Functional Analysis
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.
\[c_{00} \subset c_0 \subset l^\infty \]
Definition
($\ell^2$)
\[\ell^2 = \set{(x_n):\sum_{n=1}^\infty |x_n|^2 < \infty }\]
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 \(\ell^2\).
(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.