hcju/proofwikiqa huggingface dataset:
By the Heine-Borel Theorem, $\mathcal K^n \subseteq \mathcal C^n$. Thus from Generated Sigma-Algebra Preserves Subset, $\sigma \left({\mathcal K^n}\right) \subseteq \sigma \left({\mathcal C^n}\right)$. Next, let, for all $n \in \N$, $B^- \left({\mathbf 0; n}\right)$ be the closed ball of radius $n$ around $\mathbf 0$ in $\R^n$. Observe that $\R^n = \displaystyle \bigcup_{n \mathop \in \N} B^- \left({\mathbf 0; n}\right)$. Now let $U \in \mathcal C^n$ be a closed subset of $\R^n$. Then from Intersection with Subset is Subset and Intersection Distributes over Union: :$\displaystyle U = U \cap \R^n = U \cap \bigcup_{n \mathop \in \N} B^- \left({\mathbf 0; n}\right) = \bigcup_{n \mathop \in \N} \left({U \cap B^- \left({\mathbf 0; n}\right)}\right)$ From Intersection of Closed Sets is Closed in Topological Space, $U \cap B^- \left({\mathbf 0; n}\right)$ is closed for all $n \in \N$. By definition, $B^- \left({\mathbf 0; n}\right)$ is bounded. Thus, by the Heine-Borel Theorem, $U \cap B^- \left({\mathbf 0; n}\right)$ is compact. Thus, any closed set is the countable union of compact sets. By the third axiom for a $\sigma$-algebra, this means that $\mathcal C^n \subseteq \sigma \left({\mathcal K^n}\right)$. Now the definition of generated $\sigma$-algebra ensures that $\sigma \left({\mathcal C^n}\right) \subseteq \sigma \left({\mathcal K^n}\right)$. Hence the result, by definition of set equality. {{qed}}
By definition of $x$ to the $n$ rising: :$x^{\overline n} = x \left({x + 1}\right) \cdots \left({x + n - 1}\right)$ By definition of the binomial coefficient of a real number: :$\dbinom {x + n - 1} n = \dfrac {\left({x + n - 1}\right) \left({x + n - 2}\right) \cdots x} {n!}$ Hence the result. {{qed}} Category:Binomial Coefficients Category:Factorials Category:Rising Factorials 281ldsrexhi9ewe5sjbi9fuerv0gh89
Let $\omega := e^{2 \pi i / m}$. Then by the Binomial Theorem: :$(1): \quad \displaystyle \sum_{0 \mathop \le j \mathop < m} \paren {1 + \omega^j}^n \omega^{-j k} = \sum_t \sum_{0 \mathop \le j \mathop < m} \binom n t \omega^{j \paren {t - k} }$ By Sum of Geometric Sequence: :$\displaystyle \sum_{0 \mathop \ge j \mathop < m} \omega^{j \paren {t - k} } = \begin{cases} \dfrac {1 - \omega^{m \paren {t - k} } } {1 - \omega^{t - k} } & : t - k \not \equiv 0 \pmod m \\ m & : t \equiv k \pmod m \end{cases}$ We have that: :$\omega = \exp \dfrac {2 \pi i} m \implies \omega^m = 1$ and so: {{begin-eqn}} {{eqn | l = \sum_{0 \mathop \le j \mathop < m} \omega^{r j}       | r = \frac {1 - \omega^{r m} } {1 - \omega^r}       | c = Sum of Geometric Sequence }} {{eqn | r = m \sqbrk {r \equiv 0 \pmod m}       | c = where $\sqbrk {\, \cdots \,}$ is Iverson's convention }} {{end-eqn}} Thus the summation on the {{RHS}} of $(1)$ is: :$m \displaystyle \sum_{t \bmod m \mathop = k} \binom n t$ The summation on the {{LHS}} of $(1)$ is: {{begin-eqn}} {{eqn | l = \sum_{0 \mathop \le j \mathop < m} \paren {1 + \omega^j}^n \omega^{-j k}       | r = \sum_{0 \mathop \le j \mathop < m} \paren {\omega^{-j / 2} + \omega^{j / 2} }^n \omega^{j \paren {n / 2} - k}       | c =  }} {{eqn | r = \sum_{0 \mathop \le j \mathop < m} \paren {2 \cos \dfrac {j \pi} m}^n \omega^{j \paren {n / 2} - k}       | c =  }} {{end-eqn}} Because the {{RHS}} is wholly real, so must the {{LHS}} be. So, taking the real parts of the {{LHS}} and equating it to the {{RHS}} reveals the result. {{qed}}
