Added completeness proofs in functional analysis.

docs/en/mathematics/functional-analysis/metric-spaces/completeness.md

@@ -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.