TITLE: Integral of $\frac{1}{1+x^{2/3}}$
QUESTION [1 upvotes]: Can you explain how to integrate $ \frac{1}{1+x^{2/3}} $ for $ x \in [0,1]$ , we cannot use any formula directly to get the solution. Is there any simple approach to integrate such kind of functions?
Thanks in advance.

REPLY [3 votes]: Substitution works here: $$u =x^{1/3} \implies du = \frac 13 x^{-2/3}\,dx \implies dx = 3x^{2/3}\,du = 3u^2\,du$$
This gives us $$\int_0^1 \frac {3u^2}{1+u^2}\,du $$
Now do polynomial division to get $$\int_0^1 3\,du - 3\int \frac{du}{1+u^2}$$
Now, simply put $u = \tan \theta$ and you're good to go.


TITLE: A set S is a subspace of an inner product space $V$ iff $(S^{\perp})^{\perp}=S$?
QUESTION [1 upvotes]: $S$ is any set of a finite-dimensional inner product space $V$. Prove or disprove that $S$ is a subspace of $V$ iff $(S^{\perp})^{\perp}=S$.

My approach was to consider a minimum set $B$ of linearly independent vectors which spans $S$. If such $B$ exists then there exists some $v \in B$ and $v \notin S$, but $v \in (S^{\perp})^{\perp}=(B^{\perp})^{\perp}=B$ so contradiction.
What I can't deal with is how can I consider such $B$. Can I just say "Try every set of $V$ and find the minimum?"
Any help is appreciated.

REPLY [3 votes]: First of all orthogonal complement is a subspace, so if $S=(S^{\perp})^{\perp},$ then $S$ must be a subspace.
Conversely, it is easy to see that $S \subseteq (S^{\perp})^{\perp}.$ For the other containment,
Hint: First show $$V= S \oplus S^{\perp},$$ so for $u \in (S^{\perp})^{\perp},$ there exists $u_1 \in S,u_2 \in S^{\perp}$ such that $u=u_1+u_2.$ It follows $u-u_1 \in S^{\perp}.$ You can also show $u-u_1 \in(S^{\perp})^{\perp}.$ Conclude $u =u_1 \in S.$


TITLE: An equality in the proof of Krull theorem
QUESTION [1 upvotes]: I'm reading a proof of the following theorem:

Let $A$ be a noetherian local ring, let $\mathfrak{m}$ be its maximal ideal and let $\mathfrak{a}$ be an ideal of $A$. Then, it holds:
  $$ \bigcap_{n \in \mathbb{N}}\left(\mathfrak{a}+\mathfrak{m}^n\right)=\mathfrak{a}.$$

During the proof it is states that it holds
$$\pi^{-1}\left(\bigcap_{n \in \mathbb{N}}(\mathfrak{m}/\mathfrak{a})^n\right)=\bigcap_{n \in \mathbb{N}} \pi^{-1}\left((\mathfrak{a} + \mathfrak{m}^n)/\mathfrak{a}\right)$$
where $\pi: A \to A/\mathfrak{a}$ is the natural quotient map.
This equality seems true to me, but I can't give a rigorous proof. Any hint?

REPLY [1 votes]: This comes from a more   general formula: if $\,\mathfrak{a,b,c}\,$ are ideals in a commutative ring $A$ such that $\,\mathfrak a\subset  \mathfrak{b}$ and $\,\mathfrak a\subset  \mathfrak{c} $, then in the quotient ring $A/\mathfrak a$,
$$ \mathfrak{b/a}\cdot\mathfrak{c/a}=(\mathfrak{b\cdot c+a})/\mathfrak{a},$$
which is just an extension to quotient ideals of the definition of the product of elements in $A/\mathfrak a$:
$$(x+\mathfrak a)(y+\mathfrak a)=xy+ \mathfrak a.$$


TITLE: Nice example where $D^{\alpha}\Lambda_{f}\neq\Lambda_{D^{\alpha}f}$?
QUESTION [1 upvotes]: Let $\Omega\subset\mathbb{R}^n$ be open and $f$ be a locally integrable function. The distribution associated with $f$, $\Lambda_{f}\in D'(\Omega)$, is defined via \begin{equation}
\Lambda_{f}(\phi)=\int f\phi,
\end{equation} where $\phi\in D(\Omega)$ is a test function.
For $\mu$, a regular measure, $\Lambda_{\mu}$ is defined to be $\Lambda_{\mu}(\phi)=\int\phi d\mu$.
The distribution derivative of $f$ is the distribution defined by \begin{equation}
(D^{\alpha}\Lambda_{f})(\phi)=(-1)^{|\alpha|}\Lambda_{f}(D^{\alpha}\phi),
\end{equation} where $D^{\alpha}$ is the derivative operator with multi-index $\alpha$.
One natural question is the commutability of $D^{\alpha}$ and $\Lambda$, that is, whether \begin{equation}
D^{\alpha}\Lambda_{f}=\Lambda_{D^{\alpha}f}.
\end{equation}
If $f\in C^{N}(\Omega)$ for some $N>|\alpha|$, then this is true. Rudin says in general this is false by giving a counterexample on $\mathbb{R}$. He takes $f$ to be a left-continuous function of bounded variation, then $Df$ exists a.e. Then he shows $D\Lambda_{f}(\phi)= \Lambda_{\mu}$, where $\mu([a,b])=f(b)-f(a)$.
This shows $D$ and $\Lambda$ commute only if $f$ is absolutely continuous.
However, this example is not quite satisfactory in the sense Rudin interprets $D$ as the a.e. derivative, rather than a regular derivative. In fact, if $f$ is differentiable then it is absolutely continuous.
A good counterexample would be one in which $f\in C^{N-1}(\Omega)$ with $N=|\alpha|$ and the commutativity fails.
Thanks very much!