The proof proceeds by induction on $n$. For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \sum_k \left[{n \atop k}\right] = n!$ $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | l = \sum_k \left[{0 \atop k}\right]       | r = \sum_k \delta_{0 k}       | c = Unsigned Stirling Number of the First Kind of 0 }} {{eqn | r = 1       | c = all terms vanish but for $k = 0$ }} {{eqn | r = 0!       | c = {{Defof|Factorial}} }} {{end-eqn}} Thus $P \left({0}\right)$ is seen to hold. === Basis for the Induction === $P \left({1}\right)$ is the case: {{begin-eqn}} {{eqn | l = \sum_k \left[{1 \atop k}\right]       | r = \sum_k \delta_{1 k}       | c = Unsigned Stirling Number of the First Kind of 1 }} {{eqn | r = 1       | c = all terms vanish but for $k = 1$ }} {{eqn | r = 1!       | c = {{Defof|Factorial}} }} {{end-eqn}} Thus $P \left({1}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({m}\right)$ is true, where $m \ge 2$, then it logically follows that $P \left({m + 1}\right)$ is true. So this is the induction hypothesis: :$\displaystyle \sum_k \left[{m \atop k}\right] = m!$ from which it is to be shown that: :$\displaystyle \sum_k \left[{m + 1 \atop k}\right] = \left({m + 1}\right)!$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \sum_k \left[{m + 1 \atop k}\right]       | r = \sum_k \left({m \left[{m \atop k}\right] + \left[{m \atop k - 1}\right]}\right)       | c = {{Defof|Unsigned Stirling Numbers of the First Kind}} }} {{eqn | r = m \sum_k \left[{m \atop k}\right] + \sum_k \left[{m \atop k - 1}\right]       | c =  }} {{eqn | r = m \sum_k \left[{m \atop k}\right] + \sum_k \left[{m \atop k}\right]       | c = Translation of Index Variable of Summation }} {{eqn | r = \left({m + 1}\right) \sum_k \left[{m \atop k}\right]       | c =  }} {{eqn | r = \left({m + 1}\right) m!       | c = Induction Hypothesis }} {{eqn | r = \left({m + 1}\right)!       | c = {{Defof|Factorial}} }} {{end-eqn}} So $P \left({m}\right) \implies P \left({m + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{\ge 0}: \sum_k \left[{n \atop k}\right] = n!$ {{qed}}
The proof proceeds by induction on $n$. For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$\displaystyle \sum_k \paren {-1}^k {n \brack k} = \delta_{n 0} - \delta_{n 1}$ $\map P 0$ is the case: {{begin-eqn}} {{eqn | l = \sum_k \paren {-1}^k {0 \brack k}       | r = \sum_k \delta_{0 k}       | c = Unsigned Stirling Number of the First Kind of 0 }} {{eqn | r = 1       | c = all terms vanish but for $k = 0$ }} {{eqn | r = \delta_{0 0} - \delta_{0 1}       | c = {{Defof|Kronecker Delta}} }} {{end-eqn}} Thus $\map P 0$ is seen to hold. $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = \sum_k \paren {-1}^k {1 \brack k}       | r = \sum_k \paren {-1}^k \delta_{1 k}       | c = Unsigned Stirling Number of the First Kind of 1 }} {{eqn | r = -1       | c = all terms vanish but for $k = 