Functional Analysis: Lecture Notes

1 Background

Functional analysis is a brunch of mathematical analysis that deals with linear structure. In particular, we consider the meta-perspective on mathematical functions and how they interact under summation and scalar multiplication. The most famous example of the application of functional analysis is Fourier series (theory of decomposing any function into an infinite sum of cosines/sines).

The word “functional” was first used in the context of calculus of variation, implying a function whose argument is a function. Vito Volterra furst generalised the idea of linear functional, and then Hadamard was first to use the term in the context. Hadamard founded the modern school of linear functional analysis further developed by Riesz and a group of polish mathematicians around Stefan Banach.

1.1 Metric spaces and metric topology

Definition 1.1. Suppose $𝑋$ is a given non-empty set where we want to define a notion of distance. A metric/distance on $𝑋$ is a function $𝑑 : 𝑋 \times 𝑋 \to [0, \infty)$ where the following properties hold true for all $𝑥, 𝑦, 𝑧 \in 𝑋$ :

$𝑑 (𝑥, 𝑦) = 0 \Leftrightarrow 𝑥 = 𝑦$
$𝑑 (𝑥, 𝑦) = 𝑑 (𝑦, 𝑥)$
$𝑑 (𝑥, 𝑦) \leq 𝑑 (𝑥, 𝑧) + 𝑑 (𝑧, 𝑦)$

$𝑑 (𝑥, 𝑦)$ is said to be the distance from $𝑥$ to $𝑦$ . (M3) is the triangle inequality. The set $ℝ$ has the metric given by $𝑑 (𝑥, 𝑦) = | 𝑥 - 𝑦 |$ , called the standard metric.

Definition 1.2. Let $(𝑋, 𝑑)$ be a metric space.

Given $𝑥 \in 𝑋$ and $𝑟 > 0$ , the open ball of radius $𝑟$ centered at $𝑥$ is the set

𝐵_{𝑟} (𝑥) = {𝑦 \in 𝑋 | 𝑑 (𝑦, 𝑥) < 𝑟}

Suppose $𝐺 \subset 𝑋$ Then, we say that $𝑥 \in 𝐺$ is an interior point of $𝐺$ if there is some $𝑟$ such that $𝐵_{𝑟} (𝑥) \subset 𝐺 .$
$𝐺$ is called open if every $𝑥 \in 𝐺$ is an interior point of $𝐺$
A set $𝐹$ is closed if its compliment $𝐹^{𝑐}$ is closed.

1.2 Dense subset

Definition 1.3. Let $(𝑋, 𝑑)$ be a metric space. $𝐷 \subset 𝑋$ is called dense in $𝑋$ if for any $𝑥 \in 𝑋$ there is $𝑟 > 0$ such that

𝐷 \cap 𝐵_{𝑟} (𝑥) \neq \emptyset .

In words, $𝐷$ is dense if it intersects any open ball.

1.3 Sequences, Cauchy Sequences, and completeness

Definition 1.4. Let $(𝑋, 𝑑)$ be a metric space. A sequence in $𝑋$ is an ordered set of elements in $𝑋$ infinite subset of the natural numbers.

Sequence $(𝑥_{𝑛})$ converges to $𝑥$ if ${lim}_{𝑛 \to \infty} 𝑑 (𝑥_{𝑛}, 𝑥) = 0$ .

Definition 1.5. A sequence $(𝑥_{𝑛})$ is a Cauchy sequence if ${lim}_{𝑛, 𝑚 \to \infty} 𝑑 (𝑥_{𝑛}, 𝑥_{𝑚}) = 0$ .

All convergent sequences are Cauchy, but not all Cauchy sequences are convergent as we may simply remove the point of convergence from our space.

Definition 1.6. A complete metric space is one where all Cauchy sequences are convergent.

Theorem 1.7. Let $(𝑋, 𝑑)$ be a metric space. Then let $(𝑋_{1}, 𝑑_{1})$ also be a metric space with following properties

$(𝑋_{1}, 𝑑_{1})$ is complete
There is a function $𝜑 : 𝑋 \to 𝑋_{𝑛}$ such that for all $𝑥, 𝑦 \in 𝑋$
$𝑑 (𝑥, 𝑦) = 𝑑_{1} (𝜑 (𝑥), 𝜑 (𝑦))$
For any point $𝑥 \in 𝑋_{1}$ there is Cauchy sequence $(𝑥_{𝑛})$ in $𝑋$ such that ${lim}_{𝑛 \to \infty} 𝑥_{𝑛} = 𝜑 (𝑥)$

$𝑋_{1}$ is called a completion of X. Moreover, it is essentially unique.

1.4 Continuity and uniform continuity

For this section we let $(𝑋_{1}, 𝑑_{1})$ and $(𝑋_{2}, 𝑑_{2})$ be metric spaces, and a function $𝑓 : 𝑋_{1} \to 𝑋_{2}$ .