REPLY [1 votes]: The Cantor function belongs to $C^0$ and its distributional derivative of order $1$ is not described by the pointwise derivative (which exists a.e. This qualifies as "good counterexample" for $N=1$ under the terms of your question.
Counterexamples for higher values of $N$ can be obtained by taking the indefinite integral of the Cantor function $N-1$ times.


TITLE: Determining simulation threshold
QUESTION [0 upvotes]: Let's suppose we have some sort of game (for instance tic-tac-toe) with a limited amount of moves. I want to simulate that game, and with that I will use a random number generator (RNG) to calculate the moves.
In the case of tic-tac-toe, there are at most $9!$ moves (if we forget about the games that end before the last move). If I were to run that number of simulations, there is no guarantee that I've ran all possible scenarios due to the inherent randomness of the process. The only way I would have ran all scenarios is if the RNG would magically pop out a straight sequence of moves.
Assuming my reasoning to be correct, I ask if there is some way to determine a threshold or a probability of cases run that would allow me to set a representative number of simulations?
Storing the moves and looking at what was made is not, for this purpose, an option.

REPLY [0 votes]: It's not so simple as mjqxxxx said, because the way you would presumably use the random numbers is to decide the move at each step, and not to choose an entire game out of the whole set of possibilities. In that case it's not trivial at all. Consider a game whose tree looks like a completely biased binary tree. If each move is randomly selected from among the two possibilities with equal probability, then you will get the same one-move game half the time, and the same two-move game $\frac{1}{4}$ the time, and so on.. It's much harder to cover all the possible games in this case, as compared to a game with a balanced tree.


TITLE: Is spin of elementary particles same as the rotation of a planet?
QUESTION [4 upvotes]: By the word spin of elementary particle, one would imagine the particle to be rotating around its own axis, just like a planet rotates, but is it actually true? While spinning does an elementary particle actually rotate around its own axis just like a planet?

REPLY [1 votes]: Just as linear momentum is associated with the number of wavelengths you can fit into a given distance of a wave repeating in space, so angular momentum is associated with the number of 'wavelengths' you can fit into the 360 degrees of a wave repeating in angle, around a point.
If you look at diagrams of the orbitals of an electron around an atom of defined angular momentum, you find spherical harmonics, which fit sinusoidal waves around a sphere. These are waves in angular space, just as the states of defined linear momentum are complex sinusoidal waves in linear space.
We can see how the orbital angular momentum is related to that of a macroscopic body if we compare the wave patterns of set of six particles moving linearly around a circle, and a single particle at the centre with spherical harmonic waves arranged around it. The linear waves in the particles of the macroscopic rigid body extend the radial pattern of the waves surrounding the central particle.

Particles with intrinsic angular momentum localised to a point have to have such a wave around them. A particle of spin half is represented by a spinor, and has to flip sign on rotating 360 degrees, and fit a whole number of wavelengths into 720 degrees. It can't be constant over all angles, and still manage to flip sign, unless it has zero amplitude. However, it's still just an angular wave, like orbital angular momentum.
Whether having an angular wavelength implies it is 'rotating' is a tricky question, like whether having a linear wavelength means a particle is 'moving' is. With linear waves, the phase velocity (the motion of the wave peaks) is completely different to the group velocity (the motion of a wave packet), which is what we normally think of as the particle 'moving'. A particle in a pure momentum state has no well-defined position, no observable distinguishing features anywhere which could be said to 'move' around. We can sum different pure momentum states to get a confined wavepacket that can 'move', and we can do the same sort of thing by combining orbital angular momentum states, but the minimal intrinsic angular momentum is fixed, so we probably don't have the freedom to play around with it in the same way.


text:
\subsection{Invariants and measurements for filtered chain
  complexes} \label{s:filt-ch}

As a supplement to the previous material we describe
here a number of numerical invariants of filtered chain complexes that
will be useful in~\S\ref{s:main-geom} when we prove our main geometric
results.

