text:
\begin{document}

\maketitle

\begin{abstract}
\noindent
We derive inequalities for time-discrete and time-continuous martingales
that are similar to the well-known Burkholder inequalities. For the time-discrete
case arbitrary martingales in $L^p(\Omega)$ are treated, whereas in the
time-continuous case martingales defined by It{\^o} integrals w.r.t.\
a multi-dimensional Wiener process are considered. The estimates for the
time-discrete martingales are related to the more general results by I.~Pinelis (1994) and are proved to be sharp by a different and more
elementary proof for this special setting.
Further, for time-continuous martingales the presented inequalities are
generalizations of similar estimates
proved by M.~Zakai (1967) and E.~Rio (2009) to the general multi-dimensional case.
Especially, these inequalities possess smaller constants compared to the ones
that result if the original Burkholder inequalities would be applied for such
estimates. Therefore, the presented inequalities are highly
valuable in, e.g., stochastic analysis and stochastic numerics.

\end{abstract}

\section{Burkholder inequalities for martingales}

In stochastic analysis the Burkholder inequalities give powerful estimates
for martingales, and they are used frequently.
In the following, let $(\Omega, \mathcal{F}, \Prob)$ be a complete probability space.

Let $p \in {[2,\infty[}$, $d, N \in \mathbb{N}$,
and let $(\mathcal{G}_n)_{n \in \{0,1, \ldots,N\}}$
be a filtration.
Consider a time-discrete martingale $(M_n)_{n\in\{0,1,\dots,N\}}$
in $L^{p}(\Omega;\mathbb{R}^d)$ w.r.t.\ the filtration
$(\mathcal{G}_{n})_{n\in\{0,1,\dots,N\}}$.
Define $d_0 = M_0$ and $d_k=M_k-M_{k-1}$ for $k \in \{1,\ldots,N\}$ such
that $M_n = \sum_{k=0}^{n} d_k$. According to \cite{Bur1988a}, the Burkholder
inequalities state
\begin{align} \label{eq:dBurkholder1}
	\| M_n \|_{L^p(\Omega; \mathbb{R}^d)}^2 \le (p-1)^2
	\bigg\| \sum_{k=0}^{n} \| d_k \|^2 \bigg\|_{L^{\frac{p}{2}}(\Omega; \mathbb{R})}
\end{align}
and
\begin{align} \label{eq:dBurkholder2}
	\bigg\| \sup_{\nu \in\{0,1,\dots,n\}} \| M_{\nu} \|
	\bigg\|_{L^p(\Omega; \mathbb{R})}^2
	\le p^2 \bigg\| \sum_{k=0}^{n} \| d_k \|^2 \bigg\|_{L^{\frac{p}{2}}(\Omega;
	\mathbb{R})}
\end{align}
for all $n \in \{0,1,\dots,N\}$ where the constants are best possible.

These inequalities carry over to time-continuous martingales defined by
It{\^o} stochastic integrals~\cite{Bur1988a}.
Let $(\mathcal{F}_t)_{t \in [0,T]}$ for some $T>0$ be a filtration that fulfills
the usual conditions.
For $m\in\mathbb{N}$, let $W \colon [0,T] \times \Omega \to \mathbb{R}^m$
be an $m$-dimensional Wiener process w.r.t.\ the filtration
$(\mathcal{F}_t)_{t \in [0,T]}$.
For $j \in \{1,\dots,m\}$, let $f^j \colon [0,T] \times \Omega \to \mathbb{R}^d$
be an $(\mathcal{F}_t)_{t\in[0,T]}$-adapted and measurable stochastic process with

$\Erw \big[ \big( \int_0^T  \| f^j_u \|^2 \, \mathrm{d}u \big)^{\frac{p}{2}} \big] < \infty$.

Then, it holds

\begin{align} \label{eq:cBurkholder1}
	\bigg\| \sum_{j=1}^m \int_{0}^t f^{j}_u \, \mathrm{d} W^{j}_u \bigg\|_{L^{p}(\Omega;
	\mathbb{R}^d)}^2 &\le (p-1)^2 \bigg\| \int_{0}^t \sum_{j=1}^m \big\| f^{j}_u
	\big\|^2 \, \mathrm{d} u \bigg\|_{L^{\frac{p}{2}}(\Omega; \mathbb{R})}
\end{align}

and

\begin{align} \label{eq:cBurkholder2}
	\bigg\| \sup_{s\in [0,t]} \bigg\| \sum_{j=1}^m \int_{0}^s f^{j}_u \, \mathrm{d}
	W^{j}_u \bigg\| \bigg\|_{L^{p}(\Omega; \mathbb{R})}^2
	&\le p^2 \bigg\| \int_{0}^t \sum_{j=1}^m \big\| f^{j}_u \big\|^2 \, \mathrm{d} u
	\bigg\|_{L^{\frac{p}{2}}(\Omega; \mathbb{R})}
\end{align}

for all $t\in[0,T]$, where the constants are best possible, see \cite{Bur1988a}.

Considering for example the convergence analysis of numerical methods for
stochastic
(partial) differential equations, the expressions on the right-hand
sides of inequalities~\eqref{eq:dBurkholder1}-\eqref{eq:cBurkholder2} are
usually not needed explicitly.
This is because the triangle inequality is often applied to the
$L^{\frac{p}{2}}(\Omega; \mathbb{R})$-norms in order to obtain suitable
upper bounds.
In the following, we show that for such inequalities the constants can be
reduced in case of $p>2$ compared to the constants given in
\eqref{eq:dBurkholder1}-\eqref{eq:cBurkholder2}.

