avewright/proofwiki-math huggingface dataset:
Theorem: Complex Plane is Banach Space

The complex plane, along with the complex modulus, forms a Banach space over $\C$.

Proof:
The complex plane $\C$ is a vector space over $\C$.
That the norm axioms are satisfied is proven in Complex Modulus is Norm.
Then we have Complex Plane is Complete Metric Space.
Hence the result.
{{qed}}
Theorem: Rational Number Space is not Complete Metric Space

Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean metric $d$.
Then $\struct {\Q, \tau_d}$ is not a complete metric space.

Proof:
Proof by Counterexample:
First note that Rational Numbers form Metric Space.
It remains to be shown that $\struct {\Q, \tau_d}$ is not complete.
Consider the sequence $\sequence {a_n}$ given by:
:$a_n := \dfrac {f_{n + 1} } {f_n}$
where $\sequence {f_n}$ is the sequence of Fibonacci numbers.
By Ratio of Consecutive Fibonacci Numbers:
:$\ds \lim_{n \mathop \to \infty} a_n = \phi := \dfrac {1 + \sqrt 5} 2$
But Square Root of Prime is Irrational.
That means $\dfrac {1 + \sqrt 5} 2$ is likewise irrational.
Thus $\sequence {a_n}$ is a Cauchy sequence of rational numbers which converges to a number which is not in $\Q$.
The result follows by definition of complete metric space.
{{qed}}
In fact, any sequence of rational numbers that converges to an irrational number (in the metric space $\R$) is a Cauchy sequence that does not converge in $\Q$.
Theorem: Limit of Subsequence equals Limit of Sequence/Real Numbers

Let $\sequence {x_n}$ be a sequence in $\R$.
Let $l \in \R$ such that:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
Let $\sequence {x_{n_r} }$ be a subsequence of $\sequence {x_n}$.
Then:
:$\ds \lim_{r \mathop \to \infty} x_{n_r} = l$

Proof:
Let $\epsilon > 0$.
As $\ds \lim_{n \mathop \to \infty} x_n = l$, it follows that:
:$\exists N: \forall n > N: \size {x_n - l} < \epsilon$
Now let $R = N$.
Then from Strictly Increasing Sequence of Natural Numbers:
:$\forall r > R: n_r \ge r$
Thus $n_r > N$ and so:
:$\size {x_n - l} < \epsilon$
The result follows.
{{qed}}
Theorem: Distance Function of Metric Space is Continuous

Let $M = \struct {A, d}$ be a metric space.
Let $\tau_A$ be the topology on $A$ induced by $d$.
Let $\struct {A \times A, \tau}$ be the product space of $\struct {A, \tau_A}$ and itself.
Then the distance function $d: A \times A \to \R$ is a continuous mapping.

Proof:
Let $d_\infty: \paren {A \times A} \times \paren {A \times A} \to \R$ be the metric on $A \times A$ defined by:
:$\map {d_\infty} {\tuple {x, y}, \tuple {x', y'} } = \max \set {\map d {x, x'}, \map d {y, y'} }$
By P-Product Metric Induces Product Topology, $\tau$ is the topology on $A \times A$ induced by $d_\infty$.
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
Let $\tuple {x_0, y_0} \in A \times A$.
Suppose that $\tuple {x, y} \in A \times A$ and $\map {d_\infty} {\tuple{x, y}, \tuple{x_0, y_0} } < \dfrac 1 2 \epsilon$.
Then:
{{begin-eqn}}
{{eqn | l = \size {\map d {x, y} - \map d {x_0, y_0} }
      | o = \le
      | r = \size {\map d {x, y} - \map d {x_0, y} } + \size {\map d {x_0, y} - \map d {x_0, y_0} }
      | c = Triangle Inequality for Real Numbers
}}
{{eqn | o = \le
      | r = \map d {x, x_0} + \map d {y, y_0}
      | c = Reverse Triangle Inequality
}}
{{eqn | o = <
      | r = \epsilon
      | c = by the definition of $d_\infty$
}}
{{end-eqn}}
The result follows from the definition of a continuous mapping.
{{qed}}
Category:Continuous Mappings
Category:Metric Spaces
Category:Product Spaces
m9ugkhahebr3jngjq8jzm2gnbbi2afs
Theorem: Sequence Converges to Within Half Limit/Real Numbers

