## real analysis – An inequality involving fractional laplacian

I have to prove that for $$sin(0,1)$$, $$uinmathcal{S}(mathbb{R}^n)$$, ($$u$$ is a Schwartz’s function):
$$|(-Delta)^su(x)|leq c_{n,s}|x|^{-n-2s},quadforall xinmathbb{R}^nsetminus B_1(0),$$
for some $$c_{n,s}>0$$, where:
$$(-Delta)^su(x):=-frac{C(n,s)}{2}int_{mathbb{R}^n}frac{u(x+y)+u(x-y)-2u(x)} {|y|^{n+2s}},dy,quadforall xinmathbb{R}^n,$$
is the fractional laplacian. I have no idea. Any help would be appreciated.

## triangle inequality – Can any norm be used in a product metric

Given a norm $$|cdot|:Bbb{R}^2 to Bbb{R}$$ and metric spaces $$(X,d_X), (Y,d_Y)$$, we define $$D((x_1,y_1),(x_2,y_2)) = |(d_X(x_1,x_2),d_Y(y_1,y_2))|$$. Is $$D$$ always a metric? I know it is when working with the 1-norm, 2-norm and $$infty$$-norm.

Currently, I am aware that the norms on $$Bbb{R}^2$$ are strongly equivalent, in the sense that for any two norms $$|cdot|,||cdot||$$, there exists positive reals $$c_1,c_2$$ such that $$c_1|v|le ||v||le c_2|v|$$. However, if $$(X,d)$$ is a discrete metric space, there exist functions that are strongly equivalent to $$d$$ but violate the triangle inequality, so I’m not sure how useful norms being equivalent are.

Is there some trick I am missing which can prove that $$D$$ always satisfices the triangle inequality, or is there a counterexample to my claim?

## real analysis – Prove inequality in metric spaces

We prove that the statement: Let $$(mathbb{R}^n,d)$$ is a
euclidean space, $$X$$ is a non-empty closed subset of $$mathbb{R^n}$$. If $$xin mathbb{R}^n-X$$, then exist $$yin X$$ such that for all $$zin X$$, $$d(x,y)leq d(x,z)$$.

I thought going through a contradiction, it is not easy for me.
On the other hand, i tried to find the existence, but
i failed it. I am interested in this problem, can you give me a hint?

## inequality – \$sum frac {1} {alpha + log_a {b}} le frac {2}{alpha}\$

This is another olympiad-problem my teacher gave us to train.
$$sum frac {1}{alpha + log_a{b}} le frac {2}{alpha}$$ with $$alpha in (0, 2)$$ and $$a, b, c in (0, 1) cup (1, infty)$$. After working a little bit with it, we get $$sum log_{a ^ alpha b} {a} le frac {2}{alpha}$$, but I don’t know how to go on.

## trigo inequality prove for all x between -pi to pi

$$forall xinleft(-pi,piright):sin(pi-x)leqpi-x$$

## inequality – Is the following property true of any repeating cycle of \$n\$ real values?

I have found it challenging to state this inductive argument about a cycle of reals clearly and concisely.

I would greatly appreciate if someone could show me how to make the same argument in a more standard way or offer suggestions for tightening up the argument.

Let:

• $$%$$ be the modulo operation
• $$c_1, c_2, dots, c_n$$ form a repeating cycle of $$n$$ reals with:
$$c_{i+n} = c_i$$
• $$M(c_1, c_2, dots, c_n) = sumlimits_{i=1}^n c_i$$
• $$f_{k,n}(c_i) = c_{(i+k) % n}$$

Claim:

There exists $$i$$ such that for all $$j ge 0$$:
$$sumlimits_{t=0}^jf_{t,n}(c_i) le dfrac{(j+1)M(c_1,dots,c_n)}{n}$$

Argument:

(1) Base Case: $$n=2$$

• This follows since min$$(c_1,c_2) le dfrac{c_1+c_2}{2}$$

(2) Assume that it is true for any such cycle consisting of up to $$n ge 2$$ reals.

(3) Inductive Case: Let $$c_1, c_2, dots, c_{n+1}$$ form a cycle of $$n+1$$ reals with:
$$c_{i+n+1} = c_i$$

(4) There exists $$k$$ where $$c_k le dfrac{M(c_1,dots,c_{n+1})}{n+1}$$

(5) Define $$d_1, d_2, dots, d_n$$ such that:

$$d_i = begin{cases} c_i, & i < k\ c_{i+1} & i ge k\ end{cases}$$

so that:

$$frac{M(d_1, dots, d_n)}{n} le frac{M(c_1, dots, c_{n+1})}{n+1}$$

(6) By assumption, there exists $$i$$ such that for all $$j ge 0$$:

$$sumlimits_{t=0}^j f_{t,n}(d_i) le frac{(j+1)M(d_1, dots, d_n)}{n}$$

(7) Case 1: $$i < k$$

$$sumlimits_{t=0}^j f_{t,n}(c_i) = sumlimits_{t=0}^j f_{t,n}(d_i) le frac{(j+1)M(d_1, dots, d_n)}{n} le frac{(j+1)M(c_1, dots, c_{n+1})}{n+1}$$

$$sumlimits_{t=0}^j f_{t,n}(c_i) = c_k + sumlimits_{t=0}^{j-1} f_{t,n}(d_i) le frac{M(c_1,dots,c_{n+1})}{n+1} + frac{(j)M(d_1, dots, d_n)}{n} le frac{(j+1)M(c_1,dots,c_{n+1})}{n+1}$$