\section{Generalized Burkholder-type inequalities}

In this section, we generalize some Burkholder-type inequalities proved
by E.~Rio~\cite{Rio2009a} and M.~Zakai~\cite{Zak1967a}, respectively,
to the multi-dimensional case. The resulting constants for these inequalities are
smaller than the ones that one would get from
inequalities~\eqref{eq:dBurkholder1}-\eqref{eq:cBurkholder2}.
The proofs for these inequalities are postponed to Section~\ref{sec:proofs}.
Firstly, we consider the time-discrete case.

\begin{prop} \label{thm:discreteBurkholder}
	Let $p\in {[2,\infty[}$ and $N \in \mathbb{N}$. Further, let
	$(M_n)_{n\in\{0,1,\dots,N\}}$ with
	$M_n = \sum_{k=0}^{n} d_k$ be a time-discrete
	martingale in $L^{p}(\Omega; \mathbb{R}^d)$
	w.r.t.\ the filtration $(\mathcal{G}_{n})_{n \in \{0,1,\dots,N\}}$.
	Then, it holds

	\begin{align} \label{eq:discreteBurkholder}
		\| M_n \|_{L^{p}(\Omega; \mathbb{R}^d)}^2 \le (p-1) \sum_{k=0}^{n}
		\| d_k \|_{L^{p}(\Omega; \mathbb{R}^d)}^2
	\end{align}

	and

	\begin{align*}
		\bigg\| \sup_{\nu \in\{0,1,\dots,n\}} \| M_{\nu} \|
		\bigg\|_{L^{p}(\Omega; \mathbb{R})}^2 \le \frac{p^2}{p-1} \sum_{k=0}^{n}
		\| d_k \|_{L^{p}(\Omega; \mathbb{R}^d)}^2
	\end{align*}

	for all $n \in \{0,1,\dots,N\}$. The constants are best possible.
\end{prop}

Here, it has to be pointed out that the result
\eqref{eq:discreteBurkholder} is a special case of \cite[Prop.~2.5]{MR1331198}

by Pinelis if $d=1$, and the proof given in \cite{MR1331198}
can be easily generalized to the case $d \geq 1$ that we consider. Nevertheless,
our poof for Proposition~\ref{thm:discreteBurkholder} is rather
elementary compared to the one given for the more general setting in
\cite{MR1331198}. Thus, the presented proof may be of some interest on its own.

In the case of $d=1$, an inequality similar to~\eqref{eq:discreteBurkholder} and its
sharpness are proved in~\cite{MR1331198}. In Section~\ref{sec:proofs},
we extent this proof to general $d \in \mathbb{N}$.

Next, we consider the time-continuous case.
Already in 1967, Zakai proved inequality~\eqref{eq:Zakai}
of the following proposition in the case of $d=m=1$, see
\cite[Theorem~1]{Zak1967a}.
We generalize this idea to $\mathbb{R}^d$-valued integrands of It{\^o}
stochastic integrals w.r.t.\ an $m$-dimensional Wiener process.

\begin{prop} \label{thm:Zakai}
	Let $p\in {[2,\infty[}$, and for $j \in \{1,\dots,m\}$, let
	$f^j \colon [0,T] \times \Omega \rightarrow \mathbb{R}^d$ be an
	$(\mathcal{F}_t)_{t\in[0,T]}$-adapted, measurable stochastic process
	with

	$\Erw \big[ \big( \int_0^T \| f^j_u \|^2 \, \mathrm{d} u \big)^{\frac{p}{2}}
	\big] <	\infty$.
	Then, it holds
	\begin{align} \label{eq:Zakai}
		\bigg\| \sum_{j=1}^m \int_{0}^t f^{j}_u \, \mathrm{d} W^{j}_u
		\bigg\|_{L^{p}(\Omega; \mathbb{R})}^2
		&\le (p-1) \int_{0}^t \bigg\| \sum_{j=1}^m \big\| f^{j}_u \big\|^2
		\bigg\|_{L^{\frac{p}{2}}(\Omega; \mathbb{R})} \, \mathrm{d} u
	\end{align}
	and
	\begin{align*}
		\bigg\| \sup_{s\in [0,t]} \bigg\| \sum_{j=1}^m \int_{0}^s f^{j}_u
		\, \mathrm{d} W^{j}_u \bigg\| \bigg\|_{L^{p}(\Omega; \mathbb{R})}^2
		&\le \frac{p^2}{p-1} \int_{0}^t \bigg\|
		\sum_{j=1}^m \big\| f^{j}_u \big\|^2
		\bigg\|_{L^{\frac{p}{2}}(\Omega; \mathbb{R})} \, \mathrm{d} u
	\end{align*}
	for all $t\in[0,T]$.
\end{prop}

Here, it is an open problem whether the constants in Proposition~\ref{thm:Zakai}
are best possible. Since we only have one-sided estimates, we cannot
transfer the results from the time-discrete case in
Proposition~\ref{thm:discreteBurkholder} to the time-continuous case
as Burkholder did in \cite{Bur1988a}, cf.\ \cite{Dol1969a}.
However, for $p>2$ the constants in Proposition~\ref{thm:discreteBurkholder}
and Proposition~\ref{thm:Zakai} are of order $\mathcal{O}(p)$ as
$p \to \infty$, and they are smaller than the ones in
inequalities~\eqref{eq:dBurkholder1}-\eqref{eq:cBurkholder2},
where the constants are of order $\mathcal{O}(p^2)$ as $p \to \infty$.
This makes the inequalities in Proposition~\ref{thm:discreteBurkholder} and
Proposition~\ref{thm:Zakai} highly valuable for, e.g., the convergence analysis of
numerical methods for stochastic (partial) differential equations where
constants matter.

\section{Proofs} \label{sec:proofs}

\begin{proof}[Proof of Proposition~\ref{thm:discreteBurkholder}]
	In the case of $p=2$, the assertion follows from the discrete Burkholder
	inequalities~\eqref{eq:dBurkholder1} and~\eqref{eq:dBurkholder2}.
	Therefore, we assume $p\in {]2,\infty[}$ in the following.
	An inequality similar to the one in~\eqref{eq:discreteBurkholder} and
	its sharpness was proved by E.~Rio in \cite[Section~2]{Rio2009a} for
	the case of $d=1$.

