fourier analysis – How is the Cauchy-Schwarz inequality used to bound this derivative?

In “Hardy’s Uncertainty Principle, Convexity and Schrödinger Evolutions” (link) on page 5, the authors state that they are using the Cauchy-Schwarz inequality to bound the derivative of the $L^2(mathbb{R}^n)$ norm of a solution to a certain differential equations, but I am not sure how exactly they applied it.

Some context: Let $v$, $phi$, $V$, $F$ be nice enough functions of $x$ and $t$ so that the following integrals are well-defined, $A>0, Binmathbb{R}$ be constants, and $u=e^{-phi} v$ solve
$$partial_t u = (A+iB)left( Delta u + Vu+Fright).$$
Denote the $L^2$ inner product on $mathbb{R}^n$ between some $f$ and $g$ as $(f, g) = int f g^{dagger} dx$, where $g^dagger$ is the complex conjugate of $g$, and define $f^+=mathrm{max}{f, 0}$.

We know the equality
$$partial_t vert !vert v vert!vert^2_{L^2} = 2,mathrm{Re}left(Sv,vright) + 2,mathrm{Re}left((A+iB)e^phi F, vright),$$
where
$$mathrm{Re}left(Sv,vright) = -Aint |nabla v|^2 + left(A|nabla phi|^2+partial_t phi right) |v|^2 + 2B ,mathrm{Im}, v^{dagger} nablaphicdotnabla v + left( A,mathrm{Re},V – B,mathrm{Im}, Vright)|v|^2dx,$$
holds true. The authors go on to conclude that the Cauchy-Schwarz inequality implies that
$$partial_t vert !vert v(t) vert!vert^2_{L^2} le 2vert !vert A , left(mathrm{Re},V(t)right)^+ – B,mathrm{Im}, V(t) vert !vert_{infty}vert !vert v(t) vert !vert^2_{L^2} + 2 sqrt{A^2+B^2} vert !vert F e^phi vert !vert_{L^2} vert !vert v(t) vert !vert_{L^2}$$
when
$$left(A+frac{B^2}{A}right)|nablaphi|^2 + partial_t phile 0, ,,,,mathrm{in}, mathbb{R}_+^{n+1}.$$
However, I am not sure how the authors used the C.S. inequality to arrive at this conclusion, and am especially confused as to where the factor of $B^2/A$ came from, and why we only need the constraint to hold over $mathbb{R}_+^{n+1}$ when we are integrating over all of $mathbb{R}^{n}$, though I understand why we only care about positive time.

Does anyone have any insight here?

equation solving – Reduce for a simply inequality

Clear("Global`*")

g = P*(S - s)/(w*x*nl);

sqre = Sqrt((1 - x*g)^2 - (4*g));

In the Reduce the inequalities have to be a single expression: either a List or joined by And (&&). Your xg must be either x g or x*g

sol = Reduce({(3 + x*g > -sqre), P > 0, S > 0, w > 0, x > 0, nl > 0, 
   0 <= s <= S}, s, Reals)

w > 0 && S > 0 && 
 P > 0 && ((0 < nl < (P S)/
      w && ((0 < x < (4 nl P S w)/(
          P^2 S^2 - 2 nl P S w + nl^2 w^2) && (-2 nl w + P S x - nl w x)/(
           P x) + 2 Sqrt((nl^2 w^2 + nl^2 w^2 x)/(P^2 x^2)) <= s <= 
          S) || (x == (4 nl P S w)/(
          P^2 S^2 - 2 nl P S w + 
           nl^2 w^2) && (s == 
            0 || (-2 nl w + P S x - nl w x)/(P x) + 
             2 Sqrt((nl^2 w^2 + nl^2 w^2 x)/(P^2 x^2)) <= s <= S)) || (x > (
          4 nl P S w)/(
          P^2 S^2 - 2 nl P S w + 
           nl^2 w^2) && (0 <= 
            s <= (-2 nl w + P S x - nl w x)/(P x) - 
             2 Sqrt((nl^2 w^2 + nl^2 w^2 x)/(P^2 x^2)) || (-2 nl w + P S x - 
              nl w x)/(P x) + 2 Sqrt((nl^2 w^2 + nl^2 w^2 x)/(P^2 x^2)) <= 
            s <= S)))) || (nl == (P S)/w && 
     x > 0 && (-2 nl w + P S x - nl w x)/(P x) + 
       2 Sqrt((nl^2 w^2 + nl^2 w^2 x)/(P^2 x^2)) <= s <= S) || (nl > (P S)/
      w && ((0 < x < (4 nl P S w)/(
          P^2 S^2 - 2 nl P S w + nl^2 w^2) && (-2 nl w + P S x - nl w x)/(
           P x) + 2 Sqrt((nl^2 w^2 + nl^2 w^2 x)/(P^2 x^2)) <= s <= 
          S) || (x >= (4 nl P S w)/(P^2 S^2 - 2 nl P S w + nl^2 w^2) && 
         0 <= s <= S))))