Definition 1.8. Function $𝑓$ is continuous at $𝑥 \in 𝑋_{1}$ if for all $𝜀 > 0$ there is $𝛿 > 0$ such that

𝑑_{1} (𝑥, 𝑦) < 𝛿 ⟹ 𝑑_{2} (𝑓 (𝑥), 𝑓 (𝑦)) < 𝜀 .

Definition 1.9. Function $𝑓$ is uniformly continuous if for all $𝜀 > 0$ there is $𝛿 > 0$ such that, for any $𝑥, 𝑦 \in 𝑋_{1}$ with $𝑑 (𝑥, 𝑦) < 𝛿$ , we have $𝑑 (𝑓 (𝑥), 𝑓 (𝑦)) < 𝜀$ .

Theorem 1.10. Suppose $𝑓 : [𝑎, 𝑏] \to ℝ$ is continuous, then it is also uniformly continuous.

1.5 Bair’s Theorem

Dense subsets are nice because we can use them to approximate any point in a metric space. Hence, it would be very nice for us if the density was preserved under intersection. It is easy to show that if $𝐴_{1}, 𝐴_{2}, \dots, 𝐴_{𝑛}$ are dense and open, then $\cap_{𝑖 = 1}^{𝑛} 𝐴_{𝑖}$ is open and dense in $𝑋$ . However, the result is much more tricky if we let $𝑛 \to \infty$ .

Theorem 1.11. Let $(𝑋, 𝑑)$ be a complete metric space. Let ${𝐴_{𝑘}}_{𝑘 \geq 1}$ be a countable collection of dense open sets, then $⋂_{𝑖 = 1}^{𝑛} 𝐴_{𝑖}$ is dense in $𝑋$ .

Proof. TODO $□$

Here is another version of Bair’s theorem

Theorem 1.12. Let $(𝑋, 𝑑)$ be a complete metric space with

𝑋 = ⋃_{𝑖 = 1}^{\infty} 𝐹_{𝑛}

where each $𝐹_{𝑛}$ is closed. Then at least one of the $𝐹_{𝑛}$ contains an interior point, that is there is some integer $𝑚$ , some point $𝑥 \in 𝐹_{𝑚}$ , and some radius $𝑟 > 0$ , such that $𝐵_{𝑟 (𝑥)} \in 𝐹_{𝑚}$ .

Proof. Let $𝐴_{𝑛} = 𝐹_{𝑛}^{𝑐}$ be the compliment of $𝐹_{𝑛}$ . Since, $𝐹_{𝑛}$ is closed, we have $𝐴_{𝑛}$ open. Then

𝑋 = ⋃_{𝑛 = 1}^{\infty} 𝐹_{𝑛} ⟹ \emptyset = ⋂_{𝑛 = 1}^{\infty} 𝐴_{𝑛},

which by the Theorem 1.11 implies that one of the $𝐴_{𝑛}$ is not dense. Hence, there is some point $𝑥 \in 𝑋$ and radius $𝑟 > 0$ such that $𝐵_{𝑟 (𝑥)} ⋂ 𝐴_{𝑛} = \emptyset$ , which implies that $𝐵_{𝑟 (𝑥)} \subset 𝐹_{𝑛}$ , as desired. $□$

The two versions are equivalent to each other, but we require a more general result.

Definition 1.13. Let $𝑋$ be a metric space. A set $𝑆 \subset 𝑋$ is nowhere dense if its closure $\bar{𝑆}$ contains no interior points.

By definition, each point in $\bar{𝑆}$ can be approximate by a Cauchy sequence in $𝑆$ . $\bar{𝑆}$ lacking interior points implies lack of subset balls, hence, there is no ball that has all of its points approximated by a Cauchy sequence in $𝑆$ , in other words $𝑆$ is not dense in any ball.

Definition 1.14. A set is of the first category if it is a countable union of nowhere dense sets. Otherwise, it is of the second category.

Theorem 1.15. (Bair's Theorem) Any complete metric space $𝑋$ is of the second category.

2 Linear spaces

2.1 Spaces and subspaces

Definition 2.1. A linear space over the scalar field $𝐾$ is a non-empty set $𝑉$ where sum $+ : 𝑉 \times 𝑉 \to 𝑉$ , and scalar multiplication $\cdot : 𝐾 \times 𝑉 \to 𝑉$ are defined. $𝑉$ must be closed under these operations.

Sum must be commutative and associative, and the usual distributive laws between the sum and multiplicatoin apply.

Definition 2.2. Let $𝑉$ be a linear space over $𝐾$ . Then $𝑊 \subset 𝑉$ is said to be linear subspace, or just subspace of $𝑉$ if it is closed under the sum and scalar multiplication.