	We adapt his proof to the case of general $d\in\mathbb N$.
	For this, we generalize \cite[Proposition~2.1]{Rio2009a}.
	Let $X, Y \in L^{p}(\Omega; \mathbb{R}^d)$ and let $\mathcal{G} \subseteq
	\mathcal{F}$ be some sub-$\sigma$-algebra such that $X$ is
	$\mathcal{G}$-measurable and
	$\Erw [Y \vert \mathcal{G}]=0$ $\Prob$-a.s. Then, we first prove that
	\begin{align} \label{eq:Rio}
		\| X+Y \|_{L^{p}(\Omega; \mathbb{R}^d)}^2
		\le \| X \|_{L^{p}(\Omega; \mathbb{R}^d)}^2
		+(p-1) \| Y \|_{L^{p}(\Omega; \mathbb{R}^d)}^2.
	\end{align}
	If $X=0$ or $Y=0$, this inequality is clearly true, so we assume
	$\| X \|_{L^{p}(\Omega; \mathbb{R}^d)} > 0$ and $\| Y
	\|_{L^{p}(\Omega; \mathbb{R}^d)} > 0$. Define the
	function $\varphi \colon [0,1] \to \mathbb{R}$ by $\varphi(t)=\| x + ty \|^p$
	for $x,y \in \mathbb{R}^d$. Using Taylor expansion, it holds

	$\varphi(1) = \varphi(0) + \varphi'(0) + \int_0^1 \varphi''(t) (1-t) \, \mathrm{d} t$

	and thus
	\begin{align*}
		\| X+Y \|^p &= \| X \|^p + p \| X \|^{p-2} \sum_{i=1}^d X^i Y^i
		+ p \int_0^1 \| X+tY \|^{p-2} \| Y \|^2 (1-t) \, \mathrm{d} t \\
	&
	\phantom{=\| X \|^p}~+ p(p-2) \int_0^1 \| X+tY \|^{p-4} \bigg( \sum_{i=1}^d
	(X^i+tY^i) Y^i \bigg)^2 (1-t) \, \mathrm{d} t
	\end{align*}
	$\Prob$-a.s. Considering the integrand of the last
	integral in the Taylor expansion above, it holds
	\begin{align*}
		\| X+tY \|^{p-4} \bigg( \sum_{i=1}^d (X^i+tY^i) Y^i \bigg)^2
		\le \| X+tY \|^{p-2} \| Y \|^2
	\end{align*}
	$\Prob$-a.s.\ by Cauchy-Schwarz inequality. Then, it follows

	\begin{align} \label{eq:normXYtaylor1}
		\begin{split}
			\| X+Y \|^p &\le \| X \|^p + p \| X \|^{p-2} \sum_{i=1}^d X^i Y^i

			+ p(p-1)\! \int_0^1 \| X+tY \|^{p-2} \| Y \|^2 (1-t) \, \mathrm{d} t
	\end{split}
	\end{align}

	$\Prob$-a.s. Due to the assumptions, it holds
	\begin{align*}
		\Erw \big[ \Erw[ \| X \|^{p-2} X^i Y^i \vert \mathcal{G} ] \big]
		= \Erw \big[ \| X \|^{p-2} X^i \Erw [ Y^i \vert \mathcal{G} ] \big]
		= 0
	\end{align*}
	and because
	\begin{align*}
		\Erw \big[ \| X+tY \|^{p-2} \| Y \|^2 \big]
		\le \big( \Erw [ \| X+tY \|^p ] \big)^{\frac{p-2}{p}} \big( \Erw [ \| Y \|^p ]
		\big)^{\frac{2}{p}}
	\end{align*}
	by H\"older's inequality with $\frac{p-2}{p}+\frac{2}{p}=1$,
	taking the expectation on both sides of
	inequality~\eqref{eq:normXYtaylor1} and applying Fubini's theorem,
	we obtain
	\begin{align*}
		&\Erw [ \| X+Y \|^p ] \le \Erw [ \| X \|^p ] + p(p-1) \int_0^1 \big(
		\Erw [ \| X+tY \|^p ] \big)^{\frac{p-2}{p}} \big( \Erw [ \| Y \|^p ]
		\big)^{\frac{2}{p}} (1-t) \, \mathrm{d} t .
	\end{align*}
	This is a multi-dimensional version of \cite[Inequality~(2.1)]{Rio2009a}.
	Now, we use a Gronwall-type inequality, i.e., we apply
	\cite[Lemma on p.~171]{Zak1967a} with $\alpha=\frac{2}{p}$.
	Then, it follows

	\begin{align*}
		&\Erw [ \| X+Y \|^p ] \le \bigg( \Erw [ \| X \|^p ]^{\frac{2}{p}}
		+ \frac{2}{p}p(p-1) \int_0^1 (1-t) \, \mathrm{d} t \,
		\big( \Erw [ \| Y \|^p ] \big)^{\frac{2}{p}} \bigg)^{\frac{p}{2}},
	\end{align*}

	and, since $\int_0^1(1-t) \, \mathrm{d} t=\frac{1}{2}$, inequality~\eqref{eq:Rio}
	holds by raising both sides of the inequality above to the power of $\frac{2}{p}$.
	Due to \cite[Remark~2.1]{Rio2009a}, the constant $p-1$ in
	inequality~\eqref{eq:Rio} is best possible.

	We remark that the considerations after \cite[inequality~(2.1)]{Rio2009a}
	on \cite[p.~150]{Rio2009a} prove essentially Zakai's Gronwall-type
	inequality in \cite[Lemma on p.~171]{Zak1967a}.

	Now, we consider inequality~\eqref{eq:discreteBurkholder}. Since
	$(M_n)_{n\in\{0,1,\dots,N\}}$ is a martingale, it holds that
	$\Erw [M_{n} \vert \mathcal{G}_{n-1} ] = M_{n-1}$ $\Prob$-a.s.\ for
	$n \in \{1,\dots,N\}$, i.e., $\Erw [d_{n} \vert \mathcal{G}_{n-1} ] = 0$ $\Prob$-a.s.
	Thus, inequality~\eqref{eq:discreteBurkholder}
	follows from applying inequality~\eqref{eq:Rio} to $M_n= M_{n-1}+d_{n}$
	by induction on $n \in \{1,\dots,N\}$, cf.\ \cite[Theorem~2.1]{Rio2009a}.