Let $\sequence {x_n}$ be a sequence in $\R$.
Let $\sequence {x_n}$ be convergent to the limit $l$.
That is, let $\ds \lim_{n \mathop \to \infty} x_n = l$.
Suppose $l > 0$.
Then:
:$\exists N: \forall n > N: x_n > \dfrac l 2$
Similarly, suppose $l < 0$.
Then:
:$\exists N: \forall n > N: x_n < \dfrac l 2$

Proof:
Suppose $l > 0$.
From the definition of convergence to a limit:
:$\forall \epsilon > 0: \exists N: \forall n > N: \size {x_n - l} < \epsilon$
That is, $l - \epsilon < x_n < l + \epsilon$.
As this is true for ''all'' $\epsilon > 0$, it is also true for $\epsilon = \dfrac l 2$ for some value of $N$.
Thus:
:$\exists N: \forall n > N: x_n > \dfrac l 2$
as required.
Now suppose $l < 0$.
By a similar argument:
:$\forall \epsilon > 0: \exists N: \forall n > N: l - \epsilon < x_n < l + \epsilon$
Thus it is also true for $\epsilon = -\dfrac l 2$ for some value of $N$.
Thus:
:$\exists N: \forall n > N: x_n < \dfrac l 2$
as required.
{{qed}}
Theorem: Sequence Converges to Within Half Limit/Complex Numbers

Let $\sequence {z_n}$ be a sequence in $\C$.
Let $\sequence {z_n}$ be convergent to the limit $l$.
That is, let $\ds \lim_{n \mathop \to \infty} z_n = l$ where $l \ne 0$.
Then:
:$\exists N: \forall n > N: \cmod {z_n} > \dfrac {\cmod l} 2$

Proof:
Suppose $l > 0$.
Let us choose $N$ such that:
:$\forall n > N: \cmod {z_n - l} < \dfrac {\cmod l} 2$
Then:
{{begin-eqn}}
{{eqn | l = \cmod {z_n - l}
      | o = <
      | r = \frac {\cmod l} 2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \cmod l - \cmod {z_n}
      | o = \le
      | r = \cmod {z_n - l}
      | c = Reverse Triangle Inequality
}}
{{eqn | o = <
      | r = \frac {\cmod l} 2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \cmod {z_n}
      | o = >
      | r = \cmod l - \frac {\cmod l} 2
      | c = 
}}
{{eqn | r = \frac {\cmod l} 2
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:Limits of Sequences
asix4g1cp1vfjt3d3s0q6karg0dwwiw
Theorem: Number to Reciprocal Power is Decreasing

The real sequence $\sequence {n^{1/n} }$ is decreasing for $n \ge 3$.

Proof:
We want to show that $\paren {n + 1}^{1 / \paren {n + 1} } \le n^{1/n}$.
Thus:
{{begin-eqn}}
{{eqn | l = \paren {n + 1}^{1 / \paren {n + 1} }
      | o = \le
      | r = n^{1/n}
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \paren {n + 1}^n
      | o = \le
      | r = n^{n + 1}
      | c = raising both sides to the power of $n \paren {n + 1}$
}}
{{eqn | ll= \leadstoandfrom
      | l = \paren {n \paren {1 + \frac 1 n} }^n
      | o = \le
      | r = n^{n + 1}
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \paren {1 + \frac 1 n}^n
      | o = \le
      | r = \frac {n^{n + 1} } {n^n} = n
      | c = 
}}
{{end-eqn}}
But from One Plus Reciprocal to the Nth:
:$\paren {1 + \dfrac 1 n}^n < 3$
Thus the reversible chain of implication can be invoked and we see that $\paren {n + 1}^{1 / \paren {n + 1} } \le n^{1/n}$ when $n \ge 3$.
So $\sequence {n^{1 / n} }$ is decreasing for $n \ge 3$.
{{qed}}
Theorem: Inverse of Diagonal Matrix

Let:
:<nowiki>$\mathbf D = \begin{bmatrix}
a_{11} & 0 & \cdots & 0 \\
0 & a_{22} & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & a_{nn} \\
\end{bmatrix}$</nowiki>
be an $n \times n$ diagonal matrix.
Then its inverse is given by:
:<nowiki>$\mathbf D^{-1} = \begin{bmatrix}
\dfrac 1 {a_{11}} & 0 & \cdots & 0 \\
0 & \dfrac 1 {a_{22}} & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & \dfrac 1 {a_{nn}} \\
\end{bmatrix}$</nowiki>
provided that none of the diagonal elements are zero.
If any of the diagonal elements are zero, $\mathbf D$ is singular. 
{{expand|If any diagonal element isn't a unit}}