1$ }} {{eqn | r = \delta_{1 0} - \delta_{1 1}       | c = {{Defof|Kronecker Delta}} }} {{end-eqn}} Thus $\map P 1$ is seen to hold. === Basis for the Induction === $\map P 2$ is the case: {{begin-eqn}} {{eqn | l = \sum_k \paren {-1}^k {2 \brack k}       | r = - {2 \brack 1} + {2 \brack 2}       | c = {{Defof|Summation}} }} {{eqn | r = - {2 \brack 1} + 1       | c = Unsigned Stirling Number of the First Kind of Number with Self }} {{eqn | r = -\binom 2 2 + 1       | c = Unsigned Stirling Number of the First Kind of n with n-1 }} {{eqn | r = -1 + 1       | c = Binomial Coefficient with Self }} {{eqn | r = 0       | c = Binomial Coefficient with Self }} {{eqn | r = \delta_{2 0} - \delta_{2 1}       | c = {{Defof|Kronecker Delta}} }} {{end-eqn}} Thus $\map P 2$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P m$ is true, where $m \ge 2$, then it logically follows that $\map P {m + 1}$ is true. So this is the induction hypothesis: :$\displaystyle \sum_k \paren {-1}^k {m \brack k} = \delta_{m 0} - \delta_{m 1} = 0$ from which it is to be shown that: :$\displaystyle \sum_k \paren {-1}^k {m + 1 \brack k} = \delta_{\paren {m + 1} 0} - \delta_{\paren {m + 1} 1} = 0$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \sum_k \paren {-1}^k {m + 1 \brack k}       | r = \sum_k \paren {-1}^k \paren {m {m \brack k} + {m \brack k - 1} }       | c = {{Defof|Unsigned Stirling Numbers of the First Kind}} }} {{eqn | r = m \sum_k \paren {-1}^k {m \brack k} + \sum_k \paren {-1}^k {m \brack k - 1}       | c =  }} {{eqn | r = m \sum_k \paren {-1}^k {m \brack k} - \sum_k \paren {-1}^k {m \brack k}       | c = Translation of Index Variable of Summation }} {{eqn | r = \paren {m - 1} \sum_k {m \brack k}       | c =  }} {{eqn | r = \paren {m + 1} \times 0       | c = Induction Hypothesis }} {{eqn | r = 0       | c =  }} {{end-eqn}} So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{\ge 0}: \sum_k \paren {-1}^k {n \brack k} = \delta_{n 0} - \delta_{n 1}$ {{qed}}
By definition: {{begin-eqn}} {{eqn | l = \Beta \left({x, 1}\right)       | r = \int_0^1 t^{x - 1} \left({1 - t}\right)^{1 - 1} \rd t       | c = {{Defof|Beta Function}} }} {{eqn | r = \int_0^1 t^{x - 1} \rd t       | c =  }} {{eqn | r = \int_0^1 t^{x - 1} \rd t       | c =  }} {{eqn | r = \left[{\frac {t^x} x}\right]_0^1       | c =  }} {{eqn | r = \frac 1 x - \frac 0 x       | c =  }} {{eqn | r = \frac 1 x       | c =  }} {{end-eqn}} {{qed|lemma}} {{begin-eqn}} {{eqn | l = \Beta \left({1, x}\right)       | r = \int_0^1 t^{1 - 1} \left({1 - t}\right)^{x - 1} \rd t       | c = {{Defof|Beta Function}} }} {{eqn | r = \int_0^1 \left({1 - t}\right)^{x - 1} \rd t       | c =  }} {{eqn | r = \int_1^0 -\left({z}\right)^{x - 1} \rd z       | c = Integration by Substitution }} {{eqn | r = -\left[{\frac {z^x} x}\right]_1^0       | c =  }} {{eqn | r = -\frac 0 x + \frac 1 x       | c =  }} {{eqn | r = \frac 1 x       | c =  }} {{end-eqn}} {{qed}}
Let $y \in x^\ll, z \in S$ such that: :$z \preceq y$ By definition of way below closure: :$y \ll x$ By definition of reflexivity: :$x \preceq x$ By Preceding and Way Below implies Way Below: :$z \ll x$ Thus by definition of way below closure: :$z \in x^\ll$ Thus by definition: :$x^\ll$ is a lower set. {{qed}}