We begin with basic definitions. Fix a commutative ring $\mathcal{R}$
with unity. By a filtered chain complex we mean a chain complex
$(C, d^C)$ of $\mathcal{R}$-modules endowed with an increasing
filtration by sub-chain complexes $C^{\leq \alpha} \subset C$, indexed
by the real numbers $\alpha \in \mathbb{R}$. An $\mathcal{R}$-linear
map $f: C \longrightarrow D$ between two filtered chain complexes
$(C, d^C)$, $(D, d^D)$ is called {\em filtered} if there exists
$\rho \in \mathbb{R}$ such that
$f(C^{\leq \alpha}) \subset D^{\leq \alpha + \rho}$ for every
$\alpha$. In that case we also say that $f$ shifts action by
$\leq \rho$.  In case $f$ preserves the filtrations (i.e.~it shifts
filtration by $\leq 0$) we say that $f$ is {\em strictly filtered}.

Let $C, D$ be two chain complexes and assume $D$ is filtered. Every
chain map $\phi:C \longrightarrow D$ induces a filtration on $C$ by
$C^{\leq \alpha} := \phi^{-1}(D^{\leq \alpha})$ which we call the {\em
  pull-back filtration} by $\phi$. Endowing $C$ with this filtration
makes the map $\phi$ strictly filtered.

Let $C$ be a filtered chain complex, and $x \in C$. Define
$A(x) \in \mathbb{R} \cup \{-\infty, \infty\}$ to be the infimal
filtration level of $C$ which contains $x$, i.e.
$A(x) := \inf \{\alpha \in \mathbb{R} \mid x \in C^{\leq \alpha}\}$.
We call $A(x)$ the {\em action level} of $x$. Sometimes we will write
$A(x;C)$ instead of $A(x)$ in order to keep track of the chain complex
that $x$ belongs to. By our conventions we have $A(0) = -\infty$ and
if $\cap_{\alpha \in \mathbb{R}} C^{\leq \alpha} = \{0\}$ then
$A(x)=-\infty$ iff $x=0$. Also, if the filtration on $C$ is
exhaustive, i.e. $\cup_{\alpha \in \mathbb{R}} C^{\leq \alpha} = C$,
then $A(x) < \infty$ for every $x \in C$.

Another measurement relevant for our considerations is the following.
\label{pp:delta-f}
Let $(C,d^{C})$, $(D, d^{D})$ be filtered chain complexes and
$f: C \longrightarrow D$ a strictly filtered $\mathcal{R}$-linear
map. Define the ``{\em action drop}'' of $f$ as
$\delta_{f} = \sup\{r\in [0,\infty) \mid \forall a\in \R, \ f(C^{\leq
  a}) \subset D^{\leq a-r}\}$. An important special case is when $C=D$
and $f = d^C$ the differential of $C$:
\begin{equation} \label{eq:delta-dC} \delta_{d^C} = \sup\{r\in
  [0,\infty) \mid \forall a\in \R, \ d^C(C^{\leq a}) \subset C^{\leq
    a-r}\}~.~
\end{equation}

A central measurement in our framework is the following. Let
$\psi: C \longrightarrow D$ be a filtered chain map and assume that
$\psi$ is null-homotopic. Define its {\em homotopical boudary level}
$B_h(\psi)$ to be the infimal action shift needed for a chain homotopy
between $\psi$ and $0$. More precisely:
\begin{equation} \label{eq:Bh-1}
  \begin{aligned}
    B_h(\psi) = \inf \bigl\{ \rho \in \mathbb{R} \mid \, & \exists \;
    \text{an} \; \mathcal{R} \text{-linear map} \; h: C
    \longrightarrow D\; \text{which shifts action by} \, \leq \rho, \\
    & \text{and such that} \; \psi = hd^C + d^Dh \bigr\}.
  \end{aligned}
\end{equation}
This notion is closely related to the boundary depth measurement
introduced by Usher~\cite{Usher1, Usher2}. In order to put things in
the right context we will explain this relation and further notions in
the next section.

\subsubsection{Boundary depth and related algebraic
  notions} \label{sec:bdry-depth}

Let $(C,d^C)$, $(D, d^D)$ be filtered chain complexes and
$\phi: C \longrightarrow D$ a {\em strictly} filtered chain
map. Denote by $Z_C^{\leq \alpha} \subset C^{\leq \alpha}$ the cycles
of $C^{\leq \alpha}$ and by $B_D \subset D$ the boundaries of $D$.
\begin{dfn}\label{def:b-depth1}
  The boundary depth of $\phi$ is defined by:
  \begin{equation}
    \begin{aligned}
      \beta (\phi)=\inf \Bigl\{ r \geq 0 \ \bigm| \ & \forall \,
      \alpha \in \R, \forall \, x \in Z_C^{\leq \alpha} \; \text{with}
      \;
      \phi(x) \in B_D, \\
      & \exists \, b \in D^{\leq \alpha+r} \; \text{such that} \;
      \phi(x) = d^D(b)\Bigr \}.
  \end{aligned}
  \end{equation}
\end{dfn}

This notion was introduced and studied extensively in symplectic
topology (in a slightly different formulation) by Usher~\cite{Usher1,
  Usher2}.

