Lecture 3: Quotient Spaces, the Baire Category Theorem and the Uniform Boundedness Theorem

MIT 18.102 Introduction to Functional Analysis, Spring 2021
Instructor: Dr. Casey Rodriguez

We finish introducing the basic notions of functional analysis including subspaces, quotient spaces, and seminorms (almost norms). Then we prove some necessary fundamental theorems for moving forward: the Baire Category Theorem and the Uniform Boundedness Theorem. The former theorem is the only category theory known to the instructor.

License: Creative Commons BY-NC-SA

[akrishna]

Timestamps and Summary for Lecture 3: Quotient Spaces, the Baire Category Theorem and the Uniform Boundedness Theorem

0:00 - Recap of Lecture 2
Last lecture: linear operators between normed vector spaces are continuous if and only if they are bounded. The space of bounded operators between two such NVS's itself forms an NVS, under the operator norm. When the target is a Banach space, this space of bounded operators also becomes Banach: this is a way to identify new Banach spaces from existing ones, by considering all bounded operators into them.

0:59 - Subspaces
A subspace of a given vector space is internally closed under addition and scaling (linear combinations). Theorem: a subspace W of a Banach space V is itself Banach if and only if it is a closed subspace of V with respect to the metric induced by the norm. Closure amounts to showing that W contains all of its limit points (every convergent sequence in W has its limit in W). We can take any sequence in W; by hypothesis (completeness) it is Cauchy and converges in W. Conversely, if the subspace is closed, we can take a Cauchy sequence in W -- regarding it as a sequence in the ambient Banach space V, it converges to an element in W by closure. This shows every Cauchy sequence in W converges in W, so W is Banach.

5:14 - Quotient Spaces
Another crucial way to form new spaces from old by passing to substructures involves the process of taking quotients. Given a subspace W of V, introduce (and verify) an equivalence relation on vectors in V by identifying two vectors together if their difference lies in W. This allows us to form the quotient set V/W under that equivalence relation; furthermore, it naturally inherits the operations from V and becomes a vector space in its own right.

11:31 - From Seminorms to Norms
Recall that a seminorm satisfies homogeneity and the triangle inequality, but not necessarily definiteness -- consider the norm of the derivative of a function; it exhibits the former two properties but resolves to zero on constant functions. Theorem: a seminorm on a vector space descends to a norm on its quotient taken with the subspace of all vectors on which the seminorm is zero. Intuitively: the presence of vectors with seminorm zero represent an "obstruction" of the seminorm from becoming a norm. By collapsing the obstruction in the quotient, we obtain a true norm.

23:11 - Baire Category Theorem
A cornerstone of general topology with vital applications in analysis. The version given here states that if a complete metric space can be expressed as the countable union of closed sets, then at least one of the closed sets contains an open ball (eq. has an interior point/nonempty interior). It can be reformulated to state that the countable intersection of open, dense sets in a complete metric space remains dense. (Note: a second result, not as relevant to functional analysis, states that the same holds true for locally compact regular spaces. This is especially important in the study of locally compact Hausdorff spaces. The version proven here requires some form of the Axiom of Choice; over ZF it turns out equivalent to a weaker form of Choice that is sufficient to build up most of real analysis).

50:39 - The Uniform Boundedness Theorem
Our first major functional analysis result: it states that a sequence of pointwise-bounded operators on a Banach space is also bounded uniformly. More precisely, a sequence of linear operators from a Banach space to a normed vector space is bounded uniformly in the operator norm as soon as each individual operator is bounded. By showing that each of these operators have bounds that might vary with each point, this theorem claims that a bound can be chosen for all operators in the sequence that is independent of the point, and can be applied globally.

[amonal]

There is a confusion between balls in the whole space and balls in a metric space (with same metric but different set). At 58:10 he states that C_k contains an open ball and assumes that it means a ball in B. But by the Baire theorem it is only have to be a ball in metric space M (i.e. intersection of open ball with closed ball B(0, 1)=M). So it is possible that ||b_0|| = 1 and latter at 59:31 b_0+b is in B(b_0, \delta_0) but not in C_k which forbids conclusions about sup(T_n(b_0+b))<=k.

I do not know why he did not kept it simpler with M=B and dropping ||b|| <= 1 requirement in C_k.

[aaronrobertcattell]

very interesting every time i watch i learn thank you

[pascalpascal]

30 :00 Proof of the Baire’s theorem is wrong “Since M is complete it contains an open ball…”
The correct proof of this version of the theorem is given in the textbook “Functional Analysis, Spectral Theory, and Applications” by Einsiedler & Ward, Springer, Graduate Texts in Mathematics, page 127.

[meaningseeker]

for this part, it seems that there is flaw in the proof of Uniform boundedness principle. I think the proof on wikipedia is flawless, but in general they are similar.

[enricods]

One question about Baire's theorem proof. In the proof is claimed: X contains an open ball. But, why is this the case? In Royden's Real Analysis first Baire's theorem is stated and proved, then as a corollary we obtain that a complete space has to be "second cathegory"; i.e., contains an open ball.

[enterprixe]

I do not understand the claim at 30:00. Why does M clearly contains an open ball?

[NickZhaoable]

For Blair Category Theorem, is that equivalent to saying that any complete metric space has a dense subset?
And so, for the complete set of linear bound operators, inside its dense subset, piecewise bound infers uniform bound.

[queenpost]

I want to see what's on the board. Not the lecturer.

[carl]

From (*) at 53:07, what is wrong with this...
for all b, sup_{n} || T_n b || = C_b < \infty
=> for all b, for all n, || T_n b || ≤ C = sup_b(C_b) < \infty
=> for all n, for all ||b||=1, || T_n b || ≤ C
=> for all n, sup_{||b||=1} || T_n b || ≤ C
=> for all n, || T_n || ≤ C
=> sup_{n} || T_n || ≤ C < \infty
I'm guessing there must something subtly wrong with the re-ordering of things, but not sure why

[gaulindidier]

When non algebraist talk about category theory 23:56