=== Sufficient Condition === Let $L$ be continuous. By Continuous iff Way Below Closure is Ideal and Element Precedes Supremum: :$\forall x \in S: x^\ll$ is an ideal in $L$ and $x \preceq \sup \left({x^\ll}\right)$ and ::for every ideal $I$ in $L$: $x \preceq \sup I \implies x^\ll \subseteq I$ where $x^\ll$ denotes the way below closure of $x$. Thus :for every element $x$ of $S$ there exists ideal $I$ in $L$: ::$x \preceq \sup I$ and for every ideal $J$ in $L: x \preceq \sup J \implies I \subseteq J$ {{qed|lemma}} === Necessary Condition === Assume that :for every element $x$ of $S$ there exists ideal $I$ in $L$: ::$x \preceq \sup I$ and for every ideal $J$ in $L: x \preceq \sup J \implies I \subseteq J$ Let $x \in S$. There exists ideal $I$ in $L$: :$x \preceq \sup I$ and for every ideal $J$ in $L: x \preceq \sup J \implies I \subseteq J$ We will prove that :$I \subseteq x^\ll$ where $x^\ll$ denotes the way below closure of $x$. Let $y \in I$. Let $J$ be an ideal in $L$ such that :$x \preceq \sup J$ Then $I \subseteq J$ Thus by definition of subset: :$y \in J$ By Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal: :$y \ll x$ Thus by definition of way below closure: :$y \in x^\ll$ {{qed|lemma}} By Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal :$\forall y \in x^\ll: y \in I$ By definition of subset: :$x^\ll \subseteq I$ Thus by definition of set equality: :$\forall x \in S: x^\ll$ is an ideal in $L$ and $x \preceq \sup \left({x^\ll}\right)$ and ::for every ideal $I$ in $L$: $x \preceq \sup I \implies x^\ll \subseteq I$ Hence by Continuous iff Way Below Closure is Ideal and Element Precedes Supremum: :$L$ is continuous {{qed}}
By definition of Beta function: :$\displaystyle \Beta \left({x + 1, y}\right) = \int_0^1 t^x \left({1 - t}\right)^{y - 1} \rd t$ With a view to expressing the primitive in the form: :$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$ let: {{begin-eqn}} {{eqn | l = u       | r = t x       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d u} {\d x}       | r = x t^{x - 1}       | c = Power Rule for Derivatives }} {{end-eqn}} and let: {{begin-eqn}} {{eqn | l = \frac {\d v} {\d x}       | r = \paren {1 - t}^{y - 1}       | c =  }} {{eqn | ll= \leadsto       | l = v       | r = -\frac {\paren {1 - t}^y} y       | c = Primitive of Power }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | o =        | r = \int_0^1 t^{x - 1} \paren {1 - t}^{y - 1} \rd t       | c =  }} {{eqn | r = \sqbrk {t^x \frac {-\paren {1 - t}^y} y}_0^1 - \int_0^1 x t^{x - 1} \frac {-\paren {1 - t}^y} y \rd t       | c = Integration by Parts }} {{eqn | r = \frac x y \int_0^1 t^{x - 1} \paren {1 - t}^y \rd t       | c = as the left hand term vanishes }} {{eqn | r = \frac x y \map \Beta {x, y + 1}       | c = {{Defof|Beta Function}} }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \map \Beta {x, y}       | r = \map \Beta {x + 1, y} + \map \Beta {x, y + 1}       | c = Beta Function of x+1 with y plus Beta Function of x with y+1 }} {{eqn | r = \frac x y \map \Beta {x, y + 1} + \map \Beta {x, y + 1}       | c = from