A special case of interest is the following. Let $c\in (C,d^{C})$ be a
boundary. Denote by $\langle c \rangle$ be the chain complex with a
single generator $c$ and zero differential and let
$j : \langle c \rangle \longrightarrow C$ be the inclusion. Endow
$\langle c \rangle$ with the pull-back filtration by $j$. Note that
with this filtration $j$ is a strictly filtered map.  Define now
\begin{equation} \label{eq:beta-c}
  \beta(c;C) := \beta \big(\langle c \rangle \stackrel{j}{\longrightarrow}
  C \bigr)~.~
\end{equation}
It is easy to see that
$\beta(c;C) = \inf \{ r \geq 0 \mid c \; \text{is a boundary in} \;
C^{\leq A(c)+r} \}$.  Note also that for every boundary $c\in C$ we
have $\beta(c;C)\geq \delta_{d^C}$, where $\delta_{d^C}$ is defined
in~\eqref{eq:delta-dC}.

Sometimes it would be more convenient to work with the following
quantity instead of $\beta(c; C)$. Let $c \in C$ be a boundary. Define
\begin{equation} \label{eq:B-c} B(c;C) = \inf \bigl\{ \alpha \in
  \mathbb{R} \mid \, \exists \, b \in C^{\leq \alpha} \; \text{such
    that} \; c=d^Cb \bigr\}.
\end{equation}
Clearly, $B(c;C) = A(c;C) + \beta(c;C)$. We call $B(c;C)$ the {\em
  boundary level} of the element $c$.

Returning to the boundary depth measurement, another special case is
the boundary depth of a chain complex $C$. This is the boundary depth
of the identity of $C$, i.e. $\beta(C)=\beta(id_{C})$. This
measurement is especially useful when $C$ is acyclic.

Yet another instance of the same notion appears in the following
setting. Let $(C,d^C)$ and $(D,d^D)$ be two filtered chain
complexes. Denote by $\hom_{\mathcal{R}}(C,D)$ the
$\mathcal{R}$-linear maps $C \longrightarrow D$.  Then
$\hom_{\mathcal{R}}(C,D)$ is a chain complex with differential
$d^{\hom}(f) = d^D \circ f - f \circ d^C$. (In case $C$ and $D$ are
graded we take $\hom_{\mathcal{R}}(C,D)$ to consist only of linear
maps that preserve degree up to a shift, and the differential is
modified to be $d^{\hom}(f) = d^D \circ f - (-1)^{|f|} f \circ d^C$,
where $|f|$ is the degree-shift of $f$.) The cycles in
$\hom_{\mathcal{R}}(C,D)$ are chain maps and the boundaries are the
null-homotopic chain maps. The chain complex $\hom_{\mathcal{R}}(C,D)$
is filtered where $\hom_{\mathcal{R}}^{\leq \gamma}(C,D)$ is the
subcomplex consisting of all $\mathcal{R}$-linear maps
$C \longrightarrow D$ that shift action by $\leq \gamma$.

Now let $\psi: C \longrightarrow D$ be a filtered chain map and assume
that $\psi$ is null-homotopic (i.e. $\psi$ is a boundary in
$\hom_{\mathcal{R}}(C,D)$). We define the {\em homotopical boundary
  depth} $\beta_h(\psi)$ of $\psi$ by
\begin{equation} \label{eq:betta-h}
  \beta_{h}(\psi) := \beta(\psi;
  \hom_{\mathcal{R}}(C,D))~.~
\end{equation}
More explicitly, $\beta_{h}(\psi)$ is the infimal $r \geq 0$ for which
$\psi$ is null-homotopic via a chain homotopy that shifts filtration
by $\leq A(\psi) + r$. As before, we define also the {\em homotopical
  boundary level} $$B_h(\psi) = B(\psi; \hom_{\mathcal{R}}(C,D)).$$

Note that if $\psi: C \longrightarrow D$ is a strictly filtered chain
map which is null homotopic then:
\begin{equation} \label{eq:beta-beta_h} \beta(\psi) \leq B_h(\psi) =
  A(\psi) + \beta_h(\psi) \leq \beta_h(\psi),
\end{equation}
where the last inequality holds because $\psi$ is strictly filtered,
hence $A(\psi)\leq 0$.

\begin{rem} \label{r:beta-filtered} We have defined the boundary depth
  $\beta(\psi)$ only for {\em strictly} filtered chain maps. However,
  the homotopical boundary depth $\beta_h(\psi)$ is defined for all
  filtered (null-homotopic) chain maps $\psi$, not only for the
  strictly filtered ones. This is because, in contrast to
  $\beta(\psi)$, the homotopical boundary depth $\beta_h(\psi)$ is
  defined as the boundary depth of the element $\psi$ inside the
  filtered chain complex $\hom_{\mathcal{R}}(C,D)$.
\end{rem}

Finally, here is another variant of the same measurement.  Let
$\mathcal{A}$ be a weakly filtered $A_{\infty}$-category with
discrepancy $\leq \bmea$ (see~\S\ref{s:wf-ai-theory}). Let
$\bmemm = (\epsilon_1^m=0, \epsilon_2^m, \ldots, \epsilon_d^m,
\ldots)$ be a sequence of non-negative real numbers, and let
$\mathcal{M}_0, \mathcal{M}_1$ be two weakly filtered
$\mathcal{A}$-modules with discrepancy $\leq \bmemm$. Let $\bmeh$ be
another sequence of non-negative real numbers, and assume that
$\bmeh \in~\hyperlink{h:asmp-e}{\lbe(\bmemm,
  \bmea)}$. (See~\S\ref{sb:mod}, page~\pageref{pp:assump-E}.)

