛼 whenever 𝑓 ≤ 𝑔 on [𝑎, 𝑏].
𝑏 𝑏
iv. |𝑓| ∈ 𝑅𝛼 [𝑎, 𝑏] and | ∫𝑎 𝑓𝑑𝛼 | ≤ ∫𝑎 |𝑓|𝑑𝛼 ≤ ‖𝑓‖∞ [𝛼(𝑏) − 𝛼(𝑎)].
𝑏 𝑏 2 1 𝑏 2 1
v. 𝑓𝑔 ∈ 𝑅𝛼 [𝑎, 𝑏] and | ∫𝑎 𝑓𝑔𝑑𝛼| ≤ (∫𝑎 𝑓 𝑑𝛼) (∫𝑎 𝑔 𝑑𝛼 )2 .
2


This is called the Cauchy-Schwarz inequality.


Proof. If 𝑃 is any partition of [𝑎, 𝑏] and 𝑐 ≥ 0 then the supremum of 𝑐𝑓 on
𝑥𝑖−1 ≤ 𝑥 ≤ 𝑥𝑖 is 𝑐𝑀𝑖 and the infimum of 𝑐𝑓 is 𝑐𝑚𝑖 .
Thus 𝑈𝛼 (𝑐𝑓, 𝑃 ) = 𝑐𝑈𝛼 (𝑓, 𝑃 ), 𝐿𝛼 (𝑐𝑓, 𝑃) = 𝑐𝐿𝛼 (𝑓, 𝑃 ) and
𝑈𝛼 (𝑐𝑓, 𝑃 ) − 𝐿𝛼 (𝑐𝑓, 𝑃 ) = 𝑐(𝑈𝛼 (𝑓, 𝑃 ) − 𝐿𝛼 (𝑓, 𝑃 )) = |𝑐 |(𝑈𝛼 (𝑓, 𝑃 ) − 𝐿𝛼 (𝑓, 𝑃 )).


If 𝑐 < 0 then 𝑐𝑓 = −|𝑐|𝑓 and:

𝑈𝛼 (𝑐𝑓, 𝑃 ) = |𝑐 |𝑈𝛼 (−𝑓, 𝑃 ) = −|𝑐 |𝐿𝛼 (𝑓, 𝑃) = 𝑐𝐿𝛼 (𝑓, 𝑃)
Similarly: 𝐿𝛼 (𝑐𝑓, 𝑃) = −|𝑐|𝑈𝛼 (𝑓, 𝑃)
So 𝑈𝛼 (𝑐𝑓, 𝑃 ) − 𝐿𝛼 (𝑐𝑓, 𝑃) = −|𝑐|(𝐿𝛼 (𝑓, 𝑃 ) − 𝑈𝛼 (𝑓, 𝑃 ))
= |𝑐|(𝑈𝛼 (𝑓, 𝑃) − 𝐿𝛼 (𝑓, 𝑃)).

, 2


But since 𝑓 ∈ 𝑅𝛼 [𝑎, 𝑏] given any 𝜖 > 0, there exists a partition, 𝑃, such that
𝜖
(𝑈𝛼 (𝑓, 𝑃 ) − 𝐿𝛼 (𝑓, 𝑃 )) <
|𝑐 |

Thus for that partition:
𝜖
𝑈𝛼 (𝑐𝑓, 𝑃 ) − 𝐿𝛼 (𝑐𝑓, 𝑃 ) = |𝑐 |(𝑈𝛼 (𝑓, 𝑃) − 𝐿𝛼 (𝑓, 𝑃)) < |𝑐 | (|𝑐|) = 𝜖 .

So 𝑐𝑓 ∈ 𝑅𝛼 [𝑎, 𝑏].


𝑏 𝑏̅ 𝑏̅
∫𝑎 𝑐𝑓𝑑𝛼 = ∫𝑎 𝑐𝑓 𝑑𝛼 = 𝑐 ∫𝑎 𝑓 𝑑𝛼 if 𝑐 ≥ 0 since 𝑈𝛼 (𝑐𝑓, 𝑃 ) = 𝑐𝑈𝛼 (𝑓, 𝑃 )
𝑏
= 𝑐 ∫_𝑎 𝑓𝑑𝛼 if 𝑐 < 0 since 𝑈𝛼 (𝑐𝑓, 𝑃) = 𝑐𝐿𝛼 (𝑓, 𝑃 )
𝑏̅ 𝑏 𝑏
and ∫𝑎 𝑓 𝑑𝛼 = ∫_𝑎 𝑓𝑑𝛼 = ∫𝑎 𝑓𝑑𝛼.
𝑏 𝑏
So ∫𝑎 𝑐𝑓𝑑𝛼 = 𝑐 ∫𝑎 𝑓𝑑𝛼 .



ii. Notice that for any partitions 𝑃, 𝑄 if 𝑓, 𝑔 ∈ 𝑅𝛼 [𝑎, 𝑏]:
𝐿𝛼 (𝑓, 𝑃) + 𝐿𝛼 (𝑔, 𝑄) ≤ 𝐿𝛼 (𝑓, 𝑃 ∪ 𝑄) + 𝐿𝛼 (𝑔, 𝑃 ∪ 𝑄)
≤ 𝐿𝛼 (𝑓 + 𝑔, 𝑃 ∪ 𝑄)
(since inf⁡(𝑓) + inf⁡(𝑔) ≤ inf(𝑓 + 𝑔))

≤ 𝑈𝛼 (𝑓 + 𝑔, 𝑃 ∪ 𝑄)
≤ 𝑈𝛼 (𝑓, 𝑃 ∪ 𝑄) + 𝑈𝛼 (𝑔, 𝑃 ∪ 𝑄)
(since sup(𝑓 + 𝑔) ≤ sup⁡(𝑓) + sup⁡(𝑔))

≤ 𝑈𝛼 (𝑓, 𝑃 ) + 𝑈𝛼 (𝑔, 𝑄).