Functional Analysis

2026-01-09

notes

these are some more informal notes that I have made after having taken the rigorous Analysis course.

  • \(c_{00}\) can be thought of as ‘finite vectors’. all \(\mathbb{R}^n\) tuples can be written as elements of \(c_{00}\) with infinitely many zeros after the $n$th term.
  • \(c_0\) are all sequences that converge to zero. thus they will include all those in \(c_{00}\) and more.
  • \(\ell^2\) are the vectors that fade fast enough for their energy 𐃏 to stay finite
  • \(\ell^\infty\) are the infinite vectors that never blow up – their entries are bounded.

\[c_{00} \subset c_0 \subset l^\infty \]

Example (halving numbers)

These numbers are mildly interesting. We know that their sum equates to 2: \[1+\frac12+\frac14+\frac18+\ldots = 2\]

We know this from school where the multiplicative rate is \(r = \frac12\), \(a=1\) and so \(S_n = \frac{a}{1-r} = \frac{1}{1-\frac{1}{2}} = 2\).

this sequence \((1,\frac12,\frac14,\frac18,\ldots)\) is in \(\ell^2\)

Proof

We just need to check the definition. We glossed over the closed-form of this sequence, but it is not difficult to verify that: \[(1,\frac12,\frac14,\frac18,\ldots) = \sum^\infty_{n=1} \frac{1}{2^{n-1}}\]

from this we have the $x_n$th term to be \(\frac{1}{2^{n-1}}\) and can thus check that the absolute value of this term squared in fact converges:

\[\sum^\infty_{n=1} \left(\frac{1}{2^{n-1}}\right)^2 = \sum_{n=1}^\infty \frac{1}{4^{n-1}} = \frac{1}{1-\frac14} = \frac43 < \infty\]

Example (K2)

an example of a sequence in this space is \(\large \sum_{k=1}^{\infty} \frac{1}{k^2} = (1,\frac14, \frac19,\frac1{16},\ldots)\).

Proof

To show \((1,\tfrac14,\tfrac19,\tfrac1{16},\ldots)\in \ell^2\) we must verify \[ \sum_{n=1}^{\infty} |x_n|^2 \;<\;\infty \quad\text{where } x_n=\frac{1}{n^2}. \] Thus \[ \sum_{n=1}^{\infty} |x_n|^2 \;=\; \sum_{n=1}^{\infty} \frac{1}{n^4}. \]

Integral test: Let \(f(x)=x^{-4}\). Then \(f\) is positive, continuous, and strictly decreasing on \([1,\infty)\). The integral test applies and gives \[ \int_{1}^{\infty} x^{-4}\,dx \;=\; \left[ -\frac{1}{3}x^{-3}\right]_{1}^{\infty} \;=\; \frac{1}{3} \;<\;\infty. \] Hence \(\sum_{n=1}^{\infty} \frac{1}{n^4}\) converges, so \((x_n)\in\ell^2\).

(alternatively) Cauchy criterion with an explicit tail bound.

For \(1< N < M\),\[ \sum_{n=N}^{M} \frac{1}{n^4} \;\le\; \int_{N-1}^{M-1} x^{-4}\,dx \;\le\; \int_{N-1}^{\infty} x^{-4}\,dx \;=\; \frac{1}{3}\,(N-1)^{-3}. \] Given \(\varepsilon>0\), choose \(N>1+\big(\tfrac{1}{3\varepsilon}\big)^{1/3}\). Then every tail \(\sum_{n=N}^{M}\frac{1}{n^4}<\varepsilon\), proving the series is Cauchy, hence convergent.

Therefore \(\sum_{n=1}^{\infty}\frac{1}{n^4}<\infty\), and the sequence belongs to \(\ell^2\).

Definition ($c_{00}$)

\[ c_{00} \;=\; \big\{\, x=(x_n)_{n\ge 1} \;:\; \exists\,N\in\mathbb{N}\ \text{with}\ x_n=0\ \ \forall\,n>N \big\}. \] All sequences with only finitely many non-zero terms.

Equivalently: sequences with finite support (only finitely many non-zero terms).

Definition ($c_{0}$)

\[ c_{0} \;=\; \big\{\, x=(x_n)_{n\ge 1} \;:\; \lim_{n\to\infty} x_n = 0 \big\}, \qquad \|x\|_\infty := \sup_{n\ge 1}|x_n|. \] All sequences that converge to 0. (This norm makes \(c_0\) a Banach space and \(c_{00}\subset c_0\).)

Definition ($\ell^\infty$)

\[ \ell^\infty \;=\; \big\{\, x=(x_n)_{n\ge 1} \;:\; \sup_{n\ge 1}|x_n| < \infty \big\}, \qquad \|x\|_\infty := \sup_{n\ge 1}|x_n|. \] (All bounded sequences; \(c_0\subset \ell^\infty\).)

Definition ($\ell^2$)

\[ \ell^2 \;=\; \big\{\, x=(x_n)_{n\ge 1} \;:\; \sum_{n=1}^\infty |x_n|^2 < \infty \big\}, \qquad \|x\|_2 := \Big(\sum_{n=1}^\infty |x_n|^2\Big)^{1/2}. \] All sequences whose squares are summable.