Lecture 2: Bounded Linear Operators

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

An extremely important example of a Banach space is the space of bounded linear operators. We introduce this space with the corresponding operator norm, allowing us to define the notion of a functional (the same “functional” in the title of the course)!

License: Creative Commons BY-NC-SA

[akrishna]

Timestamps and Summary of Lecture 2: Bounded Linear Operators

0:00 - Review of Banach Spaces
In the previous lecture, we started with vector spaces (algebraic structures defined over a field, closed by operations of addition & scaling). We examined normed vector spaces, i.e. those with a notion of "size". Norms naturally give rise to metrics, furnishing an analytic context. Those normed vector spaces that are complete (every Cauchy sequence converges) with respect to the metric induced by their norm, are called Banach spaces -- central objects of study in functional analysis.

1:13 - Summability:
Definition: A series is summable if its sequence of partial sums converges. It is absolutely summable if the series of its norms (nonnegative numbers) converges, which is stronger than regular summability. Useful theorem: a space is Banach if and only if the converse holds.

19:05 - Operators and Linear Functionals.
Linear operators (or linear maps) between vector spaces are described here as the analog of matrices; there is a correspondence between linear maps of finite-dimensional vector spaces and matrices that represent these transformations on a given basis. Functionals are linear operators from a vector space to its base field.

19:56 - Example of a Linear Operator:
Consider a continuous map from the unit square to the complex numbers that takes in a function in C([0, 1]) and "convolves it" by integrating its product with the aforementioned map over the compact interval [0, 1]. This is linear by the usual properties of integration.

23:16 - Linear operators in full generality from V -> W.
A linear operator between two vector spaces satisfies additivity and scalar homogeneity. As algebraic structures, vector spaces are essentially characterized by their closure under addition and scalar multiplication from a base field. The carrying-over of these properties from V to W indicates that linear operators are the "correct" structure-preserving maps to study between vector spaces (abstractly, K-linear maps are the morphisms in the category of vector spaces over K).

26:30 - Continuous Linear Operators
We describe a linear operator as continuous if it preserves convergent sequences. Equivalently, we can formulate continuity in a topological manner: inverse images of open sets under continuous maps remain open. This is important to distinguish in the infinite-dimensional case -- all norms on finite-dimensional vector spaces are equivalent, hence all operators between f.d. spaces are continuous, but this need not hold for Banach spaces (as we've seen, spaces like C([0, 1]) are infinite-dimensional). Unlike general or f.d. vector spaces, Banach spaces express nontrivial analytic information: to study them appropriately, we must restrict to a suitable class of continuous linear maps that do convey this data.

30:06 - Continuity and Boundedness
Theorem: a linear operator between normed vector spaces is continuous if and only if it is bounded. Informally, knowing that an operator between normed vector spaces "maps arbitrarily close points to arbitrarily close points" tells us the same information that it "maps bounded subsets to bounded subsets". It turns out that it's enough to show this criterion for continuity at a single point, e.g., the zero vector.

44:10 - Example of a Bounded Operator
Interpreting the earlier example as a continuous linear operator from C([0, 1]) -> C([0, 1]), it is clearly continuous when acting on continuous functions. Regarding it as a map between normed vector spaces (in fact, Banach spaces) allows us to view it as a bounded operator, specifically under the supremum norm as discussed in the prior lecture.

49:30 - The Space B(V, W)
Given two normed vector spaces V and W, we define B(V, W) to consist of all bounded (eq. continuous) linear functionals V -> W. Define on it the operator norm, which returns the maximal norm of a vector in the image of a given bounded linear operator; by linearity we don't lose any information by normalizing & considering the supremum over the images of just unit vectors. This is checked to satisfy the norm axioms, turning B(V, W) into a normed vector space in its own right.

58:44 - If W is Banach, so is B(V, W)
Strengthening the previous observation that B(V, W) is a normed vector space, if we assume that W is Banach (V need only be an NVS) then B(V, W) itself becomes a Banach space. Completeness follows from the initial lemma that a space is Banach when a series is absolutely summable if and only if it is summable.

1:21:08 - The (Continuous) Dual Space
For any normed vector space V, define its continuous/topological dual space V' to consist of all bounded linear functionals from V to its base field K. (i.e. B(V, K) with the above notation). These are always Banach - for the l^p sequence spaces, examples include (l^1)' = l^∞, but (l^∞)' is not l^1 (not symmetric!). In general (l^p)' = l^q, where 1/p + 1/q = 1 (conjugate exponents); l^2 is special since it is self-dual. Note: for finite-dimensional vector spaces, V' coincides with the usual (algebraic) dual space of all linear functionals V -> K because continuity is immediate. However, it is strictly a subspace of the algebraic dual in infinite dimension, since one can always construct discontinuous linear functionals on infinite-dimensional vector spaces.

[hausdorffm]

6:02
Theorem A. Let V be a normed space. Then, the following two are equivalent.
1)V is Banach space.
2)If a series of V is absolutely summable, then the series is summable.


1:05:26
Let V, W be normed spaces, and denote by B(V, W) the bounded linear oparators from V to W.
Using the above equivalences, the theorem follows:
Theorem B.
If W is a Banach space, then B(V, W) is also a Banach space.


1:22:17
In the theorem B, if we take W = K, K is, e.g., complex numbers or reals, then
W satisfies the condition of theorem B, so B(V, K) is also a Banach space, which is called
the dual space of V. Here, V is assumed as a vector space over K.
Examples.
dual of L^1 = L^inf
dual of L^2 = L^2
But
dual of L^inf is NOT L^1. !? I expect that this will be proved in later lectures.

[leedanilek]

Finally taking my "first adult analysis class"!

[azizx]

thank you so much for making this so glad

[dneary]

Are you teaching to an empty classroom for this class? Your "I'm going to assume you're laughing" comment led me to believe you were! Which is very impressive.

[foli]

Thanks for these lectures, they are great!

[jhanhadderhernandeztunubal]

1:11:38: There is a typo. It should be T_nV_1 and T_nV_2 instead of TV_1 and TV_2.

[randalllionelkharkrang]

Does this have solutions to assignments to compare solutions?

[thomasjefferson]

See how he was real slow to write example? He wrote exam ... wait 3 seconds ple.

man this teacher is awesome isnt he.

[scientiaetveritas]

Excellent lecture but I am surprised MIT is still using chalkboards. My school switched to whiteboards years ago.

[hobit]

Why not Leipzig continuous and just bounded?

[thomasjefferson]

anyone got a place to find solutions for these exercises? I know theres non released, but it would be super to see if my logic is right or not.

[phaecops]

This is not teaching. This is just copying the textbook to the blackboard and reciting the textbook text. A real teacher would give concrete examples which showed with integers how these functions worked.