Denote by $\hom^{\bmeh}(\mathcal{M}_0, \mathcal{M}_1)$ the weakly
filtered pre-module homomorphisms
$\mathcal{M}_0 \longrightarrow \mathcal{M}_1$ with discrepancy
$\leq \bmeh$ (and arbitrary action shift). As explained
in~\S\ref{sb:mod}, $\hom^{\bmeh}(\mathcal{M}_0, \mathcal{M}_1)$ is a
chain complex when endowed with the differential $\mu_1^{\text{mod}}$
of the dg-category of $\mathcal{A}$-modules. Moreover, this chain
complex is filtered by
$\hom^{\leq \rho; \bmeh}(\mathcal{M}_0, \mathcal{M}_1)$,
$\rho \in \mathbb{R}$.

Now let $\psi: \mathcal{M}_0 \longrightarrow \mathcal{M}_1$ be a
weakly filtered module homomorphism with discrepancy $\leq \bmeh$, and
assume that $\psi$ is a boundary in
$\hom^{\bmeh}(\mathcal{M}_0, \mathcal{M}_1)$ (i.e. $\psi$ is chain
homotopic to $0$ via a chain homotopy of pre-module maps with
discrepancy $\leq \bmeh$. Then we can define
$\beta_h(\psi; \bmeh) := \beta \bigl(\psi; \hom^{\bmeh}(\mathcal{M}_0,
\mathcal{M}_1)\bigr)$. Now, let $X \in \textnormal{Ob}(\mathcal{A})$
and denote by
$\psi_1^X: \mathcal{M}_0(X) \longrightarrow \mathcal{M}_1(X)$ the
chain map which is the 1'st order component of $\psi$ (corresponding
to the object $X$). By the preceding discussion we can associate to
$\psi_1^X$ its homotopical boundary depth
$\beta_h(\psi_1^X) = \beta \bigl(\psi_1^X;
\hom_{\mathcal{R}}(\mathcal{M}_0(X), \mathcal{M}_1(X)) \bigr)$ by the
recipe~\eqref{eq:betta-h}.  The two versions of the boundary depth
$\beta_h(\psi; \bmeh)$ and $\beta_h(\psi_1^X)$ are related as follows:
$$\beta_h(\psi_1^X) +
A(\psi_1^X) \leq \beta_h(\psi; \bmeh) + A(\psi) + \epsilon_1^h.$$ Here
$A(\psi_1^X)$ is the action-level of $\psi_1^X$ viewed as an element
of the filtered chain complex
$\hom_{\mathcal{R}}(\mathcal{M}_0(X), \mathcal{M}_1(X))$ while
$A(\psi)$ is the action level of $\psi$, viewed as an element of the
filtered chain complex $\hom^{\bmeh}(\mathcal{M}_0, \mathcal{M}_1)$.

In case $\psi_1^X$ is strictly filtered, then we also have the
following two inequalities:
$$\beta(\psi_1^X) \leq \beta_h(\psi; \bmeh) + A(\psi) + \epsilon_1^h,
\quad \beta(\psi_1^X) \leq \beta_h(\psi_1^X),$$ where the 2'nd
inequality comes from~\eqref{eq:beta-beta_h}.

\subsubsection{Algebraic implications} \label{sb:alg-impl}

We begin with a simple algebraic approximation lemma which says that a
altering a chain isomorphism by a chain homotopy yields an injective
map provided that the chain homotopy shifts action by a small enough
amount.

\begin{lem} \label{l:rig-cplx-n} Let $(C, d^C)$ and $(D, d^D)$ be
  filtered chain complexes, and assume that the filtration on $C$ is
  exhaustive (i.e.
  $\cup_{\alpha \in \mathbb{R}} C^{\leq \alpha} = C$) and separated
  (i.e.  $\cap_{\alpha \in \mathbb{R}} C^{\leq \alpha} = 0$). Let
  $f, g: C \longrightarrow D$ be chain maps with the following
  properties:
  \begin{enumerate}
  \item $g$ is an isomorphism.
  \item Both $g$ and $g^{-1}$ are strictly filtered.
  \item $f-g$ is null-homotopic and
    $B_h(f-g) < \min\{\delta_{d^C}, \delta_{d^D}\}$.
  \end{enumerate}
  Then $f$ is strictly filtered and moreover $f$ is injective.
\end{lem}
In our geometric applications $D=C$, $g$ will be the identity, and $f$
will be the composition of two chain morphisms
$C\stackrel{f_{1}}{\longrightarrow}
C'\stackrel{f_{2}}{\longrightarrow} C$ that are constructed
geometrically. The Lemma shows in this case that the middle complex
$C'$ contains $C$ as a retract. Results of this sort are familiar in symplectic topology  since
\cite{Co-Ra:Morse-Novikov}.

\begin{proof}[Proof of Lemma~\ref{l:rig-cplx-n}]
  Since the filtartion on $C$ is both exhaustive and separated, we
  have $-\infty < A(x) < \infty$ for every $x \neq 0$, and
  $A(0) = -\infty$.

