Notes while learning about functional analysis.

**Metric Spaces**: It is defined as the space which has a set of points, say X, and a distance function, d, where the distance function obeys three rules:

- d(x, y) > 0
- d(x, y) = d(y, x)
- d(x, y) <= d(z, x) + d(z, y) (triangle inequality)
- Examples:
*R*and the Eucilidean distance,^{n}*Complex*plane and |x-y| norm distance.

**ε-ball**: Also denoted by B_{ε}(x), are defined as {y ∈ X | d(x, y) < ε}.

**Open sets**: If *A* is a set, and *A* ⊂ *X*, it is open if every point inside B_{ε}(x) belongs to it (for all x ∈ A).

**Boundary Points**: If any B_{ε}(x) has points outside of *A*, no matter how small the ε value, it is the boundary piont of the set *A*. Here x ∈ *X*. Than B_{ε}(x) is said to have points from both *A* and it’s compliment *A ^{c}*. All the boundary points of

*A*are denoted as

*∂A*.

**Closed Sets**: A set, *A*, is defined to be closed, if it’s compliment is open.

**Closure**: The smallest closed set, is simply, *A* ∪ *∂A*. Remember :

*A*is open if,*A*∩*∂A*= ∅.*A*is closed if,*A*∩*∂A*=*A*.

In the example, where *X* = (1, 3] ∪ (4, ∞), the set *A* = (1, 3] is both open AND closed. (Since *A ^{c}* is also open, by definition,

*A*is closed).

**Convergence**: A sequence is defined as a set of points arranged in increasing order, such as x_{1} < x_{2} < x_{3} … x_{n}, also denoted by {x_{i}}. A sequence is said to be *convergent* when, in the limit, it approaches a single value, for example, a sequence, {a}^{N}_{i} converges to a single value à. Given a metric space (X, *d*), we can define the ε-ball, B_{ε}(x̃), for x ∈ X, such that,

- ∀ε > 0, ∃N ∈ ℕ, ∀n >= N : d(x, x̃) < ε

which means that for any convergent sequence, the ε-ball around it contains all the sequence points (as the ε increases, it engulfs more points from the sequence hence for every ε > 0, every point in sequence is covered).

- x̃ = lim
_{n->∞}x_{n}

Proposition: A characterstic of a closed set is: any limit of a convergent sequence is inside the closed set itself.

Proof(by contraposition)

- Let the sequence be {a
_{n}} and the limit be ã. Let’s assume that A is not closed.- It implies that the compliment of A, A
^{c}is not open either.- If ã ∈ A
^{c}, then there exists at least one such point in B_{ε}(ã) that overlaps with A, which means, B_{ε}(ã) ∩ A ≠ ∅.- Which means that there is a sub-sequence, {a
_{n}}, which lies in A, but it’s limit does not (since ã ∈ A^{c}).- Hence lim
_{n->∞}a_{n}= ã ∉ A.

**Cauchy Sequence**: A sequence which is defined by as

- ∀ε > 0, ∃N ∈ ℕ, ∀ n, m >= N : d(x
_{n}, x_{m}) < ε - It is a generalization of a convergent sequence, in which we shift our focus from the distance between the limit and all the points minimising to just the distance between the points to be minimising.
- Which mean after a certain index,
*N*, the distance between two points is always less than ε, for any ε greater than zero.

**Complete Metric Space** : A metric space is said to be complete when all of it’s Cauchy sequences converge. This simply means that for a complete metric space, Cauchy sequnces = Convergent sequences.

- Example: (0, 3), where d := |x - y| is incomplete when with an index of n, m ∈
*N*, the distance -> 0 as n, m -> ∞ but zero is excluded from this set, hence it’s incomplete - [0, 3] is complete, since zero is included in this set.
- (0, 3) under the metric, d := {1 if x ≠ y and 0 if x = y} is complete. Take a Cauchy sequence {x
_{n}}_{n∈N}, where by definition, d (x_{n}, x_{m}) < ε. For any ε, the only way this is possible is if x_{n}= x_{m}(which makes d = 0), hence there is only one point, the limit after the index n (since all the points after it must be the same), hence this Cauchy sequence is convergent, which makes this metric space complete.

**Norm** : Let X be a 𝔽-vector space. (Where 𝔽 = {ℝ, ℂ}). The norm, denoted by ||.|| is a map that transforms a vector to [0, ∞). The norm fulfils the three properties of the distance function as well.

- ||x|| = 0, only when x = 0 (positive definite)
- ||λx|| = λ||x|| (absolutely homogeneous)
- ||x+y|| <= ||x|| + ||y|| where x, y ∈ X (triangle inequality) (Here a since you cannot pull out just the addition, and always get an inquality, it is non-linear)

Hence the space defined by (X, ||.||) is called a **normed space**. A normed space is a special case of a metric space (by taking the end points of a vector, it can be seen as the distance between those two points, hence the distance function of a metric space, d_{||.||}).

**Banach Space** : If a given metric space with a possible norm, (X, d_{||.||}) is also a complete one, then the underlying norm space, (X, ||.||) is called the banach space. All the properties of a metric space apply as well. The norm function is what ties the real-complex vector space and a complete metric space, hence also defining a possible Banach space.

**l ^{P} (ℕ,𝔽)**: It is defined as a collection of all the sequences in 𝔽, which follow the condition: ∑

_{n=1}∞|x|

_{p}< ∞. Here x = {x

_{n}}, which is a sequence in 𝔽. l

^{P}becomes a vector space if the summation condition is ignored. We define a norm over such vector space, ||.||

_{p}: l

^{P}-> [0, ∞). The norm is given by the formula, (∑

_{n=1}∞|x|

_{p})

^{1/P}, again, x is a sequence.

Assumption: l^{P}is a 𝔽-vector space, and ||.||_{p}is a norm on it.