Exercise 2.3. TODO

Definition 2.4. Let $𝑊_{1}, 𝑊_{2} \subset 𝑉$ . We call the subspace of

2.2 Sequence spaces

Definition 2.5. Define sequence spaces $𝐾^{\infty} (ℕ)$ and $𝐾^{\infty} (ℤ)$ as

𝐾^{\infty} (ℕ) = {𝑥 | 𝑥 : ℕ \to 𝐾} and 𝐾^{\infty} (ℤ) = {𝑥 | 𝑥 : ℤ \to 𝐾} .

These sequence spaces can be turned into linear spaces. Let sequences $𝑥$ and $𝑦$ be sequences and $𝑘 \in 𝐾$ , then

(𝑥 + 𝑦) (𝑛) = 𝑥 (𝑛) + 𝑦 (𝑛) and (𝑘 𝑥) (𝑛) = 𝑘 𝑥 (𝑛) .

However, these sequence spaces are too large and rarely studied, hence, we will define a more practical sequence space.

Definition 2.6. Fix a number $𝑝$ with $1 \leq 𝑝 < \infty$ . The space $ℓ^{𝑝} (ℕ, 𝐾)$ (pronounced little ell) is the space of sequences $𝑥 : ℕ \to 𝐾$ such that

\sum_{𝑛 = 0}^{\infty} {| 𝑥 (𝑛) |}^{𝑝} < \infty .

Notice that for finite $𝑝$ the definition is equivalent to

{(\sum_{𝑛 = 0}^{\infty} {| 𝑥 (𝑛) |}^{𝑝})}^{\frac{1}{𝑝}} < \infty .

We will later define this as a $𝑝$ -norm. Notice that

lim_{𝑝 \to \infty} {(\sum_{𝑛 = 0}^{\infty} {| 𝑥 (𝑛) |}^{𝑝})}^{\frac{1}{𝑝}} = sup | 𝑥 (𝑛) | < \infty,

or in other words $ℓ^{\infty}$ is simply a bounded sequence.

To show that $ℓ^{𝑝}$ is a sequence space we need to show that it is closed under addition and scalar multiplication. The cases for $ℓ^{1}$ and $ℓ^{\infty}$ are simple and left out. Let us consider $ℓ^{𝑝}$ for $1 < 𝑝 < \infty$ .

Exercise 2.7. Let $1 < 𝑝 < \infty$ and show that $ℓ^{𝑝}$ is a sequence space.

By definition, we have

\sum_{𝑛 = 0}^{\infty} {| 𝑥 (𝑛) |}^{𝑝} < \infty ⟹ \sum_{𝑛 = 0}^{\infty} {| 𝑎 𝑥 (𝑛) |}^{𝑝} < \infty

where $𝑎 \in 𝐾$ , hence, it is closed under scalar multiplication. The addition is a bit more tricky. Firstly, consider $𝑓 (𝑡) = 𝑡^{𝑝}$ where $𝑡 > 0$ , taking the second derivative

𝑓^{″} (𝑡) = 𝑝 (𝑝 - 1) 𝑡^{𝑝 - 2} \geq 0,

thus, $𝑓$ is a convex function and we can apply the formal definition of convex functions

𝑓 (𝜆 𝑎 + (1 - 𝜆) 𝑏) \leq 𝜆 𝑓 (𝑎) + (1 - 𝜆) 𝑓 (𝑏) .

$𝜆 = \frac{1}{2}$ yields

𝑓 (\frac{𝑎 + 𝑏}{2}) \leq \frac{𝑓 (𝑎) + 𝑓 (𝑏)}{2} .

Next, we take $𝑥, 𝑦 \in ℓ^{𝑝}$ , hence, from definition we have

\sum_{𝑛 = 0}^{\infty} {| 𝑥 (𝑛) |}^{𝑝} < \infty and \sum_{𝑛 = 0}^{\infty} {| 𝑦 (𝑛) |}^{𝑝} < \infty .

Applying our identities for $𝑓$

\begin{matrix} {(\frac{| 𝑥 (𝑛) + 𝑦 (𝑛) |}{2})}^{𝑝} \leq \frac{{| 𝑥 (𝑛) |}^{𝑝} + {| 𝑦 (𝑛) |}^{𝑝}}{2} \\ ⟹ {| 𝑥 (𝑛) + 𝑦 (𝑛) |}^{𝑝} \leq 2^{𝑝 - 1} ({| 𝑥 (𝑛) |}^{𝑝} + {| 𝑦 (𝑛) |}^{𝑝}) \\ ⟹ \sum_{𝑛 = 0}^{\infty} {| 𝑥 (𝑛) + 𝑦 (𝑛) |}^{𝑝} \leq 2^{𝑝 - 1} \sum_{𝑛 = 0}^{\infty} ({| 𝑥 (𝑛) |}^{𝑝} + {| 𝑦 (𝑛) |}^{𝑝}) < \infty . \end{matrix}

2.3 Function spaces

TODO