(8) Case 2: $$ige k$$

$$c_i le frac{M(c_1, dots, c_{n+1})}{n+1}$$

$$sumlimits_{t=0}^j f_{t,n}(c_i) = c_k + sumlimits_{t=0}^{j-1} f_{t,n}(d_i) le frac{M(c_1,dots,c_{n+1})}{n+1} + frac{(j)M(d_1, dots, d_n)}{n} le frac{(j+1)M(c_1,dots,c_{n+1})}{n+1}$$

## Obtaining an inequality for cauchy criterion for series.

I have the following sum but I cannot bound it nicely, here is what l have tried but I’m not sure if it is correct. Here $$m>n$$
$$sum_{k=n+1}^{m}frac{frac{k}{2}+(-1)^k+sin(k)}{k!}$$

$$sum_{k=n+1}^{m}frac{frac{k}{2}+(-1)^k+sin(k)}{k!}

$$sum_{k=n+1}^{m}frac{frac{k}{2}+(-1)^k+sin(k)}{k!}

$$sum_{k=n+1}^{m}frac{frac{k}{2}+(-1)^k+sin(k)}{k!}

$$sum_{k=n+1}^{m}frac{frac{k}{2}+(-1)^k+sin(k)}{k!}

$$sum_{k=n+1}^{m}frac{frac{k}{2}+(-1)^k+sin(k)}{k!}

## Explanation to an unmotivated step in an inequality.

Follows the original problem:

Let $$a, b, c$$ be non-negative real numbers. Prove that
$$frac{a^2}{a^2 + 2left(a + bright)^2} + frac{b^2}{b^2 + 2left(b + cright)^2} + frac{c^2}{c^2 + 2left(c + aright)^2} geqslant frac{1}{3}$$

This is my approach:
The inequality is equivalent to
$$frac{1}{1 + 2left(1 + frac{b}{a}right)^2} + frac{1}{1 + 2left(1 + frac{c}{b}right)^2} + frac{1}{1 + 2left(1 + frac{a}{c}right)^2} geqslant frac{1}{3}$$
Hence, Let
$$x$$ be $$displaystyle frac{b}{a}$$,
$$y$$ be $$displaystyle frac{c}{b}$$ and
$$z$$ be $$displaystyle frac{a}{c}$$.
Now the inequality is
$$frac{1}{1 + 2left(1 + xright)^2} + frac{1}{1 + 2left(1 + yright)^2} + frac{1}{1 + 2left(1 + zright)^2} geqslant frac{1}{3} quadtextrm{with}quad xyz = 1$$
$$frac{xyz}{xyz + 2(x + xyz)^2} + frac{xyz}{xyz + 2(y + xyz)^2} + frac{xyz}{xyz + 2(z + xyz)^2} geqslant frac{1}{3}$$
$$frac{xyz}{xyz + 2x^2(1 + yz)^2} + frac{xyz}{xyz + 2y^2(1 + zx)^2} + frac{xyz}{xyz + 2z^2(1 + xy)^2} geqslant frac{1}{3}$$
$$frac{yz}{yz + 2x(1 + yz)^2} + frac{zx}{zx + 2y(1 + zx)^2} + frac{xy}{xy + 2z(1 + xy)^2} geqslant frac{1}{3}$$
$$frac{y^2z^2}{y^2z^2 + 2(1 + yz)^2} + frac{z^2x^2}{z^2x^2 + 2(1 + zx)^2} + frac{x^2y^2}{x^2y^2 + 2(1 + xy)^2} geqslant frac{1}{3}$$
By $$T_2$$‘s Lemma,
$$frac {left(xy + yz + zxright)^2} {x^2y^2 + y^2z^2 + z^2x^2 + 2left((1 + yz)^2 + (1 + zx)^2 + (1 + xy)^2right)} geqslant frac{1}{3}$$
Hence, after simplification,
$$3(x + y + z) geqslant 2(xy + yz + zx) + 3$$
Substitute back $$a,b,c$$.
$$3left( frac {a} {b} + frac {b} {c} + frac {c} {a} right) geqslant 2left( frac{a}{c} + frac{b}{a} + frac{c}{b} right) + 3$$
I think that it looks easy, but if someone has a way, it might become a solution.

Book’s Approach:
It is the same till the second inequality in my approach, but then without any motivation goes on to assume $$x = frac{np}{m^2}, y = frac{mp}{n^2}, z = frac{mn}{p^2}.$$ And then it is just $$T_2$$‘s Lemma and some simplification to get squares on the greater side.
I don’t see anything natural in assuming that and neither there is a motivation to take such step, besides it gives proof.
But my question is how does the author assume such a thing? And is there a way to complete my proof?

Thanks!

## real analysis – Inequality involving sigmoid function

Let $$sigma$$ denote the sigmoid function $$sigma(x) = frac{1}{1+e^{-x}}$$, let $$x,y in mathbb{R}$$. Given the following two conditions: $$|sigma(-x) – sigma(y)| < epsilon$$ and $$x – y > c > 0,$$ where $$epsilon$$ can be regarded as a small positive number and $$c$$ as a large positive number. Can we draw the conclusion that $$x > frac{c}{2} – f(epsilon), y <-frac{c}{2} + f(epsilon)$$, where $$f(epsilon)$$ is some function of $$epsilon$$.

## assumptions – Why a complementary inequality for function values does not simplify to False?

I have a general assumption of a function to be positive. The direct inequality (f(x)>0) does simplify to True, but a complementary one (f(x) < 0) does not simplify at all despite the fact that I explcitly assumed that funciton is not zero.

``````\$Assumptions = K(_) > 0 && K(_) != 0

FullSimplify(K(15) > 0)
True
FullSimplify(K(12) == 0)
False
FullSimplify(K(15) < 0)
K(15) < 0
``````