	Finally, Doob's maximal inequality
	implies the second inequality of this proposition. Since Doob's
	inequality is sharp, the constant is best possible, cf.\
	\cite[p.~87]{Bur1988a} and \cite[Theorem~2]{DG1978a}.
\end{proof}

\begin{proof}[Proof of Proposition~\ref{thm:Zakai}]
	The first inequality is proven by Zakai \cite[Theorem~1]{Zak1967a}
	in case of $d=m=1$. We extend this proof to the case of a
	general $m$-dimensional Wiener processes and $\mathbb{R}^d$-valued integrands.
	Due to Burkholder's inequality~\eqref{eq:cBurkholder1} and
	the assumption on processes $f^{j}$, $j \in \{1,\dots,m\}$,
	we have
	$\big\| \sum_{j=1}^m \int_{0}^t f^{j}_u \, \mathrm{d} W^{j}_u
	\big\|_{L^{p}(\Omega; \mathbb{R}^d)} <\infty$.
	Let $\delta >0$ and $\varphi \in C^2(\mathbb{R}^d; \mathbb{R})$
	with $\varphi(x) = (\delta + \| x \|^2)^{\frac{p}{2}}$ for $x \in \mathbb{R}^d$.
	Applying It{\^o}'s formula for the function $\varphi$ to the It{\^o} process
	$(X_t)_{t \in [0,T]}$ with $X_t = \smash{ \sum_{j=1}^m \int_0^t f_u^j \,
	\mathrm{d} W_u^j}$,

	then \cite[Equation~(6)]{Zak1967a} reads in the multi-dimensional case as

	\begin{align*}
		&\bigg( \delta+ \bigg\| \sum_{j=1}^m \int_{0}^t f^{j}_u \, \mathrm{d} W^{j}_u
		\bigg\|^2 \bigg)^{\frac{p}{2}} - \delta^{\frac{p}{2}} \\
		&\quad= \frac{p}{2} \int_{0}^t \bigg( \delta + \bigg\|
		\sum_{j=1}^m \int_{0}^s f^{j}_u \, \mathrm{d} W^{j}_u \bigg\|^2
		\bigg)^{\frac{p}{2}-1} \sum_{j=1}^m \sum_{i=1}^d (f^{i,j}_s)^2
		\, \mathrm{d} s \\
		&\qquad+ \frac{p(p-2)}{2} \int_{0}^t \bigg( \delta + \bigg\|
		\sum_{j=1}^m \int_{0}^s f^{j}_u \, \mathrm{d} W^{j}_u \bigg\|^2
		\bigg)^{\frac{p}{2}-2} \\
		&\qquad \qquad \times
		\sum_{l=1}^m \sum_{i,k=1}^d \bigg(\sum_{j=1}^m \int_{0}^s f^{i,j}_u \,
		\mathrm{d} W^{j}_u \bigg) \bigg( \sum_{j=1}^m \int_{0}^s f^{k,j}_u \,
		\mathrm{d} W^{j}_u \bigg) f^{i,l}_s f^{k,l}_s \, \mathrm{d} s \\
		&\qquad + p \sum_{l=1}^m \int_{0}^t \! \bigg( \delta + \bigg\|
		\sum_{j=1}^m \int_{0}^s f^{j}_u \, \mathrm{d} W^{j}_u \bigg\|^2
		\bigg)^{\!\frac{p}{2}-1} \sum_{i=1}^d \! \bigg( \sum_{j=1}^m \int_{0}^s \!\!
		f^{i,j}_u \, \mathrm{d} W^{j}_u \bigg) f^{i,l}_s \, \mathrm{d} W^l_s
	\end{align*}

	$\Prob$-a.s.
	Taking the expectation and using the Cauchy-Schwarz inequality, we obtain

	\begin{align*}
	&\Erw \bigg[ \bigg( \delta + \bigg\| \sum_{j=1}^m \int_{0}^t f^{j}_u \,
	\mathrm{d} W^{j}_u \bigg\|^2 \bigg)^{\frac{p}{2}} \bigg] - \delta^{\frac{p}{2}} \\
	&\quad = \frac{p}{2} \int_{0}^t \Erw \bigg[ \bigg( \delta + \bigg\|
	\sum_{j=1}^m \int_{0}^s f^{j}_u \, \mathrm{d} W^{j}_u \bigg\|^2
	\bigg)^{\frac{p}{2}-1} \sum_{j=1}^m \| f^{j}_s \|^2 \bigg] \, \mathrm{d} s \\
	&\qquad + \frac{p(p-2)}{2} \int_{0}^t \Erw \bigg[ \bigg( \delta +
	\bigg\| \sum_{j=1}^m \int_{0}^s f^{j}_u \, \mathrm{d} W^{j}_u \bigg\|^2
	\bigg)^{\frac{p}{2}-2} \sum_{l=1}^m \sum_{i=1}^d \bigg(
	\sum_{j=1}^m \int_{0}^s f^{i,j}_u \, \mathrm{d} W^{j}_u \,f^{i,l}_s \bigg)^2
	\bigg] \, \mathrm{d} s \\
	&\quad \le \frac{p(p-1)}{2} \int_{0}^t \Erw \bigg[ \bigg( \delta + \bigg\|
	\sum_{j=1}^m \int_{0}^s f^{j}_u \, \mathrm{d} W^{j}_u \bigg\|^2
	\bigg)^{\frac{p}{2}-1} \sum_{j=1}^m \| f^{j}_s \|^2 \bigg] \, \mathrm{d} s ,
	\end{align*}

	which corresponds to the multi-dimensional variant of
	\cite[inequality~(8)]{Zak1967a}. Then, inequality~\eqref{eq:Zakai}
	follows from the same arguments as in \cite[pp.~171--172]{Zak1967a}
	by applying a Gronwall-type inequality \cite[Lemma on p.~171]{Zak1967a}
	and letting $\delta \to 0$.

	Applying Doob's submartingale inequality
	finally yields the second inequality of the proposition.
\end{proof}

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\bibliography{Bibfile}

\end{document}


TITLE: What does inductance mean?
QUESTION [2 upvotes]: I have the following notion of inductance.
When a current flows through a conductor a magnetic field is set up. Depending on the shape of the conductor we get a magnetic flux through the conductor. If the current changes, the magnetic field changes, and the flux changes. This change causes an opposite induced emf in the conductor. The emf induced divided by $\frac {di}{dt}$ gives the inductance.
As you see above my notion of conductance is only quantitative. i.e It talks about how we calculate the inductance. It doesn't explain what physically the inductance is.
For example, resistance is how much a substance can oppose the flow of current. It can be imagined as the collisions of electrons with the molecules of the conductor. It can be seen physically without any calculations. Could someone develop the same notion of inductance.
Sometimes even my understanding of inductance (as little as I know) doesn't hold up. For example:

Consider two parallel wires of unit length, each of radius a, whose centers are a distance d apart and carry equal current in opposite directions.

How will you define inductance of such a system?
If we assume the current in the first wire changes. So we calculate the flux using the magnetic field of the first wire and the area of projection of the second wire? Something just tells me this isn't all that simple.