  Set $\rho := B_h(f-g) + \epsilon$, where $\epsilon>0$ is small
  enough such that $\rho < \min \{\delta_{d^C}, \delta_{d^D}\}$.
  Write $f = g + \eta d^C + d^D \eta$, where
  $\eta : C \longrightarrow D$ is an $\mathcal{R}$-linear map that
  shifts action by $\leq \rho$.  Since
  $\rho < \min\{\delta_{d^C}, \delta_{d^D}\}$ and $g$ is strictly
  filtered we have
  $$A(f(x)) = A\bigl(g(x) + \eta d^C(x) + d^D \eta(x)\bigr) \leq A(x),
  \; \forall x \in C,$$ hence $f$ is strictly filtered.

  For the injectivity of $f$, assume that $f(x)=0$ for some
  $x \neq 0$. Then $$g(x) = - (\eta d^C(x) + d^D \eta(x)),$$ and using
  again that $\rho < \min\{\delta_{d^C}, \delta_{d^D}\}$ we obtain
  that
  $$A(g(x)) = A(\eta d^C(x) + d^D \eta(x)) < A(x).$$
  The last inequality together with the assumption that $g^{-1}$ is
  strictly filtered imply that
  $$A(x) = A(g^{-1}g(x)) \leq A(g(x)) < A(x).$$ A contradiction.
\end{proof}

Under additional assumptions we can obtain a somewhat stronger
result. Before we state it, here are a couple of relevant notions.
The filtration $C^{\leq \alpha} \subset C$, $\alpha \in \mathbb{R}$
induces a topology on $C$ which is generated by the cosets of
$C^{\leq \alpha}$, $\alpha \in \mathbb{R}$, as basic open subsets. The
assumption that the filtration is separated (i.e.
$\cap_{\alpha \in \mathbb{R}} C^{\leq \alpha} = 0$) implies that $C$
is Hausdorff in this topology.

The filtration on $C$ is called complete if the obvious map
$C \longrightarrow \displaystyle{\varprojlim_{\alpha}} (C / C^{\leq
  \alpha})$ is surjective. This assumption implies that the previously
mentioned topology on $C$ turns $C$ into a complete topological space
(in the sense that every Cauchy sequence converges).

\begin{lem} \label{lem:rig-cplx-c} Let $(C,d^C)$, $(D,d^D)$, $f$, $g$
  be as in Lemma~\ref{l:rig-cplx-n} and assume in addition that the
  filtration on $C$ is complete. Then $f$ is a strictly filtered
  isomorphism and moreover $f^{-1}$ is also strictly filtered.
\end{lem}

\begin{proof}
  In view of Lemma~\ref{l:rig-cplx-n} we only need to show that $f$ is
  an isomorphism and that $f^{-1}$ is strictly filtered.

  We will use a well-known inversion trick, that has already been used
  in a similar setting in~\cite{Usher1, Usher2}. Fix
  $0< \epsilon <(\min\{\delta_{d^C}, \delta_{d^D}\}-B_{h}(f-g))/2$. By
  the definition of $B_{h}$ there is an $\mathcal{R}$-linear map
  $\eta: C \longrightarrow D$ that shifts actions by
  $\leq B_h(f-g)+\epsilon$ such that $f-g=d^{C}\eta + \eta d^{C}$.
  Note that $f-g$ {\em decreases} action by at least $\epsilon$.

  Now write $f = g + (f - g) = g (\id+g^{-1}(f-g)) = g(\id - k)$,
  where $k: C \longrightarrow C$ is defined by $k=-g^{-1}(f-g)$. Since
  $g^{-1}$ is strictly filtered and $f-g$ decreases filtration by at
  least $\epsilon$, the same is true for $k$. As the filtration on $C$
  is complete, the series $a = \id + \sum_{n \geq 1} k^n$ converges,
  and satisfies $(\id-k)a = a(\id-k) = \id$. Therefore $f$ is
  invertible with inverse $a g^{-1}$, a strictly filtered
  chain-morphism of $\mathcal{R}$-modules.
\end{proof}

For the next result we will assume that $\mathcal{R} = \Lambda_0$ (the
positive Novikov ring over any field $R$). Recall that the Novikov
ring $\Lambda$ is the field of fractions of $\Lambda_0$.  Denote by
$\nu: \Lambda \longrightarrow \mathbb{R} \cup
\{\infty\}$ the standard valuation defined by
\begin{equation} \label{eq:val-Nov} \nu \Bigl(a_0 T^{\lambda_0} +
  \sum_{i=1}^{\infty} a_i T^{\lambda_i} \Bigr) = \lambda_0,
\end{equation}
where $a_0 \neq 0$ and $\lambda_i > \lambda_0$ for every $i\geq 1$. As
usual we set $\nu(0) = \infty$.