Proof:
{{WLOG}}, consider the right inverse of $\mathbf D$.
Suppose none of the diagonal elements are zero. 
Then by the definition of inverse, our assertion is that the matrix product of the two matrices in question is the unit matrix of order $n$.
Now, observe that:
{{begin-eqn}}
{{eqn | l = \begin {bmatrix} a_{11} & 0 & \cdots & 0 \\ 0 & a_{22} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{nn} \\ \end {bmatrix} \begin {bmatrix} \frac 1 {a_{11} } & 0 & \cdots & 0 \\ 0 & \frac 1 {a_{22} } & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \frac 1 {a_{nn} } \\ \end {bmatrix}
      | r = \begin{bmatrix} \frac{a_{11} }{a_{11} } & 0 & \cdots & 0 \\ 0 & \frac{a_{22} }{a_{22} } & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \frac{a_{nn} }{a_{nn} } \\ \end {bmatrix}
      | c = 
}}
{{eqn | r = \begin {bmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \\ \end {bmatrix}
      | c = 
}}
{{eqn | r = \mathbf I
      | c = 
}}
{{end-eqn}}
{{qed|lemma}}
Now suppose one of the diagonal elements is zero.
Then $\map \det {\mathbf D} = 0$, where $\det$ indicates the determinant of $\mathbf D$.
From Determinant of Inverse Matrix it would follow that:
:$\map \det {\mathbf D^{-1} } = \dfrac 1 {\map \det {\mathbf D} }$
But this equation has no solution, and so $\mathbf D$ cannot admit an inverse.
{{qed}}
Category:Diagonal Matrices
Category:Inverse Matrices
h9p8ql129d4qtk0o0tzc2gs95sqm4mi
Theorem: Subspace of Complete Metric Space is Closed iff Complete

Let $\struct {M, d}$ be a complete metric space.
Let $\struct {S, d}$ be a subspace of $\struct {M, d}$.
Then $S$ is closed {{iff}} $S$ is complete.

Proof:
This will be proved by demonstrating the contrapositive:
:$S$ is not complete {{iff}} $S$ is not closed.
Theorem: Set of Chains is Closed under Chain Unions under Subset Relation

Let $\struct {P, \le}$ be an ordered set.
Let $\map {\operatorname {ch} } P$ be the set of chains in $P$.
Then $\map {\operatorname {ch} } P$ is closed under chain unions.
That is, if $\CC$ is a chain in $\map {\operatorname {ch} } P$ with respect to the subset ordering, the union $\ds \bigcup \CC$ of $\CC$ is also a chain in $P$.

Proof:
Let $\ds D = \bigcup \CC$. 
Observe that any $C \in \CC$ is a chain in $P$, hence $C \subseteq P$.
Therefore, $D \subseteq P$, by Set Union Preserves Subsets.
Now suppose that $a, b \in D$.
By definition of union, there exist $A, B \in \CC$ such that $a \in A$ and $b \in B$. 
Since $\CC$ is a chain in $\struct {\map {\operatorname {ch} } P, \subseteq}$, either $A \subseteq B$ or $B \subseteq A$.
Hence either $a \in B$ or $b \in A$.
Both $A$ and $B$ are chains in $P$, and thence $a \le b$ or $b \le a$.
Therefore, $D$ is totally ordered by $\le$.
That is, it is a chain in $P$.
{{qed}}
Category:Order Theory
Category:Closure under Chain Unions
Category:Subset Relation
2t5ufydcmqsryi0p11b3088aw22f6gp
Theorem: Jensen's Inequality (Real Analysis)

Let $I$ be a real interval.
Let $\phi: I \to \R$ be a convex function.
Let $x_1, x_2, \ldots, x_n \in I$.
Let $\lambda_1, \lambda_2, \ldots, \lambda_n \ge 0$ be real numbers, at least one of which is non-zero.
Then:
:$\ds \map \phi {\frac {\sum_{k \mathop = 1}^n \lambda_k x_k} {\sum_{k \mathop = 1}^n \lambda_k} } \le \frac {\sum_{k \mathop = 1}^n \lambda_k \map \phi {x_k} } {\sum_{k \mathop = 1}^n \lambda_k}$
For $\phi$ strictly convex, equality holds {{iff}} $x_1 = x_2 = \cdots = x_n$.

