Added completeness proofs in functional analysis.
This commit is contained in:
parent
ec3ad8cb31
commit
fbdf8f4c03
1 changed files with 75 additions and 5 deletions
|
@ -120,7 +120,23 @@ The steps in a completeness proof are as follows
|
|||
|
||||
??? note "*Proof*:"
|
||||
|
||||
Will be added later.
|
||||
Let $(x_m)_{m \in \mathbb{N}}$ be a Cauchy sequence in $(\mathbb{R}^n, d)$, then we have
|
||||
|
||||
$$
|
||||
\forall \varepsilon > 0 \exists N \in \mathbb{N}: \forall m, k > N: d(x_m, x_k) = \sqrt{\sum_{j=1}^n \big(x_m(j) - x_k(j) \big)^2} < \varepsilon,
|
||||
$$
|
||||
|
||||
obtains for all $j \in \mathbb{N}$: $|x_m(j) - x_k(j)| < \varepsilon$.
|
||||
|
||||
Which shows that $(x_m(j))_{m \in \mathbb{N}}$ is a Cauchy sequence in $\mathbb{R}$. Suppose that it converged by $x_m(j) \to x(j)$ as $(m \to \infty)$ then $x \in \mathbb{R}^n$ since $x = \big(x(1), \dots, x(n)\big)$.
|
||||
|
||||
Thus for $(k \to \infty)$ we have
|
||||
|
||||
$$
|
||||
d(x_m, x) < \varepsilon \implies x_m \to x,
|
||||
$$
|
||||
|
||||
which implies that $\mathbb{R}^n$ is complete.
|
||||
|
||||
A similar proof exists for the completeness of the Unitary space $\mathbb{C}^n$.
|
||||
|
||||
|
@ -134,7 +150,23 @@ A similar proof exists for the completeness of the Unitary space $\mathbb{C}^n$.
|
|||
|
||||
??? note "*Proof*:"
|
||||
|
||||
Will be added later.
|
||||
Let $(x_n)_{n \in \mathbb{N}}$ be a Cauchy sequence in $(C,d)$, then we have
|
||||
|
||||
$$
|
||||
\forall \varepsilon > 0 \exists N \in \mathbb{N}: \forall n, m > N: d(x_n, x_m) = \max_{t \in [a,b]} |x_n(t) - x_m(t)| < \varepsilon,
|
||||
$$
|
||||
|
||||
obtains for all $t \in [a,b]$: $|x_n(t) - x_m(t)| < \varepsilon$.
|
||||
|
||||
Which shows that $(x_m(t))_{m \in \mathbb{N}}$ for fixed $t \in [a,b]$ is a Cauchy sequence in $\mathbb{R}$. Since $\mathbb{R}$ is complete the sequence converges; $x_m(t) \to x(t)$ as $m \to \infty$.
|
||||
|
||||
Thus, for $m \to \infty$ we have
|
||||
|
||||
$$
|
||||
d(x_n, x) = \max_{t \in [a,b]} | x_n(t) - x(t) | < \varepsilon,
|
||||
$$
|
||||
|
||||
hence $\forall t \in [a,b]: | x_n(t) - x(t) < \varepsilon$, obtaining convergence to $x_n \to x$ as $n \to \infty$ and $x \in C$ which implies that $C$ is complete.
|
||||
|
||||
While $C$ with a metric $d$ defined by
|
||||
|
||||
|
@ -151,14 +183,30 @@ for all $x,y \in C$ is incomplete.
|
|||
> *Proposition 6*: the space $l^p$ with $p \geq 1$ and the metric $d$ defined by
|
||||
>
|
||||
> $$
|
||||
> d(x,y) = (\sum_{j \in \mathbb{N}} | x(j) - y(j) |^p)^\frac{1}{p},
|
||||
> d(x,y) = \Big(\sum_{j \in \mathbb{N}} | x(j) - y(j) |^p\Big)^\frac{1}{p},
|
||||
> $$
|
||||
>
|
||||
> for all $x,y \in l^p$ is complete.
|
||||
|
||||
??? note "*Proof*:"
|
||||
|
||||
Will be added later.
|
||||
Let $(x_n)_{n \in \mathbb{N}}$ be a Cauchy sequence in $(l^p,d)$, then we have
|
||||
|
||||
$$
|
||||
\forall \varepsilon > 0 \exists N \in \mathbb{N}: n, m > N: d(x_n, x_m) = \Big(\sum_{j \in \mathbb{N}} |x_n(j) - x_m(j)|^p\Big)^\frac{1}{p} < \varepsilon,
|
||||
$$
|
||||
|
||||
obtains for all $j \in \mathbb{N}$: $|x_n(j) - x_m(j)| <\varepsilon$.
|
||||
|
||||
Which shows that $(x_m(j))_{m \in \mathbb{N}}$ for fixed $j \in \mathbb{N}$ is a Cauchy sequence in $\mathbb{C}$. Since $\mathbb{C}$ is complete the sequence converges; $x_m(j) \to x(j)$ as $m \to \infty$.
|
||||
|
||||
Thus, for $m \to \infty$ we have
|
||||
|
||||
$$
|
||||
d(x_n, x) = \Big(\sum_{j \in \mathbb{N}} |x_n(j) - x(j)|^p\Big)^\frac{1}{p} < \varepsilon,
|
||||
$$
|
||||
|
||||
implies that $x_n - x \in l^p$ and $x = x_n - (x_n - x) \in l^p \implies x \in l^p$ and $x_n \to x$ as $n \to \infty$ which implies that $l^p$ is complete.
|
||||
|
||||
> *Proposition 7*: the space $l^\infty$ with the metric $d$ defined by
|
||||
>
|
||||
|
@ -170,4 +218,26 @@ for all $x,y \in C$ is incomplete.
|
|||
|
||||
??? note "*Proof*:"
|
||||
|
||||
Will be added later.
|
||||
Let $(x_n)_{n \in \mathbb{N}}$ be a Cauchy sequence in $(l^\infty,d)$, then we have
|
||||
|
||||
$$
|
||||
\forall \varepsilon > 0 \exists N \in \mathbb{N}: n, m > N: d(x_n, x_m) = \sup_{j \in \mathbb{N}} | x_n(j) - x_m(j) | < \varepsilon,
|
||||
$$
|
||||
|
||||
obtains for all $j \in \mathbb{N}$: $|x_n(j) - x_m(j)| <\varepsilon$.
|
||||
|
||||
Which shows that $(x_m(j))_{m \in \mathbb{N}}$ for fixed $j \in \mathbb{N}$ is a Cauchy sequence in $\mathbb{C}$. Since $\mathbb{C}$ is complete the sequence converges; $x_m(j) \to x(j)$ as $m \to \infty$.
|
||||
|
||||
Thus, for $m \to \infty$ we have
|
||||
|
||||
$$
|
||||
d(x_n, x) = \sup_{j \in \mathbb{N}} | x_n(j) - x(j) | < \varepsilon \implies |x_n(j) = x(j)| < \varepsilon.
|
||||
$$
|
||||
|
||||
Since $x_n \in l^\infty$ there exists a $k_n \in \mathbb{R}: |x_n(j)| \leq k_n$ for all $j \in \mathbb{N}$. Hence
|
||||
|
||||
$$
|
||||
|x(j)| \leq |x(j) - x_n(j)| + |x_n(j)| < \varepsilon + k_n,
|
||||
$$
|
||||
|
||||
for all $j \in \mathbb{N}$ which implies that $x \in l^\infty$ and $x_n \to x$ as $n \to \infty$ obtaining that $ l^\infty$ is complete.
|
Loading…
Reference in a new issue