above }} {{eqn | r = \frac {x + y} y \map \Beta {x, y + 1}       | c = simplifying }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = z       | r = \frac t m       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d t} {\d z}       | r = m       | c =  }} {{end-eqn}} Recalculating the limits: {{begin-eqn}} {{eqn | l = t       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = z       | r = 0       | c =  }} {{eqn | l = t       | r = m       | c =  }} {{eqn | ll= \leadsto       | l = z       | r = 1       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \paren {1 - \frac t m}^m       | r = \paren {1 - z}^m       | c =  }} {{eqn | l = t^{x - 1}       | r = \paren {m z}^{x - 1}       | c =  }} {{eqn | r = m^{x - 1} z^{x - 1}       | c =  }} {{end-eqn}} Thus $(1)$ can be written: {{begin-eqn}} {{eqn | l = \int_0^m \paren {1 - \frac t m}^m t^{x - 1} \rd t       | r = m^x \int_0^1 \paren {1 - z}^m z^{x - 1} \rd z       | c = Integration by Substitution }} {{eqn | r = m^x \int_0^1 \paren {1 - t}^m t^{x - 1} \rd t       | c = changing the name of the dummy variable }} {{end-eqn}} {{qed}}
It is taken for granted that the Gamma function increases monotonically on $\R_{\ge 1}$. {{handwaving|Replace the above with a link to a proof}} We begin with an inequality that can easily be verified using cross multiplication. {{handwaving|... so verify it.}} Let $x$ be a real number between $0$ and $1$. Let $n$ is a positive integer. Then: :$\displaystyle \frac {\log \Gamma \left({n - 1}\right) - \log \Gamma \left({n}\right)} {\left({n - 1}\right) - n} \le \frac {\log \Gamma \left({x + n}\right) - \log \Gamma \left({n}\right)} {\left({x + n}\right) - n} \le \frac {\log \Gamma \left({n + 1}\right) - \log \Gamma \left({n}\right)}{\left({n + 1}\right) - n}$ Since n is a positive integer, we can make use of the identity: :$\Gamma \left({n}\right) = \left({n - 1}\right)!$ Simplifying, we get: :$\log \left({n - 1}\right) \le \dfrac {\log \Gamma \left({x + n}\right) - \log \left({\left({n - 1}\right)!}\right)} x \le \log \left({n}\right)$ We now make use of the identity: :$\displaystyle \Gamma \left({x + n}\right) = \prod_{k \mathop = 1}^n \left({x + n - k}\right) \Gamma \left({x}\right)$ along with the fact that the Gamma Function is Log-Convex, to simplify the inequality: :$\displaystyle \left({n - 1}\right)^x \left({n - 1}\right)! \prod_{k \mathop = 1}^n \left({x + n - k}\right)^{-1} \le \Gamma \left({x}\right) \le n^x \left({n - 1}\right)!\prod_{k \mathop = 1}^n \left({x + n - k}\right)^{-1}$ Taking the limit as $n$ goes to infinity and using the Squeeze Theorem: :$\displaystyle \Gamma \left({x}\right) = \lim_{n \mathop \to \infty} n^x n! \prod_{k \mathop = 0}^n \left({x + n - k}\right)^{-1}$ which is another representation of Euler's form. This proves equivalence for $x$ between $0$ and $1$. The result follows from the Gamma Difference Equation. {{qed}}