Proof:
The proof proceeds by mathematical induction on $n$.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds \map \phi {\frac {\sum_{k \mathop = 1}^n \lambda_k x_k} {\sum_{k \mathop = 1}^n \lambda_k} } \le \frac {\sum_{k \mathop = 1}^n \lambda_k \map \phi {x_k} } {\sum_{k \mathop = 1}^n \lambda_k}$
$\map P 1$ is true, as this just says:
:$\ds \map \phi {\frac {\lambda_1 x_1} {\lambda_1} } \le \frac {\lambda_1 \map \phi {x_1} } {\lambda_1}$
{{begin-eqn}}
{{eqn | l = \map \phi {\frac {\lambda_1 x_1} {\lambda_1} }
      | r = \map \phi {x_1}
      | c = 
}}
{{eqn | r = \frac {\lambda_1 \map \phi {x_1} } {\lambda_1}
      | c = 
}}
{{end-eqn}}
trivially.
This is our basis for the induction.
Theorem: Real Linear Subspace Contains Zero Vector

Let $\mathbb W \subseteq \R^n$ such that $\mathbb W$ is a linear subspace of $\R^n$.
Then $\mathbb W$ contains the zero vector:
:$\mathbf 0_{n \times 1} = \begin{bmatrix}
0 \\ 0 \\ \vdots \\ 0
\end{bmatrix} \in \mathbb W$

Proof:
This is a consequence of Vector Subspace of Real Vector Space.
{{qed}}
Theorem: Zero Subspace is Subspace

Let $V$ be a vector space over $K$ with zero vector $\mathbf 0$.
The zero subspace $\set {\mathbf 0}$ is a subspace of $V$.

Proof:
We use the Two-Step Vector Subspace Test.
$\set {\mathbf 0}$ is not empty, because it contains $\mathbf 0$.
$\set {\mathbf 0}$ is closed under $+$ because:
:$\forall \mathbf x, \mathbf y \in \set {\mathbf 0}, \mathbf x + \mathbf y = \mathbf 0 + \mathbf 0 = \mathbf 0 \in \set {\mathbf 0}$
$\set {\mathbf 0}$ is closed under multiplication because:
:$\forall \lambda \in K, \mathbf x \in \set {\mathbf 0}: \lambda \mathbf x = \lambda \mathbf 0 = \mathbf 0 \in \set {\mathbf 0}$
Hence the result, from the Two-Step Vector Subspace Test.
{{qed}}
Theorem: Cartesian Product of Natural Numbers with Itself is Countable

The Cartesian product $\N \times \N$ of the set of natural numbers $\N$ with itself is countable.

Proof:
This is simply a special case of Cartesian Product of Countable Sets is Countable.
{{qed}}
Theorem: Minkowski's Inequality for Sums

Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \in \R_{\ge 0}$ be non-negative real numbers.
Let $p \in \R$, $p \ne 0$ be a real number.
If $p < 0$, then we require that $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n$ be strictly positive.
If $p > 1$, then:
:$\ds \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1/p} \le \paren {\sum_{k \mathop = 1}^n a_k^p}^{1/p} + \paren {\sum_{k \mathop = 1}^n b_k^p}^{1/p}$
If $p < 1$, $p \ne 0$, then:
:$\ds \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1/p} \ge \paren {\sum_{k \mathop = 1}^n a_k^p}^{1/p} + \paren {\sum_{k \mathop = 1}^n b_k^p}^{1/p}$

Proof:
=== Proof for $p = 2$ ===
$p = 2$ is an easily proved special case:
{{:Minkowski's Inequality for Sums/Index 2}}{{qed|lemma}}
Theorem: Minkowski's Inequality for Sums

Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \in \R_{\ge 0}$ be non-negative real numbers.
Let $p \in \R$, $p \ne 0$ be a real number.
If $p < 0$, then we require that $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n$ be strictly positive.
If $p > 1$, then:
:$\ds \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1/p} \le \paren {\sum_{k \mathop = 1}^n a_k^p}^{1/p} + \paren {\sum_{k \mathop = 1}^n b_k^p}^{1/p}$
If $p < 1$, $p \ne 0$, then:
:$\ds \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1/p} \ge \paren {\sum_{k \mathop = 1}^n a_k^p}^{1/p} + \paren {\sum_{k \mathop = 1}^n b_k^p}^{1/p}$