REPLY [3 votes]: Inductance is the property of a closed circuit (circuit meaning a conductor loop) to resist changes in current, specifically due to magnetic flux through the loop. To understand inductance, we first consider when inductance applies. One might begin with Maxwell's equations, and this is indeed the most practical way given that the student is familiar with them. Assuming you are not, for generality in teaching, we will only focus on quantities of interest, and how they relate to each other.
Background
The central quantities in electromagnetism are the electric field (a representation of the electric force), and the magnetic field (a vector which is perpendicular to the 'plane of action' of the magnetic force). When looking at some 2D area in space, where the electric field $\mathbf{E}$ and the magnetic field $\mathbf{B}$ exist and are constant, we can calculate the flux of the fields at (or "through") the surface by taking only the component of $\mathbf{E}$ or $\mathbf{B}$ (or any vector field) which is normal to the surface, at the surface, and multiplying this field value by the area of the surface.  For example, the electric flux through a rectangle $2\text{m}\times 2$m square in space, when the electric field has a constant value straight through the surface of $3$ N/C (Newtons per Coulomb), is $3*(2*2) = 12 = \Phi_E$ where $\Phi_E$ is the electric field flux at the surface. To generalize this concept of flux beyond rectangular surfaces and constant fields, we actually take the surface integral of the vector field over the surface.
In practice, the surface is the area in the middle of a circuit loop. We find that, by Faraday's law of electromagnetic induction, if we have a closed loop of wire, and a changing magnetic field $\mathbf{B}$, then there is a current generated in the wire loop which is proportional to the magnetic flux (the product of the magnetic field and the loop area). Note that we are not talking about coils, or resistors, but the entire circuit area.
Inductance
Inductance is easy to define. From Faraday's law, for a circuit loop with a magnetic flux $\Phi_M$, there is an induced voltage $V$,
$$V(t) = -\frac{\partial \Phi_M}{\partial t}$$
Now it's worth noting that a closed loop of wire subjected to changing magnetic flux does not have a definable potential (voltage) in the same way as a circuit with e.g. a battery. The loop is closed, but current flows - the electric field exists, but the voltage is not well-defined. You cannot select two points and calculate or a theoretical voltage drop between them.
Even still, because of the current, a voltage drop is present across e.g. resistors, and this voltage is the induced voltage in the circuit.
Consider, now, a circuit which obeys the relation,
$$V(t) = L \frac{dI}{dt}$$
Where $V(t)$ is the induced voltage, $I$ is the current flowing in the loop, and $L$ is a constant of proportionality. We see that:
$$L\frac{dI}{dt} = \frac{d \Phi_M}{d t}$$
Where the partial derivative has been replaced with a single variable derivative for simplicity, and the right hand side is positive because the EMF under consideration is the back-EMF. Therefore the proportionality constant $L$ is:
$$L = \frac{\frac{d \Phi_M}{d t}} {\frac{dI}{dt}} = \frac{\Phi_M}{I}$$
That is, the ratio of the magnetic flux to the current in the loop. This proportionality constant exists for closed circuits, and is called the inductance of the circuit.
It's important to understand that for a circuit to have inductance, it must be closed (current flows). It's equally important to understand that components in the circuit do not have inductance, but they do contribute to the inductance of the circuit, and this contribution is called the inductance of that component.
The current in a circuit causes a magnetic field around the conductors. The magnetic field, in turn, is responsible for the magnetic flux which determines inductance of the circuit. The other contributing factor is the size of the loop - if the loop is made bigger, the flux increases, and the circuit inductance increases. Similarly, if another circuit (not connected electrically) is nearby, its magnetic field contributes to the magnetic flux in the circuit, and this increases inductance (mutual inductance).
For a circuit with current $I$, if the current remains the same, the magnetic field around the circuit will also remain fairly constant in amplitude. However, moving wires, using shorter or longer wires, coiling wires around each other so their magnetic fields add, all these affect the total magnetic flux through the circuit and thus the inductance. Hence, an inductor is a component which has a geometry that, with a certain amount of current flowing through it, has a known amount of magnetic flux through its area, contributing a known amount of inductance.
Finer Points and Remarks
The property of inductance is the tendency of a closed circuit to affect itself by the current flowing within itself. This follows very simple linear relationships, but involving time derivatives, which give inductance is current-impeding quality. No matter how convoluted the circuit geometry is, all that matters for the inductance is (a) the area enclosed by the circuit, (b) the current flowing through the circuit, and (c) the proximity of current carrying wires to one-another. Calculating inductance is not trivial, but observing what causes inductance can aid the design of simple inductors and the reduction of induction effects.
The reason inductance occurs is the fact that the induced voltage in a circuit is proportional to the magnetic flux changes, and the magnetic flux changes depend on the current flowing through the circuit. The logic is not circular, as it may appear, but rather is a direct result of Faraday's Law. In its common integral form, Faraday's law states that:
$$\oint_C \mathbf{E} \cdot d\mathbf{l} = -\frac{\partial}{\partial t} \int_S \mathbf{B} \cdot d\mathbf{s}$$
Where $C$ is the path of the circuit, and $S$ is a surface with $C$ as its boundary. Defining the magnetic flux $\Phi_M$ as:
$$\Phi_M = \int_S \mathbf{B} \cdot d\mathbf{s}$$
And the electric potential (or voltage) around a loop $C$ as:
$$V = \int_C \mathbf{E} \cdot d\mathbf{l}$$
Faradays Law reduces to:
$$V(t) = -\frac{\partial \Phi_M}{\partial t}$$
Which is the same equation as we had before, assuming $V(t)$ is the closed circuit potential (zero in the absence of changing magnetic fields).
The only relationship we desire for a circuit, then, is one of the form:
$$V(t) = L \frac{dI}{dt}$$
That is, a linear relationship between change in current and induced voltage. As we have already shown, this is how we come to the quantity of inductance. Inductance is thus a proportionality constant between voltage and changes in current, and describes the tendency of changes in current in a circuit to change the magnetic flux through that circuit. This, through a sort of feedback process, limits the rate at which the current can change.

