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real analysis – Periodical functions with $f^{(n)} = f$, but $f^{(k)} neq f$ for $kin {1,ldots,n-1}$

For any infinitely differentiable function $f: mathbb{R}to mathbb{R}$ and positive integer $kinmathbb{N}$, let $f^{(k)}$ denote the $k$-th derivative of $f$.

For which $ninmathbb{N}$, $n>1$, is there a periodical function $f:mathbb{R}to mathbb{R}$ with the property that $f^{(n)} = f$, but $f^{(k)} neq f$ for $kin {1,ldots,n-1}$?

real analysis – Finiteness of a bilinear combination

For $jinmathbb{N}$, consider continuous functions $f_j:(0,1)tomathbb{mathbb{R}^+}$ such that
namely $f_j(t)in L_t^{infty}((0,1),l_j^1(mathbb{N}))$. I would like to understand whether the quantity
is finite. Some observations:

  • If for every $j$ the function $f_j$ is increasing in $t$, then $S_f$ is finite. Indeed,

$$S_fleq sum_{jinmathbb{N}}f_j(T)sum_{kinmathbb{N}}int_0^1 f_k(t)dtleq sum_{jinmathbb{N}}f_j(T)int_0^1sup_{tin(0,1)}sum_{jinmathbb{N}}f_k(t)dt<+infty$$

Similarly, if the functions $f_j$ attains their maxima on a finite set of points in $(0,1)$, then $S_f$ is finite.

  • More generally, $S_f$ is finite if $f_j(t)in L_j^1L_t^{infty}$, but of course this is not always the case. For example, one can take (a smooth modification of) $f_j(t):=delta_{t=j^{-1}}$. Nevertheless, also in this case $S_f$ is finite.

  • I tried to consider the case when the $f_j$ have plenty of oscillations, but I was unable to find a counterexample.

So my question is the following: is it true that $S_f$ is always finite?

In the context I’m interested, there is an additional (weak) control of the derivatives of $f_j$, of the form $2^{-j}f_j’in L^1(0,1)$, so one could also try to prove the finiteness of $S_f$ under this additional assumption.

Thank you for any suggestion.

complexity classes – What is the computational class of a pushdown automaton with real values?

Say there is a push-down automata, in this example I’ll use a deadfish-like set:

+: increase x by 1

0: set x to 0

ln: set x to ln(x) <-- real valued result

With x being an infinite precision real-valued variable, does this allow said machine to have more power than if it was operating on integers? Or am I misunderstanding something?

real analysis – Homeomorphism between two metric spaces via identity

Suppose $(X,d) $ is a complete metric space with $U_1,U_2,…$ nonempty open subsets, with none equal to $X.$ Let $U=
bigcap_{n=1}^{infty } U_n neq emptyset$
and define $d_n $ on
$U_n $ as $$d_{n}(x,y) =text{min} (D_{n} (x,y),1)$$ where $$D_{n}(x,y)
=d(x,y)+lvert frac{1}{d(x,U_n^c) } – frac{1}{d(y,U_n^c)}
rvert. $$
Define $$D(x,y)=sum_{n=1}^{infty } frac{1}{2^n} d_{n}
(x,y). $$

If I want to show that $(U,d)$ is homeomorphic to $(U,D)$ by the identity function, then I can show that $(U,d)$ and $(U,D)$ have the same open sets.

$$implies $$
Let $V$ be an open set in $(U,d)$. Let $x in V$ and $r<1.$ If
$d(x,y)<r$, then $y in V.$ $f(x)=x$ so $x in f ^{-1} (V).$ If
$D(x,y)<r,$ then $f(y)=y in V$ so $y in f ^{-1} (V).$ Thefore $f
^{-1} (V) $
is an open set.

Let $x in f ^{-1} (V)$ open set. Let $r<1.$ $D(x,y)<r implies y in
f ^{-1} (V).$
$f(x)=x in V.$ If $d(x,y)<r,$ then $y in f ^{-1} (V)$
so $f(y)=y in V.$ Thus $V$ is an open set so $(U,d)$ is homeomorphic
to $(U,D).$

Is it correct to use a radius of less than 1? I did this because $D(x,y)$ is less than or equal to 1.


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real analysis – Proving the limit exists when point tends to the cluster point of a set

Let c be a cluster point of A $subset mathbb{R}$ and $f: A rightarrow mathbb{R}$ be a function. Suppose that for every sequence in {$x_{n}$} in A, such that $lim x_{n} = c$, the sequence ${f(x_{n})}_{0}^{infty}$ is Cauchy. Prove that $lim _{xrightarrow c}f(x)$ exists.

I did not know how to approach this question, but I found a skeleton answer here:

Why do I need to show uniqueness?

I still don’t understand how we show that the limit exists though.

real analysis – Definition of the flow of an ODE and its inverse

At lesson, the teacher considers a flow $Phi$ given by the solutions of the ode system for $tin(0, T)$ and $xinmathbb R^d$,
y'(s)=b(y(s), s),&sleq T\

that is $Phi(x, t, s)=y(s)$ solving $(star)$. He said that we will be mostly concerned with $Phi(cdot, 0, cdot)$. The field $b$ is assumed to be Lipschitz continuous in both variables and bounded.

Then, he intoduces the inverse $Psi$ of the above flow as follows: $Psi(x, 0, s)=y(s)$ satisfying
y'(s)=-b(y(s), t-s),&s<tleq T\

and he said that $Psi$ is such that
Phi(Psi(x, 0, s), 0, s)=x,quad Psi(Phi(x, 0, s), 0, s)=x.qquad (starstar)

I do not understand $(starstar)$. Can someone help me? Maybe is the definition of the inverse wrong?

Thank you

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