Proof:
Define:
:$q = \dfrac p {p - 1}$
Then:
:$\dfrac 1 p + \dfrac 1 q = \dfrac 1 p + \dfrac {p - 1} p = 1$
It follows that:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \paren {a_k + b_k}^p
      | r = \sum_{k \mathop = 1}^n a_k \paren {a_k + b_k}^{p - 1} + \sum_{k \mathop = 1}^n b_k \paren {a_k + b_k}^{p - 1}
      | c = 
}}
{{eqn | o = \le
      | r = \paren {\sum_{k \mathop = 1}^n {a_k}^p}^{1 / p} \paren {\sum_{k \mathop = 1}^n \paren {\paren {a_k + b_k}^{p - 1} }^q}^{1 / q} + \paren {\sum_{k \mathop = 1}^n {b_k}^p}^{1 / p} \paren {\sum_{k \mathop = 1}^n \paren {\paren {a_k + b_k}^{p - 1} }^q}^{1 / q}
      | c = Hölder's Inequality for Sums (twice)
}}
{{eqn | r = \paren {\sum_{k \mathop = 1}^n {a_k}^p}^{1 / p} \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1 / q} + \paren {\sum_{k \mathop = 1}^n {b_k}^p}^{1 / p} \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1 / q}
      | c = Power of Power, and {{hypothesis}}: $\paren {p - 1} q = p$
}}
{{eqn | r = \paren {\paren {\sum_{k \mathop = 1}^n {a_k}^p}^{1 / p} + \paren {\sum_{k \mathop = 1}^n {b_k}^p}^{1 / p} } \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1 / q}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1 - 1 / q}
      | o = \le
      | r = \paren {\sum_{k \mathop = 1}^n {a_k}^p}^{1 / p} + \paren {\sum_{k \mathop = 1}^n {b_k}^p}^{1 / p}
      | c = dividing by $\ds \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1 / q}$
}}
{{eqn | ll= \leadsto
      | l = \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1 / p}
      | o = \le
      | r = \paren {\sum_{k \mathop = 1}^n {a_k}^p}^{1 / p} + \paren {\sum_{k \mathop = 1}^n {b_k}^p}^{1 / p}
      | c = as $1 - \dfrac 1 q = p$
}}
{{end-eqn}}
Theorem: Minkowski's Inequality for Sums

Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \in \R_{\ge 0}$ be non-negative real numbers.
Let $p \in \R$, $p \ne 0$ be a real number.
If $p < 0$, then we require that $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n$ be strictly positive.
If $p > 1$, then:
:$\ds \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1/p} \le \paren {\sum_{k \mathop = 1}^n a_k^p}^{1/p} + \paren {\sum_{k \mathop = 1}^n b_k^p}^{1/p}$
If $p < 1$, $p \ne 0$, then:
:$\ds \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{1/p} \ge \paren {\sum_{k \mathop = 1}^n a_k^p}^{1/p} + \paren {\sum_{k \mathop = 1}^n b_k^p}^{1/p}$