Edit after rereading this answer to give some further information on practical inductance
Partial, Interior, and Exterior Inductance
The magnetic field around a current-carrying wire is described by the Biot-Savart law, which is an inverse-square law (often somewhat confusingly written as $\mathbf{r}/|\mathbf{r}|^3$). Therefore, the magnetic field contribution of a particular segment of wire is strongest just outside the conductor surface, and falls off very rapidly. This means that most of the contribution to inductance in the overall loop occurs in close proximity to a conductor's surface.
Working along these sorts of lines, we can write the expression for the partial inductance of an open-loop component (like a straight wire, or a coil), even though we cannot strictly define the inductance without a closed loop. Generally, when the magnetic field effects are confined to a small region in a larger circuit, calculating inductance of that region (even if it is not a closed loop) makes for a very good approximation, and gives an indication of the impact that region has on the overall inductance of the loop.
The Biot-Savart law also helps demonstrate why the shape and size of a conductor is a determining factor in inductance calculations. When analyzing a conductor configuration, we often consider a line current (sometimes called a filament), which is centered inside the wire, as the source of the magnetic field. Then, though the Biot-Savart law seems to imply an infinitely strong magnetic field as you approach the center, we only actually integrate the field outside the conductor. The difference between 30 gauge and 22 gauge wire suddenly becomes important, because the smaller the wire, the closer to infinite field strength you can get. In reality of course, currents are not filaments, but rather we have a current density as the source of the magnetic field, with some distribution within the conductor. The fields within a conductor complicate the situation further.
In fact, at low frequencies (specifically, whenever the skin effect can be neglected) we must consider not only the magnetic field around a current-carrying wire to calculate inductance, but also the magnetic field within that wire. Unlike the electric field, which must necessarily be zero in an ideal conductor, the magnetic field can exist within a conductor and (as one might expect from the discussion above) this can have a significant impact. But the interaction of the current density and the magnetic field within the conductor is not so easy to relate to the inductance of the circuit loop. For this reason, the two concepts are often divided into interior and exterior inductance, for inductance within the conductor and inductance outside of the conductor respectively.
Finally, inductance calculations are often complicated when carrying alternating currents because of the skin effect and the proximity effect, both of which can significantly alter the result. These effects are often discussed in RF and microwave design, and in the design of power distribution systems and power engineering. A good reference book (such as Grover's Inductance Calculations) should guide you through various corrections and secondary effects if you want to go further.


TITLE: Finite-dimensional version of the word problem for groups
QUESTION [14 upvotes]: The (uniform) word problem for groups can be stated in several equivalent ways:
Word Problem for Groups (WP)
Instance: A finite presentation of a group G and an element w of G as a product of generators and their inverses.
Question: Does every linear representation of G in a (not necessarily finite-dimensional) Hilbert space map w to the identity operator?
Equivalently: Does every unitary representation of G in a Hilbert space map w to the identity operator?
Equivalently: Does every normal subgroup of G contain w?
Equivalently: Is w=1 in G?
As is well-known, WP is undecidable; more specifically, it belongs to RE∖coRE, where RE is the class of recursively enumerable languages and coRE is the class of their complements.  (In fact, it is known that WP remains undecidable even if we fix the group G and its finite presentation suitably, but that is not the topic of this question.)
We can consider the finite-dimensional version of WP.
Finite-Dimensional Word Problem for Groups (FWP)
Instance: Same as WP.
Question: Does every matrix representation of G map w to the identity matrix?
Equivalently: Does every unitary matrix representation of G map w to the identity matrix?
Equivalently: Does every normal subgroup of G of finite index contain w?
(The equivalence is based on a result by Malcev; see this post by Greg Kuperberg and the comments attached to it.)
The two problems have different answers for some instances; that is, sometimes w≠1 in G but every matrix representation maps w to the identity matrix.  There even exists a finitely presented infinite group which does not have a nontrivial matrix representation.  See the answers to the question “Finitely presented sub-groups of GL(n,C)” by Dmitri.
In fact, FWP is in coRE unlike WP: if we also give the dimension as part of the input, the problem becomes decidable (because it is a system of algebraic equations over ℂ), and therefore FWP can be put in coRE by trying all dimensions.

Question: Is FWP decidable?

I do not even know whether FWP is NP-hard or not.  As far as I know, FWP can be in P or undecidable, or anywhere in between!
(Note: FWP can be viewed as a (very) special case of the problem discussed in the question “Decidability of matrix algebra” by Ricky Demer.)

REPLY [19 votes]: FWP is undecidable by a result of Slobodskoi. Slobodskoi shows that the "Universal theory" of finite groups is undecidable. What you are asking for is whether the "Q-theory" of the pseudovariety of finite groups is decidable. The universal theory and Q-theory are equivalent for the pseudovariety of finite groups (see Kharlampovich-Sapir, section 2.4 for a discussion), so the Q-theory is undecidable too.


text:
\section{Transformations Maintaining Point Symmetry}
\label{sec:maintain_symmetry}
Encoding and decoding schemes such as the ones mentioned in Section~\ref{sec:expected} can be generally used for any dynamical system with arbitrary distributions on the inputs $U$, the state vectors $\widetilde{X}_t$ and the state estimates $X_t$. However, characterizing the attained level of average distortion (using expressions~\eqref{eq::onebit_dist} and~\eqref{eq::multibit_dist}) requires the knowledge of the distribution of the state estimate. While a distribution can be obtained for the initial and target state vectors, it may be difficult to incorporate the system dynamics, the estimation method as well as the controller into the process of finding a distribution of the inputs, {states} and {states} estimate. In such cases, numerical evaluations can aid into finding the needed distribution, as was shown in Example 3 in Section~\ref{sec:expected}.

Although it is necessary to find the {{state} distribution in order to characterize the distortion, the knowledge of existing {{symmetries} in the distribution can directly give {{possible choices for} the transformation function $\alpha_t(\cdot)$ which {{may attain} high levels of distortion{; for} example, if there is a point symmetry in the distribution, mirroring across the symmetry point attains the maximum possible distortion.
In this section, we ask the following question: ``{\it under which conditions on the dynamical system, does point symmetry in the initial and target states results into point symmetry on the {states} estimate}?''