Proof: (l^{P}, ||.||_{p}) is a banach space.Let {x

^{(k)}} be a cauchy sequence in l^{P}. Since we are considering sequences of sequences,

x^{(1)}= (x^{(1)}_{1}, x^{(1)}_{2}, x^{(1)}_{3}, x^{(1)}_{4}, x^{(1)}_{5}, …)

x^{(2)}= (x^{(2)}_{1}, x^{(2)}_{2}, x^{(2)}_{3}, x^{(2)}_{4}, x^{(2)}_{5}, …)

x^{(3)}= (x^{(3)}_{1}, x^{(3)}_{2}, x^{(3)}_{3}, x^{(3)}_{4}, x^{(3)}_{5}, …)

x^{(4)}= (x^{(4)}_{1}, x^{(4)}_{2}, x^{(4)}_{3}, x^{(4)}_{4}, x^{(4)}_{5}, …)

x^{(5)}= (x^{(5)}_{1}, x^{(5)}_{2}, x^{(5)}_{3}, x^{(5)}_{4}, x^{(5)}_{5}, …)

. = (., ., ., ., ., …)

. = (., ., ., ., ., …)

. = (., ., ., ., ., …)The goal here is to prove that the sequence on the left-hand side, x

^{(k)}, has a limit, which is also in l^{P}

We start by picking a random column, say the 4th column, call it : x^{(k)}_{4}. This particular sequence is, by itself, a banach space (since it is single valued space with a norm over it, ||.||_{p}).

The normed distance between elements in this sequence will be: |x^{(k)}_{4}- x^{(l)}_{4}|^{P}.

These normed distance would be lesser than the combined such distance of all columns: ∑∞_{n=1}|x^{(k)}_{n}- x^{(l)}_{n}|^{P}

Hence it follows that: |x^{(k)}_{4}- x^{(l)}_{4}|^{P}< ∑∞_{n=1}|x^{(k)}_{n}- x^{(l)}_{n}P = ||x^{(k)}- x^{(l)}||^{P}_{p}< ε.

But, since the latter is a Cauchy sequence, the former, that is the indivisual columns, are all also a Cauchy sequence. Since these columns are a Cauchy sequence AND belong to a banach space, they have a limit as well. Let’s call the limit, x’^{(k)}.

Hence, each column has a limit. We can collect every columns limit into a sequence: x’ = (x’_{1}, x’_{2}, x’_{3}, x’_{4}, x’_{5}, …)

Now, we have to show that our original Cauchy sequence, x^{(k)}approaches it’s limit, x’ (which means it’s indeed a convergent sequence), or |x^{(k)}- x’| < ε.

We start by pulling out the infinity through taking the limit, essentially restricting the columns to N.

lim_{n->∞}|x^{(k)}_{n}- x’_{n}|^{P}.

Now, we also have a second infinite sequence, in x’. We pull that out as well:

lim_{l->∞}lim_{n->∞}|x^{(k)}_{n}- x^{(l)}_{n}|^{P}.

Now, the inner normed distance should look familiar: we know that the indivisual columns can be bounded by any arbirtary upper bound, which we canselect. Let it be ε’.

Since, |x^{(k)}_{n}- x^{(l)}_{n}|^{P}< (ε’)^{P}.

|x^{(k)}_{n}- x^{(l)}_{n}| < ε’.

Hence we simply take the limit now: |x^{(k)}- x’| < ε’, and set ε’ = ε/2.

Hence we have proved that any sequence {x^{(k)}} converges to a limit, x’ within some arbitary ε. This means that all Cauchy sequences are convergent, hence the (l^{P}, ||.||_{p}) is a banach space.(

Every statement about Cauchy sequence and it’s upper-bound being ε, is true after some kth index, where k > 0.)

**Inner Product** measures the distance, length and the angle between two vectors in 𝔽-vector space. It is denoted by

- < x, x > >= 0 and < x, x > is only equal to 0 if and only if x = 0 vector.
- < x, y > = < y, x > if 𝔽 = ℝ AND < x, y > = < y, x >
_{conjugate}if 𝔽 = ℂ - < x, y
_{1}+ y_{2}> = < x, y_{1}> + < x, y_{2}> and < x, λy > = λ< x, y >, if 𝔽 = ℝ and < x, λy > = λ_{conjugate}< x, y >, if 𝔽 = ℂ

**Hilbert Space**. If (X, ||.||_{p}) is a banach space, than (X, <., .>) is considered to be a Hilbert space, that is a space which has an inner product that measures distance, length and angle while also being complete metric space.

**Examples of Hilbert Spaces**. Here are some important examples of hilbert spaces:

- ℝ
^{n}and ℂ^{n}where the norm, < x, y > is := ∑^{n}_{i=1}x_{i}y_{i}(x-bar for complex case). - The space l
^{2}(N, 𝔽), where the norm, < x, y > is := ∑^{∞}_{i=1}x_{i}y_{i}(x-bar for complex case).

A quick check for the l

^{2}being a hilbert space, but check the norm conditions< x, x > := ∑

^{∞}_{i=1}xbar_{i}x_{i}= ∑^{∞}_{i=1}|x|^{2}which is >= 0.

but < x, x > = 0. Hence |x|^{2}= 0, so x has to be the zero vector< x, y > := (∑

^{∞}_{i=1}ybar_{i}x_{i})^{-}= ∑^{∞}_{i=1}(ybar_{i}x_{i})^{-}

which is ∑^{∞}_{i=1}y_{i}xbar_{i}= < y, x >.< x, λy > = ∑

^{∞}_{i=1}x_{i}λy_{i}= λ ∑^{∞}_{i=1}x_{i}y_{i}= λ< x, y >(

Here xbar or ybar means the conjugate)