For a simpler expression, albeit with the relationship of s obscured

sol2 = Simplify(sol, {P > 0, S > 0, w > 0, x > 0, nl > 0, 0 <= s <= S})

(nl w < P S && ((s == 0 && (4 nl P S w)/(P S - nl w)^2 == x) || ((
        4 nl P S w)/(P S - nl w)^2 < x && 
       P s x + nl w (2 + x + 2 Sqrt(1 + x)) <= P S x))) || (nl w > P S && (
    4 nl P S w)/(P S - nl w)^2 <= x) || 
 P S x + 2 nl w Sqrt(1 + x) <= P s x + nl w (2 + x)

upper lower bounds – On an inequality involving the abundancy index of the Eulerian component $p^k$ of an odd perfect number $p^k m^2$

The topic of odd perfect numbers likely needs no introduction.

Denote the classical sum of divisors of the positive integer $x$ by $sigma(x)=sigma_1(x)$, the abundancy index of $x$ by $I(x)=sigma(x)/x$, and the deficiency of $x$ by $D(x)=2x-sigma(x)$.

Euler proved that a hypothetical odd perfect number must take the form
$$n = p^k m^2$$
where the Eulerian component $p^k$ satisfies the constraints $p equiv k equiv 1 pmod 4$ and $gcd(p,m)=1$.

Since $p$ is (the special) prime, we have the formula
$$I(p^k)=frac{sigma(p^k)}{p^k}=frac{p^{k+1} – 1}{p^k (p – 1)}$$
and corresponding (sharp?) upper bound
$$I(p^k)<frac{p^{k+1}}{p^k (p – 1)}=frac{p}{p – 1}.$$
We also have the formula
$$frac{D(p^k)}{p^k}=2-I(p^k),$$
and corresponding (sharp?) upper bound
$$frac{D(p^k)}{p^k}=2-I(p^k) leq 2-frac{p+1}{p}=frac{p-1}{p}.$$

We obtain
$$I(p^k)bigg(2 – I(p^k)bigg) < frac{p}{p-1}cdotfrac{p-1}{p}=1.$$

Here are my questions:

(1) Can one substantially improve on the bound
$$I(p^k)bigg(2 – I(p^k)bigg) < 1?$$

(2) If the answer to Question (1) is YES, my next question is “How?”.

(3) If the answer to Question (1) is NO, can you explain/show why the bound cannot be substantially improved?

MY ATTEMPT

I do am aware of the fact that
$$f(k):=g(p):=I(p^k)bigg(2 – I(p^k)bigg)=frac{sigma(p^k) D(p^k)}{p^{2k}}=frac{(p^{k+1} – 1)(p^{k+1} – 2p^k + 1)}{Bigg(p^k (p – 1)Bigg)^2}$$
and that
$$frac{partial f}{partial k} = -frac{2(p^k – 1)log(p)}{Bigg(p^k (p – 1)Bigg)^2} < 0$$
while
$$frac{partial g}{partial p} = frac{2(p^k – 1)(p^{k+1} – (k+1)p + k)}{p^{2k+1} (p – 1)^3} > 0.$$

This means that
$$g(5) leq g(q) = f(k) leq f(1)$$
since the computations above show that $f(k)$ is decreasing while $g(q)$ is increasing.

In particular, the quantity $f(1)$ in the inequality $f(k) leq f(1)$ simplifies to:
$$f(k) = I(p^k)bigg(2 – I(p^k)bigg) leq f(1) = I(p)bigg(2 – I(p)bigg) = frac{p+1}{p}bigg(frac{2p – (p+1)}{p}bigg) = frac{p^2 – 1}{p^2},$$
which somehow improves on the upper bound of $1$.

fa.functional analysis – Poincare Inequality for $H^2$ function satisfying homogeneous Robin boundary conditions

Let $Omegasubsetmathbb{R}^3$ be a bounded smooth domain. In general, for a Poincare inequality of the type
$$|u|_{L^2}le C |nabla u|_{L^2}$$
to hold for all $uin Xsubset H^1(Omega)$ and $C$ independent of $u$, then $X$ needs to be such that it doesn’t contain constant translates. That is, if we consider $u+M$ for large $M>0$, the left hand side of the inequality increases indefinitely while the right hand side is unchanged, so we need some extra constraint in the definition of $X$. So common choices are $X=H^1_0(Omega)$ or $X={uin H^1(Omega)| int_Omega u,dx=0}$.