A general answer to the aforementioned question appears to be difficult. Therefore, we limit our answer in this work to the scope of linear controllers.
{For a given} initial and target state, let $X^{(\text{ref})}$ be the reference trajectory that the control system ideally wishes to follow. We assume that the system controller selects an input vector that is a linear function of $X^{(\text{ref})}$. In many cases, $X^{(\text{ref})}$ is also a linear function of the initial and target states ({\it e.g.}, when the reference trajectory is the solution of an LQR problem for the noiseless version of the system). Then we can write {$ U_t = K_t \left(X_t - X_t^{(\text{ref})}\right)$.}

Moreover, we assume that the optimal estimation function that the system uses is a linear one in the observations, {\it i.e.}, we assume $X_t$ is a linear function of $Y_1^t$, $X_{\text{init}}$ and $X_{\text{target}}$.
By incorporating the controller and estimation equations into the system dynamics, one can arrive at the following relation {$X = MQ$, where $Q =  \left[X^\prime_{\text{init}} \:\:\: X^\prime_{\text{target}} \:\:\: {w_{1}^T}^\prime \:\:\: {v_{1}^T}^\prime \right]$, and}

the matrix $M$ is a function of the matrices $A$, $B$, $C$, $K_t$ and the linear function used in the estimation of state $X_t$ from the observations. We assume that $w_1^T$ and $v_{1}^T$ are uncorrelated Gaussian random vectors. We first prove the following lemma.

\begin{lemma}
\label{lem:symmetry_map}
If {a random vector $V_1 \in \mathbb{R}^n$} has {Point Symmetry across $\mu_{V_1}$}, and $g$ is an affine function, then the {random vector} $V_2 = g(V_1)$ {has} Point Symmetry {across} $g(\mu_{V_1})$.
\end{lemma}

\begin{IEEEproof}
{If $V_1$ has Point Symmetry, then the following conditions are equivalent:}
\begin{align*}
f_{V_1}(v_1) & = f_{V_1}(2\mu_{V_1} - v_1), & \forall v_1 \in V_1, \\
f_{V_1}(v_1) & = f_{2\mu_{V_1} -V_1} (v_1),  & \forall v_1 \in V_1.
\end{align*}

{Thus}, to prove that $V_2$ also has Point Symmetry, it suffices to prove that the density of $V_2$ and $2 \mu_{V_2} - V_2$ is the same.
{Consider the} two random vectors $W_1$ and $W_2$. {If they have the same support and the same density function}, then $g(W_1)$ and $g(W_2)$ will also have the same density for any {function} $g$; {we denote this by writing $W_1 \sim W_2$}. Thus,
\begin{align*}
V_1 & \sim  2 \mu_{V_1} - V_1 \\
g(V_1) & \sim g(2 \mu_{V_1} - V_1) \\
M_1 V_1 + M_2 & \sim 2 M_1 \mu_{V_1} - M_1 V_1 + M_2 \\
V_2 & \sim  2 (M_1 \mu_{V_1} + M_2)  - (M_1 V_1 + M_2) \\
V_2 & \sim 2 (\mu_{V_2}) - V_2.
\end{align*}

Thus, $V_2$ has a point of symmetry.
\end{IEEEproof}

\begin{theorem}
 \label{thm::MaintiningSym}
 If $X_{\text{init}}$ and $X_{\text{target}}$ are independent of $w_1^T$ and $v_1^T$, and both have Point Symmetries, then the vectors $X_t$ as well as $X$ will all have Point Symmetries for any matrix $M$.
\end{theorem}

\begin{Pf}
{
 First, note that $w_1^T$ and $v_1^T$ are Gaussian random vectors, and therefore have Point Symmetries across their mean points. Since $X_{\text{init}}$ and $X_{\text{target}}$ are independent of $w_1^T$ and $v_1^T$, then the vector $Q$ also has a Point Symmetry across the mean point (which is the concatenation of the mean points of the respective components of $Q$); we denote this point by $\mu_Q$. Then, by virtue of Lemma~\ref{lem:symmetry_map}, $X$ (respectively $X_t$) will also have Point Symmetry across the point $M \mu_Q$ (respectively across the point $\mu_Q$ left-multiplied by the corresponding section of the matrix $M$).}
\end{Pf}

\textbf{{Revisiting Example-3 of Section~\ref{sec:expected}}}:
{Example 3 in Section~\ref{sec:expected} shows an example where the initial and target points exhibit Point Symmetry. In such an example, the LQR controller is a linear function of the previous states (one can find such a controller by applying the KKT conditions). Since the system is noiseless, then the estimated states are equal to the observations. Therefore, the conditions for Theorem~\ref{thm::MaintiningSym} are met, and Point Symmetry is preserved for $X_t$ and the whole trajectory $X$. Note, however, that the point of symmetry for $X_t$ changes with $t$, {\it i.e.}, it progresses along the $x$ axis as shown in Figure~\ref{fig::simulationMirroring}. }

config: arxiv


TITLE: Vietoris Topology
QUESTION [4 upvotes]: let $ X $ be a topological space and $ \operatorname{Exp}(X) $ is the set of all  closed non-empty subsets of $X$ .
If $ U , V_{1}, V_{2}\ldots V_{n}$ are the  non-empty open subset in $ X$, define:

$$ \langle U , V_{1}, V_{2}, \ldots V_{n} \rangle   = \{ F \in \operatorname{Exp}(X)\mid F\subseteq U, \forall 1 \leq i \leq n : F\cap V_{i} \neq \emptyset\}$$
families $ B  $includes all sets of the form $\langle U , V_{1}, V_{2}, \ldots, V_{n} \rangle $  is the basis for a topology   for $\operatorname{ Exp} (X) $
This topology is called the Vietoris topology.
My question:
if $ X$ is a $T_{1}$ space, then  is the Vietoris topology  $ T_{1}$?

REPLY [0 votes]: Brandsma's answer is superb, and I here only mean to add some detail in the way of clarification regarding compactness of the Vietoris Topology, and to provide a small amount of information regarding literature surrounding the question and the issue of compactness.  According to Keesling (Theorem 3 in Normality and Compactness are Equivalent in Hyperspaces), published in 1970 in the Bulletin of the AMS, the  under the assumption that the Continuum Hypothesis (CH) holds, $H(X)$ (with the Vietoris topology) is normal if and only if $X$ is compact.  The team of Costantini, Levi, and Pelant discussed various compactness results in their article Compactness and Local Compactness in Hyperspaces, published in Topology and Its Applcations in 2002, and they mention the 1951 work of Michael, Topologies on Spaces of Subsets, published in the Transactions of the AMS, in which it is shown that $H(X)$ is compact if and only if $X$ is compact.  It appears that CH is not required for this result.
Regarding the original question, the above mentioned article by Michael includes an excellent summary of several related questions (see Theorem 4.9), and in particular, Michael points out that the converse does not hold; that is, $X$ being $T_1$ does not imply that $H(X)$ is $T_1$.
Continue to enjoy theorems about Continua, Generalized Continua, their Hyperspaces, and related results for general topological spaces!