First we present a lemma: {{:Integral Form of Gamma Function equivalent to Euler Form/Lemma}} Recall the definition of the partial Gamma function: :$\displaystyle \Gamma_m \left({x}\right) := \frac {m^x m!} {x \left({x + 1}\right) \left({x + 2}\right) \cdots \left({x + m}\right)}$ We have that: {{begin-eqn}} {{eqn | r = \int_0^\infty e^{-t} t^{x - 1} \rd t - \Gamma_m \left({x}\right)       | o =        | c =  }} {{eqn | r = \int_0^m e^{-t} t^{x - 1} \rd t + \int_m^\infty e^{-t} t^{x - 1} \rd t - \Gamma_m \left({x}\right)       | c =  }} {{eqn | r = \int_0^m e^{-t} t^{x - 1} \rd t + \int_m^\infty e^{-t} t^{x - 1} \rd t - \int_0^m \left({1 - \frac t m}\right)^m t^{x - 1} \rd t       | c = Partial Gamma Function expressed as Integral }} {{eqn | r = \int_m^\infty e^{-t} t^{x - 1} \rd t + \int_0^m \left({e^t - \left({1 - \frac t m}\right)^m t^{x - 1} }\right) \rd t       | c =  }} {{end-eqn}} We have that for large $t$: :$t^{x - 1} < t^{t / 2}$ and so as $m \to \infty$: :$\displaystyle \int_m^\infty e^{-t} t^{x - 1} \rd t \to 0$ Then: {{begin-eqn}} {{eqn | o =        | r = \int_0^m \left({e^t - \left({1 - \frac t m}\right)^m t^{x - 1} }\right) \rd t       | c =  }} {{eqn | o = \le       | r = \int_0^m \frac {t^2 e^{-t} } m e^{-t} t^{x - 1} \rd t       | c = Lemma }} {{eqn | r = \frac 1 m \int_0^m t^{x + 1} e^{-t} \rd t       | c =  }} {{eqn | o = <       | r = \frac 1 m \int_0^\infty t^{x + 1} e^{-t} \rd t       | c =  }} {{end-eqn}} Now we have that as $m \to \infty$: :$\displaystyle \frac 1 m \int_0^\infty t^{x + 1} e^{-t} \rd t \to 0$ so: :$\displaystyle \int_0^\infty e^{-t} t^{x - 1} \rd t - \Gamma_m \left({x}\right) = 0$ leading to: :$\displaystyle \int_0^\infty e^{-t} t^{x - 1} \rd t = \lim_{m \mathop \to \infty} \dfrac {m^x m!} {x \left({x + 1}\right) \left({x + 2}\right) \cdots \left({x + m}\right)}$ as was to be demonstrated. {{qed}}
For ease of understanding, let Leibniz's notation be used: :$\dfrac {\d^k u} {\d x^k} := D_x^k u$ Then we have: {{begin-eqn}} {{eqn | l = D_x^2 w       | r = \map {\dfrac {\d} {\d x} } {\dfrac {\d w} {\d x} }       | c =  }} {{eqn | r = \map {\dfrac {\d} {\d x} } {\dfrac {\d w} {\d u} \dfrac {\d u} {\d x} }       | c = Derivative of Composite Function }} {{eqn | r = \dfrac {\d u} {\d x} \map {\dfrac {\d} {\d x} } {\dfrac {\d w} {\d u} } + \dfrac {\d w} {\d u} \dfrac {\d^2 u} {\d x^2}       | c = Product Rule for Derivatives }} {{eqn | r = \dfrac {\d u} {\d x} \map {\dfrac {\d} {\d u} } {\dfrac {\d w} {\d u} } \dfrac {\d u} {\d x} + \dfrac {\d w} {\d u} \dfrac {\d^2 u} {\d x^2}       | c = Derivative of Composite Function }} {{eqn | r = \map {\dfrac {\d^2 w} {\d u^2} } {\dfrac {\d u} {\d x} }^2 + \dfrac {\d w} {\d u} \dfrac {\d^2 u} {\d x^2}       | c =  }} {{eqn | r = D_u^2 w \paren {D_x^1 u}^2 + D_u^1 w D_x^2 u       | c = {{Defof|Leibniz's Notation for Derivatives}} }} {{end-eqn}} {{qed}}
$D_x^n$ can be expressed as a determinant: :$D_x^n = \begin{vmatrix} \dbinom {n - 1} 0 u_1 & \dbinom {n - 1} 1 u_2 & \dbinom {n - 1} 2 u_3 & \cdots & \dbinom {n - 1} {n - 2} u_{n - 1} & \dbinom {n - 1} {n - 1} u_n       \\                    -1 & \dbinom {n - 2} 0 u_1 & \dbinom {n - 2} 1 u_2 & \cdots & \dbinom {n - 2} {n - 3} u_{n - 2} & \dbinom {n - 2} {n - 2} u_{n - 1} \\                     0 &                    -1 & \dbinom {n - 3} 0 u_1 & \cdots & \dbinom {n - 3} {n - 4} u_{n - 3} & \dbinom {n - 3} {n - 3} u_{n - 2} \\                \vdots &                \vdots &                \vdots & \ddots &                            \vdots &                            \vdots \\                     0 &                     0 &                     0 & \cdots &                                -1 &                   \dbinom 0 0 u_1 \end{vmatrix}$ where $u_j := \paren {D_x^j u} D_u$. Both sides of this equation are differential operators which are to be applied to $w$. {{finish|Fill in the details}}