Let $C$ be a finite dimensional chain complex over $\Lambda$. Fix a
basis $\mathcal{G}$ of $C$ over $\Lambda$ and let
$A: \mathcal{G} \longrightarrow \mathbb{R}$ be a function. Similarly
to~\S\ref{sbsb:filtration-mixed} we will use $A$ to define a
filtration on $C$ by $\Lambda_0$-modules. Extend $A$ to a function
$A:C \to \R \cup \{-\infty\},$ by
$$A\Bigl(\sum \lambda_j e_j \Bigr) = \max\{-\nu(\lambda_j) + A(e_j)\},$$
where $e_j$ are the elements of the basis $\mathcal{G}$,
$0 \neq \lambda_j \in \Lambda$, $A(e_j)$ is the
pre-determined value of $A$ on the generator $e_j,$ and $\nu$ is the
preceding valuation. Define now
$$C^{\leq \alpha} := \{ x\in C \mid A(x) \leq \alpha\}.$$
It is easy to see that $C^{\leq \alpha} \subset C$,
$\alpha \in \mathbb{R}$, is an increasing filtration of $C$ by
$\Lambda_0$-modules (though not by vector space over $\Lambda$). Since
$A(x) = -\infty$ iff $x=0$, this filtration is separated. Moreover, it
is exhaustive and complete.

From now on we will make the following {\em standing assumption:}
$A(d^C x) \leq A(x)$, $\forall \, x \in C$. In other words, we assume
that each $C^{\leq \alpha} \subset C$, $\alpha \in \mathbb{R}$, is a
subcomplex of $C$ (over $\Lambda_0$).

It is important to note that the function $A$, as defined above,
coincides with the action level of the preceding filtration on $C$, as
defined at the beginning of~\S\ref{s:filt-ch}. Thus no confusion
should arise by denoting them both by $A$.

We will make use of the following defintion from~\cite{Usher-Zhang}.
\begin{dfn} \label{d:delta-robust}
  A subspace $V \subset Ker(d^C) \subset C$ is called {\em
    $\delta$-robust} if for all $v \in V$ and $w \in C$ such that
  $v = d^C(w),$ we have $A(w) \geq A(v) + \delta.$
\end{dfn}

\begin{rem} \label{rem:delta-robust-subspace-in-kernel} According to
  the above definition, a complement $W$ in $Ker(d^C)$ to $Im(d^C)$ is
  a $\delta$-robust subspace for all $\delta > 0.$ Hence if
  $V \subset Im(d^C)$ is $\delta$-robust then $V \oplus W$ is also
  $\delta$-robust. We will call a $\delta$-robust subspace
  $V \subset Im(d^C)$ a {\em proper $\delta$-robust} subspace.
\end{rem}

\begin{prop}\label{prop:rig-cplx2} Let $(C,d^{C})$ be a chain complex
  as above, and let $f:C \longrightarrow C$ be a chain map. Assume
  that $d^{C}$ splits as a sum $d^{C}=d_{0}+d_{1}$ such that $d_0$ is
  an $\Lambda$-linear differential which (like $d^C$)
  also preserves the given filtration on $C$. Furthermore, assume that
  $\dim_{\Lambda}(H_{\ast}(C,d_{0})) \geq
  \dim_{\Lambda}(H_{\ast}(C,d^C))$. If
  $B_{h}(f-id_{C})< \delta_{d_{1}}$, then
  $$\dim_{\Lambda}(Im(f))\geq
  \dim_{\Lambda}(H_{\ast}(C,d_{0}))~.~$$
\end{prop}

The proposition follows directly from the following two lemmas.

\begin{lem}\label{lem:delta-robust-existence} Let $(C, d^C)$
  be a chain complex as above, and assume that its differential splits
  as $d^C = d_0+d_1$ with $d_0$ satisfying the same assumptions as in
  Proposition~\ref{prop:rig-cplx2}. Then
  $\dim_{\Lambda}(H_{\ast}(C,d_{0})) -
  \dim_{\Lambda}(H_{\ast}(C,d^C))$ is
  even. Furthermore, denote the latter number by $2k$ and assume that
  $k \geq 0$. Then $(C,d_C)$ admits a proper $\delta_{d_1}$-robust
  subspace of dimension at least $k$.
\end{lem}

\begin{lem} \label{lem:delta-robust-injective} Let $(C, d^C)$ be a
  chain complex as in Lemma~\ref{lem:delta-robust-existence} and
  $f: C \longrightarrow C$ be a chain map. Let $0< \epsilon < \delta$
  and suppose that $B_h(f-id_{C}) = \delta - \epsilon$. Then $f$ is
  injective on each (resp. proper) $\delta$-robust subspace, and maps
  it to a (resp. proper) $\epsilon$-robust subspace.
\end{lem}