Proof:
Let $\mathbf a$ and $\mathbf b$ be real finite sequences:
{{begin-eqn}}
{{eqn | l = \mathbf a
      | r = \sequence {a_k}_{1 \mathop \le k \mathop \le n}
}}
{{eqn | l = \mathbf b
      | r = \sequence {b_k}_{1 \mathop \le k \mathop \le n}
}}
{{end-eqn}}
Let $\norm {\mathbf a}_p$ denote the $p$-norm of $\mathbf a$:
:$\norm {\mathbf a}_p := \paren {\ds \sum_{k \mathop = 1}^n {a_k}^p}^{1 / p}$
Define:
:$q = \dfrac p {p - 1}$
Then:
:$\dfrac 1 p + \dfrac 1 q = \dfrac 1 p + \dfrac {p - 1} p = 1$
Then:
{{begin-eqn}}
{{eqn | l = \paren {\norm {\mathbf a + \mathbf b}_p}^p
      | r = \norm {\paren {\mathbf a + \mathbf b}^p}_1
}}
{{eqn | r = \norm {\mathbf a \paren {\mathbf a + \mathbf b}^{p - 1} + \mathbf b \paren {\mathbf a + \mathbf b}^{p - 1} }_1
}}
{{eqn | r = \norm {\mathbf a \paren {\mathbf a + \mathbf b}^{p - 1} }_1 + \norm {\mathbf b \paren {\mathbf a + \mathbf b}^{p - 1} }_1
}}
{{eqn | o = \le
      | r = \norm {\mathbf a}_p \norm {\paren {\mathbf a + \mathbf b}^{p - 1} }_q + \norm {\mathbf b}_p \norm {\paren {\mathbf a + \mathbf b}^{p - 1} }_q
      | c = Hölder's Inequality for Sums (twice)
}}
{{eqn | r = \paren {\norm {\mathbf a}_p + \norm {\mathbf b}_p} \norm {\paren {\mathbf a + \mathbf b}^{p - 1} }_q
      | c = simplifying
}}
{{eqn | r = \paren {\norm {\mathbf a}_p + \norm {\mathbf b}_p} \norm {\paren {\mathbf a + \mathbf b}^{p - 1} }_{p / \paren {p - 1} }
      | c = {{hypothesis}}: $\dfrac p {p - 1} = q$
}}
{{eqn | r = \paren {\norm {\mathbf a}_p + \norm {\mathbf b}_p} \paren {\norm {\mathbf a + \mathbf b}_p}^{p - 1}
      | c = Transformation of P-Norm
}}
{{eqn | ll= \leadsto
      | l = \norm {\mathbf a + \mathbf b}_p
      | o = \le
      | r = \norm {\mathbf a}_p + \norm {\mathbf b}_p
      | c = dividing both sides by $\paren {\norm {\mathbf a + \mathbf b}_p}^{p - 1}$
}}
{{end-eqn}}
{{qed}}
Theorem: Minkowski's Inequality for Integrals

Let $f, g$ be (Darboux) integrable functions.
Let $p \in \R$ such that $p > 1$.
Then:
:$\ds \paren {\int_a^b \size {\map f x + \map g x}^p \rd x}^{1/p} \le \paren {\int_a^b \size {\map f x}^p \rd x}^{1 / p} + \paren {\int_a^b \size {\map g x}^p \rd x}^{1 / p}$

Proof:
Define:
:$q = \dfrac p {p - 1}$
Then:
:$\dfrac 1 p + \dfrac 1 q = \dfrac 1 p + \dfrac {p - 1} p = 1$
It follows that:
{{begin-eqn}}
{{eqn | l = \int_a^b \size {\map f x + \map g x}^p \rd x
      | r = \int_a^b \size {\map f x} \size {\map f x + \map g x}^{p - 1} \rd x + \int_a^b \size {\map g x} \size {\map f x + \map g x}^{p - 1} \rd x
      | c = 
}}
{{eqn | o = \le
      | r = \paren {\int_a^b \size {\map f x}^p \rd x}^{1 / p} \paren {\int_a^b \paren {\size {\map f x + \map g x}^{p - 1} }^q \rd x}^{1 / q} + \paren {\int_a^b \size {\map g x}^p \rd x}^{1 / p} \paren {\int_a^b \paren {\size {\map f x + \map g x}^{p - 1} }^q \rd x}^{1 / q}
      | c = Hölder's Inequality for Integrals (twice)
}}
{{eqn | r = \paren {\paren {\int_a^b \size {\map f x}^p \rd x}^{1 / p} + \paren {\int_a^b \size {\map g x}^p \rd x}^{1 / p} } \paren {\int_a^b \paren {\size {\map f x + \map g x}^{p - 1} }^q \rd x}^{1 / q}
      | c = simplifying
}}
{{eqn | r = \paren {\paren {\int_a^b \size {\map f x}^p \rd x}^{1 / p} + \paren {\int_a^b \size {\map g x}^p \rd x}^{1 / p} } \paren {\int_a^b \size {\map f x + \map g x}^p \rd x}^{1 / q}
      | c = Power of Power, and {{hypothesis}}: $\paren {p - 1} q = p$
}}
{{eqn | ll= \leadsto
      | l = \paren {\int_a^b \size {\map f x + \map g x}^p \rd x}^{1 - 1 / q}
      | o = \le
      | r = \paren {\int_a^b \size {\map f x}^p \rd x}^{1 / p} + \paren {\int_a^b \size {\map g x}^p \rd x}^{1 / p}
      | c = dividing by $\ds \paren {\int_a^b \size {\map f x + \map g x}^p \rd x}^{1 / q}$
}}
{{eqn | ll= \leadsto
      | l = \paren {\int_a^b \size {\map f x + \map g x}^p \rd x}^{1 / p}
      | o = \le
      | r = \paren {\int_a^b \size {\map f x}^p \rd x}^{1 / p} + \paren {\int_a^b \size {\map g x}^p \rd x}^{1 / p}
      | c = as $1 - \dfrac 1 q = p$
}}
{{end-eqn}}
{{Namedfor|Hermann Minkowski|cat = Minkowski}}
Theorem: Characteristic Function Determined by 1-Fiber