Recall the definition of $\Z'_n$: :$\Z'_n = \set {\eqclass k n \in \Z_n: k \perp n}$ and the definition of $\map \phi n$: :$\map \phi n = \card {\set {k: 1 \le k \le n, k \perp n} }$ The result follows from Integer is Congruent to Integer less than Modulus. {{qed}}
If $h = \mathbf 0$ there is nothing to prove. Let $h \ne \mathbf 0$. By the definition of the supremum: : $\dfrac{\left|{Lh}\right|} {\left\Vert{h}\right\Vert} \le \left\Vert{L}\right\Vert_3 = \left\Vert{L}\right\Vert$ whence: : $\left|{Lh}\right| \le \left\|{L}\right\| \left\|{h}\right\|$ {{qed}}
In the summation: :$\displaystyle \sum_{j \mathop = 0}^0 D_u^j w \sum_{\substack {\sum_{p \mathop \ge 0} k_p \mathop = j \\ \sum_{p \mathop \ge 0} p k_p \mathop = 0 \\ \forall p \ge 0: k_p \mathop \ge 0} } 0! \prod_{m \mathop = 0}^1 \dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} }$ the only element appearing is for $j = 0$, and the product is vacuous. Thus: :$\displaystyle \prod_{m \mathop = 0}^1 \dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} } = 1$ and we are left with: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 0}^0 D_u^j w       | r = D_u^0 w       | c =  }} {{eqn | r = w       | c = {{Defof|Zeroth Derivative}} }} {{end-eqn}} {{qed}}
In the summation: :$\displaystyle \sum_{j \mathop = 0}^1 D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = 1 \\ \forall p \ge 1: k_p \mathop \ge 0} } 1! \prod_{m \mathop = 1}^1 \dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} }$ we need to consider $j = 0$ and $j = 1$. Let $j = 0$. Consider the set of $k_p$ such that: :$k_1 + k_2 + \cdots = 0$ :$1 \times k_1 + 2 k_2 + \cdots = 1$ :$k_1, k_2, \ldots \ge 0$ It is apparent by inspection that no set of $k_p$ can fulfil these conditions. Therefore when $j = 0$ the summation is vacuous Let $j = 1$. Consider the set of $k_p$ such that: :$k_1 + k_2 + \cdots = 1$ :$1 \times k_1 + 2 k_2 + \cdots = 1$ :$k_1, k_2, \ldots \ge 0$ By inspection, it is seen that these can be satisfied only by: :$k_1 = 1$ and all other $k_p = 0$. When $k_m = 0$: :$\dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} } = 1$ by definition of zeroth derivative and factorial of $0$. Thus any contribution to the summation where $k_m = 0$ can be disregarded. Thus we have: {{begin-eqn}} {{eqn | o =        | r = \sum_{j \mathop = 0}^1 \sum_{\substack {k_1 \mathop = j \\ k_1 \mathop = 1 \\ k_1 \mathop \ge 0} } D_u^j w \dfrac {1!} {k_1! \paren {1!}^{k_1} } \paren {D_x^1 u}^{k_1}       | c =  }} {{eqn | r = \sum_{k_1 \mathop = 1} D_u^1 w \dfrac {1!} {k_1! \paren {1!}^{k_1} } \paren {D_x^1 u}^{k_1}       | c = simplifying: $j = 0$ is vacuous }} {{eqn | r = D_u w D_x u       | c =  }} {{end-eqn}} {{qed}}