\begin{proof}[Proof of Proposition~\ref{prop:rig-cplx2}]
  By Lemma~\ref{lem:delta-robust-existence}, there exists a proper
  $\delta_{d_{1}}$-robust subspace $V$ in $(C,d^C)$ of dimension $k$
  (where $k$ is given by that lemma). By
  Lemma~\ref{lem:delta-robust-injective}, $f(V)$ will be a proper
  $\epsilon$-robust subspace of dimension $k$. Consider a subspace
  $V' \subset C$ of dimension $k$ such that $d^C(V') = V$, and a
  complement $W$ in $Ker(d^C)$ to $Im(d^C).$ Then $d^C(f(V')) = f(V),$
  showing that $\dim d^C(f(V')) = k,$ and $f(W)$ will again be a
  complement in $Ker(d^C)$ to $Im(d^C)$. (Note that
  $f(W) \cap d^C(C) = 0$ because, by assumption, $f-\id_C$ is
  null-homotopic, so $f$ induces an isomorphism in homology.) Now, by
  Lemma~\ref{lem:delta-robust-injective} again, $f(W)$ will have the
  correct dimension. Finally the three subspaces $f(V),f(V'),f(W)$ are
  direct summands of $C$ whence
  $\dim_{\Lambda}(Im(f)) \geq
  \dim_{\Lambda}(H_{\ast}(C,d^C)) + 2k$, finishing the
  proof.
\end{proof}

\begin{proof}[Proof of Lemma \ref{lem:delta-robust-existence}]
  The identities
  \begin{equation*}
    \begin{aligned}
      & \dim_{\Lambda}(C) =
      \dim_{\Lambda} (H_{\ast}(C,d^C)) + 2
      \dim_{\Lambda} (Im(d_C)), \\
      & \dim_{\Lambda}(C) =
      \dim_{\Lambda} (H_{\ast}(C,d_{0})) + 2
      \dim_{\Lambda} (Im(d_{0})),
    \end{aligned}
  \end{equation*}
  show that
  $\dim_{\Lambda}(H_{\ast}(C,d_{0})) -
  \dim_{\Lambda}(H_{\ast}(C,d^C))$ is even. Moreover we
  obtain:
  \begin{equation} \label{eq:dimension-relation}
    \dim_{\Lambda} (Im(d^C)) =
    \dim_{\Lambda} (Im(d_{0})) + k.
  \end{equation}

  From~\cite[Proposition 7.4]{Usher-Zhang}, it is immediate to
  construct a projection $\pi : C \longrightarrow Im(d_{0})$, that
  restricts to the identity on $Im(d_0)$ and satisfies
  $A(\pi(x)) \leq A(x)$ for all $x \in C$.

  From~\eqref{eq:dimension-relation} we now have that
  $\dim(Ker(\pi|_{Im(d^C)})) \geq k.$ We claim that
  $V = Ker(\pi|_{Im(d^C)})$ is $\delta_{d_1}$-robust. Indeed, if
  $v \in V, w \in C,$ and $v = d^C w,$ then writing
  $d^C w = d_0 w + d_1 w,$ and using $\pi(v) = 0$ we obtain
  $d_0w = \pi(d_0 w) = - \pi(d_1 w),$ whence $v = (id - \pi) (d_1 w).$
  Therefore
  $A(v) = A((id - \pi) (d_1 w)) \leq A(d_1 w) \leq A(w) -
  \delta_{d_1}$.  This implies $A(w) \geq A(v) + \delta_{d_1}$,
  concluding the proof.
\end{proof}

\begin{proof}[Proof of Lemma~\ref{lem:delta-robust-injective}]
  Let $V \subset C$ be a $\delta$-robust subspace. We write
  $f = id_C + d^C h - h d^C,$ where
  $A(h(x)) \leq A(x) + (\delta - \epsilon)$, for all $x \in C$. If
  $v \in V$ is such that $f(v) = 0,$ we would have
  $v + d^C(h(v)) = 0,$ which would yield $w = - h(v),$ with $v = dw$
  and $A(w) \leq A(v) + \delta - \epsilon$. On the other hand
  $\delta$-robustness implies $A(w) \geq A(v) + \delta$. A
  contradiction.

  If $f(v) = d^C z,$ we would have $v + d^C(h(v)) = d^C z,$ which
  would yield $w = z - h(v),$ with $v = dw.$ Therefore by
  $\delta$-robustness we obtain
  $A(v) + \delta \leq A(z- h(v)) \leq \max \{A(h(v)),A(z)\}.$ Since
  $A(h(v)) \leq A(v) + \delta - \epsilon,$ we get
  $A(v) + \delta \leq A(z),$ and
  $A(f(v)) \leq \max \{ A(v), A(h(v)) \} \leq A(v) + \delta - \epsilon
  \leq A(z) - \epsilon.$ We conclude that
  $A(z) \geq A(f(v)) + \epsilon,$ which finishes the proof.
\end{proof}

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