TITLE: Any k Symmetric functions possess a common zero (Guillemin-Pollack)
QUESTION [0 upvotes]: The theorem states that (call it theorem A)

Any $k$ smooth functions $f_1,\dots,f_k$ on $S^k$ that satisfy the symmetry condition $f_i(-x) = -f_i(x)$, $i=1,\dots,k$, must possess a common zero.

The proof is based on a theorem (call it theorem B)

If $g: S^k\to R^{k+1}-\{0\}$ is symmetric about the origin ($g(-x) = -g(x)$), then $g$ intersects every line through $0$ at least once.

by making $f(x) = (f_1(x),f_2(x),\dots, f_k(x), 0)$ and pick line as the $x_{k+1}$ axis.
My naive question is that theorem B requires that the codomian of $g$ does not contain the origin; while in the proof of theorem A we invoke theorem B and find a zero point of $f$. I'm confused if it vilotates the condition?
Another question is that in theorem $A$ if we make $f_1(x)=x$ be the identity map on $S^1$, why would it have a zero point? Many thanks.
Edit: Thanks for @Mariano Suárez-Álvarez 's reply in the comments below. For the first question, I realize it's in the way we perform the proof. We should show theorem A by contradiction, that is assuming the function $f(x) = (f_1(x),f_2(x),\dots, f_k(x), 0)$ has no zero point. Then it is valid to invoke B. By picking the line as the $x_{k+1}$ axis, the assumption $f(x)$ has no zero point contracts theorem B. For the second question, the functions $f_i$ is actually $f_i:S^k\to R$, then my example is not valid.

REPLY [0 votes]: To close the question, I copy the edits from above.
For the first question, I realize it's in the way we perform the proof. We should show theorem A by contradiction, that is assuming the function $f(x) = (f_1(x),f_2(x),\dots, f_k(x), 0)$ has no zero point. Then it is valid to invoke B. By picking the line as the $x_{k+1}$ axis, the assumption $f(x)$ has no zero point contracts theorem B.
For the second question, the functions $f_i$ is actually $f_i:S^k\to R$, then my example is not valid.


TITLE: Suppose $X_{1},...\sim U[0, 1]$. Let us define $X(n)$ = $\max_{1≤i≤n} X_i$ . Prove that $X(n)$ converges in probability to $1$.
QUESTION [2 upvotes]: Suppose $X_{1},...\sim U[0, 1]$. Let us define $X(n)$ = $\max_{1≤i≤n} X_i$ . Prove that $X(n)$ converges in probability to $1$.
Is this solution enough?... I cannot understand the solution provided in the below mentioned link:  https://www.stat.cmu.edu/~larry/=stat705/Lecture4.pdf

REPLY [0 votes]: As the different commentors here agree we also need the $X_i$ to be independent for this to work. Here is the same proof as in the PDF you linked, with additional explanations.
We need to show that $\Bbb{P}(|X_{(n)} - 1|\geq\varepsilon)$ converges to 0 as $n\to \infty$. (This is the definition of convergence in probability). So let $\varepsilon>0$, then we have:
$$\Bbb{P}(|X_{(n)} - 1|\geq\varepsilon) = \Bbb{P}(1-X_{(n)}\geq\varepsilon) = \Bbb{P}(X_{(n)}\leq1-\varepsilon)$$
For the first equality we used that $X_{(n)}-1$ is non-positive, since $X_{(n)}$ is bounded by 1.
Now comes the crucial step: What is the probability that the maximum of all the random variables $X_i$ is smaller than some value? It's the same as saying: every $X_i$ must be smaller than this value (since then also the maximum of all of them are smaller. If one was bigger, then also the maximum would be bigger!). So in that sense, the following two events are the same:
$$\Bbb{P}(X_{(n)}\leq1-\varepsilon) = \Bbb{P}(X_1 \leq 1-\varepsilon, ...,X_n \leq 1-\varepsilon).$$
Finally, we use the independence assumption to get
$$\Bbb{P}(X_1 \leq 1-\varepsilon, ...,X_n \leq 1-\varepsilon) = \prod_{i=1}^{n}\Bbb{P}(X_i < 1-\varepsilon) = (1-\varepsilon)^n.$$
We see that the last term tends towards 0 as $n$ tends towards infinity.


TITLE: Is $e^{\sqrt{2}}\gt 3$ or $e^{\sqrt{2}}\lt 3$
QUESTION [6 upvotes]: $e^{\sqrt{2}}\gt 3$ or $e^{\sqrt{2}}\lt 3$  which  one  holds  true $?$   I  know  that  $2\lt e \lt 3$ and $\sqrt{2}\gt 1$. Little  help  on  how  to  use  them  to  find  the  right  inequality.
Thanks  for  any  help.

REPLY [1 votes]: If you can use that $e>2.7$ and $\sqrt 2> 1.4$, then
$$e^{\sqrt 2} > 2.7^{1.4} = (1+1.7)^{1.4} > 1+1.4\cdot 1.7 = 3.38$$
by Bernoulli's inequality.


TITLE: Prove that $w_{2n}(\gamma^n\oplus \gamma^n)\neq 0$
QUESTION [4 upvotes]: Let $\gamma^n$ be the canonical $n$-plane bundle over the infinite Grassmann manifold $G_n(\mathbb{R}^{\infty})$. I'm asked to prove that

$$w_{2n}(\gamma^n\oplus \gamma^n)\neq 0$$

(exercise 9-A from Milnor-Stasheff)
I started writing down a proof:
$w_{2n}(\gamma^n \oplus \gamma^n)=\sum_{i+j=2n} w_i(\gamma^n)\smile w_j(\gamma^n)$, but being $w_k(\gamma^n)=0$ for $k> n$, we have $$w_{2n}(\gamma^n \oplus \gamma^n)=w_n(\gamma^n)\smile w_n(\gamma^n)$$
How do I prove that this is not zero? it's enough to say that $w_n(\gamma^n)^2\neq 0 \in H^*(G_n(\mathbb{R}^{\infty})$ (because it's the polynomial algebra generated by the Stiefel-Whitney class up to degree $n$)?
It seems a little bit sloppy as an argument, but I cannot think anything else right now

REPLY [1 votes]: In order to mark this question as answered, the above reasoning is correct. (Thanks to Qiaochu Yuan for the confirm)