Here’s my question. Suppose, we’d like to say that there exists $C$ such that for all $uin X={uin H^2(Omega)|(partial_n u+u)_{|partialOmega}=0}subset H^2(Omega)$ we have
$$|u|_{L^2}le C|nabla u|_{L^2}.$$
First, is this true? If so, how does one prove such a statement? Essentially the requirement that $u$ satisfies the homogeneous Robin condition $(partial_n u+u)_{|partialOmega}=0$ should at least formally rule out constant translates, since $(partial_n u+u)_{|partialOmega}=0$ is not invariant under translation of $u$ by constants.

My guess is that it IS true, however, the usual proof I know of such statements usually relies on some compactness argument. For example, if $X$ were simply $H_0^1(Omega)$, then for the sake of contradiction, if we assume that there exists a sequence $u_nin H_0^1$ such that
$$|u_n|_{L^2}ge n|nabla u_n|_{L^2}$$
then, defining $v_n=u_n/|u_n|_{L^2}$, we have
$$frac{1}{n}ge |nabla v_n|_{L^2}.$$
Thus we have a bounded sequence in $H^1$ and a subsequence that converges strongly in $L^2$ and weakly in $H^1$ to some $vin H^1$. Because $|nabla v_n|_{L^2}to 0$, $v$ is constant. And since the trace map is continuous (and weakly continuous) from $H^1(Omega)$ to $H^frac{1}{2}(partialOmega)$ we have that $v$ is in fact in $H^1_0(Omega)$ and therefore $v=0$. Then we have a contradiction because $|v_n|_{L^2}=1$ for each $n$ implies that $|v|_{L^2}=1$.

Now this argument doesn’t work for Robin boundary conditions because now the relevant (Robin) trace operator is continuous and weakly continuous from $H^2(Omega)$ to $H^frac{1}{2}(partialOmega)$. In particular, if $v$ is the weak $H^1$ limit of a sequence $v_nin H^2$, then $v$ could be in $H^1$ but not $H^2$ and thus the notion of the normal derivative $(partial_n v)_{|partialOmega}$ may not even make sense for $v$. And without being able to say $(partial_n v+v)_{|partialOmega}=0$, we can’t necessarily say that $v=0$ like we did in the previous paragraph. So this is where I’m stuck. Any help would be appreciated.

rank inequality is actually a equality

In general, for any matrix $A(mtimes n)$ holds $$rank(A)le min(m,n).$$

Are there conditions, when this unequality is actually a equality.

When does $$rank(A)=min(m,n)$$ hold?

pr.probability – Is there exists inequality relation between KL-divergence and $L_2$ norm?

According to the pinsker inequality, we have the following inequality:
begin{equation}
delta_{TV} (p, q)^2 leq frac{1}{2} D_{KL}(p,q),
end{equation}

where $delta_{TV} (cdot, cdot)$ and $D_{KL}(cdot, cdot)$ are total variation distance and Kullback–Leibler divergence, respectively.

On the other hands, the total variation distance is related to the $L_1$ norm by the identity:
begin{equation}
delta_{TV}(p, q) = frac{1}{2} int_{mathcal{X}} |p(x)-q(x)| , dx,
end{equation}

and thus by using the Cauchy–Schwarz inequality we can obtain that
begin{equation}
delta_{TV} (p, q)^2 leq frac{1}{4} int_{mathcal{X}} (p(x)-q(x))^2 , dx.
end{equation}

I denotes the RHS by $L_2(p, q)$, i.e., $L_2(p, q) = int_{mathcal{X}} (p(x)-q(x))^2 , dx$.

My question is that is there exists some (inequality) relation between the $D_{KL}(cdot, cdot)$ and $L_2(p, q)$?

co.combinatorics – An isoperimetric inequality for a Hamming sphere

Let $S$ be subset of ${0,1}^n$ such that element of $S$ has weight (the number of $1$-coordinates) $k$.

Denote by $S_r$ the number the $r$-boundary of $S$ i.e. the set of elements $y$ such that there is $x$ in $S$ such that the Hamming distance between $x$ and $y$ is at most $r$.

Question: How small can be the cardinality of $S_r$ depend on $|S|$, $k$ and $r$?

How can I make sense of this Matrix inequality?

In https://tel.archives-ouvertes.fr/tel-02926037/document one page 35 the following Lemma is stated:

enter image description here

where the division is to be interpreted component wise. How can I make sense or even proof this inequality?

algebraic manipulation – Rearranging inequality

I’m trying to rearrange this inequality $a + b a – 1 > a^2 – c a – b c$, to appear as $-a^2 + (b+c+1)a>1-bc$ or $-a^2 + (b+c+1)a-1+bc>0$. I’ve tried to achieve that with Collect, but couldn’t make it work on both sides of the inequality.

Any ideas on how to achieve that?

From square norm to norm in an inequality

If I have an inequality involving the norm squared of kind:
$$||text{something}||^2<||phi||^2$$ then can I say that in general $$||text{something}||<a_1||phi||$$ with $0<a_1<1$?