Let $A \subseteq S$.
Let $f:S \to \left\{{0, 1}\right\}$ be a mapping.
Denote by $\chi_A$ the characteristic function on $A$.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$f {{=}} \chi_A$}}
{{item|(2):|$\forall s \in S: \map f s {{=}} 1 \iff s \in A$}}
{{end-itemize}}
Using the notion of a fiber, $(2)$ may also be expressed as:
:$(2'): \quad \map {f^{-1} } 1 = A$

Proof:
=== $(1)$ implies $(2)$ ===
Follows directly from the definition of characteristic function.
{{qed|lemma}}
Theorem: Characteristic Function of Set Difference

Let $A, B \subseteq S$.
Then:
:$\chi_{A \mathop \setminus B} = \chi_A - \chi_{A \cap B}$ 
where:
:$A \setminus B$ denotes set difference
:$\chi$ denotes characteristic function.

Proof:
Suppose that $\map {\chi_{A \mathop \setminus B} } s = 1$.
Then by definition of characteristic function:
:$s \in A \setminus B$
That is, by definition of set difference, $s \in A$ and $s \notin B$; in particular, $s \notin A \cap B$.
Hence:
:$\map {\chi_A} s = 1$ and $\map {\chi_{A \cap B} } s = 0$
Conclude that:
:$\map {\chi_{A \mathop \setminus B} } s = 1 \implies \map {\chi_A} s - \map {\chi_{A \cap B} } s = 1$
Now conversely suppose $\map {\chi_A} s - \map {\chi_{A \cap B} } s = 1$.
Considering that $\map {\chi_{A \cap B} } s \ge 0$, it must be that $\map {\chi_A} s = 1$ and $\map {\chi_{A \cap B} } s = 0$.
Hence, $s \in A$, and $s \notin A \cap B$.
By definition of set intersection, this means $s \notin B$.
Hence $s \in A \setminus B$ by definition of set difference.
Thus:
:$\map {\chi_{A \mathop \setminus B} } s = 1$
Hence, it has been established that:
:$\map {\chi_{A \mathop \setminus B} } s = 1 \iff \map {\chi_A} s - \map {\chi_{A \cap B} } s = 1$
which, by application of Characteristic Function Determined by 1-Fiber, gives the result.
{{qed}}
Theorem: Characteristic Function of Symmetric Difference

Let $A, B \subseteq S$.
Then:
:$\chi_{A \symdif B} = \chi_A + \chi_B - 2 \chi_{A \cap B}$ 
where:
:$\chi$ denotes characteristic function
:$\symdif$ denotes symmetric difference.

Proof:
By definition of symmetric difference:
:$A \symdif B = \paren {A \cup B} \setminus \paren {A \cap B}$
Thus:
:$\chi_{A \symdif B} = \chi_{A \mathop \cup B} - \chi_{\paren {A \mathop \cup B} \mathop \cap \paren {A \mathop \cap B} }$
by Characteristic Function of Set Difference.
But by Intersection is Subset of Union and Intersection with Subset is Subset:
:$\paren {A \cup B} \cap \paren {A \cap B} = A \cap B$
Hence it follows that:
:$\chi_{A \symdif B} = \chi_{A \mathop \cup B} - \chi_{A \mathop \cap B}$
which by Characteristic Function of Union: Variant 2 becomes:
:$\chi_{A \symdif B} = \chi_A + \chi_B - 2 \chi_{A \mathop \cap B}$
as desired.
{{qed}}