## oa.operator algebras – Compute norms of operator's polynomials in Hilbert space and generalize von Neumann inequality

To let $$T$$ to be an operator $$l ^ 2 ({ mathbb {Z} _ { geq 0}}) to l ^ 2 ({ mathbb {Z} _ { geq 0}})$$. $$e_n mapsto sqrt {1 – q ^ {2 (n + 1)}} e_ {n + 1}$$, Where $$0 , I would like to calculate $$| f (T, T ^ {*}) |$$ (Operator norm) for each $$f in mathbb {C} (z, bar z)$$, operator $$T ^ {*}$$Of course this is the Hilbert conjugate and $$T ^ {*} e_0 = 0$$. $$T ^ {*} e_n = sqrt {1-q ^ {2n}} e_ {n-1}$$ to the $$n> 0$$,

I am not sure if this calculation is possible at all, and I would like to show that $$| f (T, T ^ {*}) | to sup limits_ {| z | leq 1} f (z, bar z)$$ as $$q$$ goes to 1, because I actually have to calculate standards first.

I think that there is a generalization of von Neumann's inequality $$q$$analogue or so (there are many generalizations) because the usual inequality proves this for polynomials $$g in mathbb {C} (z)$$ (without $$bar z$$).

Question: Does anyone know of useful facts or inequalities related to my question? Something that could help.

## Algebra pre-calculus – Prove the inequality in the context of Pythagoras theorem

$$sqrt {A ^ 2 + (N * A) ^ 2} + sqrt {B ^ 2 + (N * B) ^ 2} < sqrt {A ^ 2 + (N * A + M) ^ 2} + sqrt {B ^ 2 + (N * BM) ^ 2}$$

Where $$A$$ and $$B$$ are positive real numbers and $$N$$ and $$M$$ are real numbers.

$$I + J$$ is equal to the left side of the inequality and has the lowest possible value given the length of the lowest segment, the length of $$A$$ and the length of $$B$$

## Inequalities – In search of a good proof of the following inequality

To let $$0 leq x <1$$. $$1 leq p < infty$$ and $$q$$ let the conjugate exponent be defined by
$$1 / p + 1 / q = 1.$$

I'm looking for a nice proof

$$frac {(1-x ^ p) ^ {1 / p} (1-x ^ q) ^ {1 / q}} {(1-x) (1 + x ^ c) ^ {1 / c }} geq 1,$$
Where $$c$$ is defined via
$$2 ^ {1 / c} = p ^ {1 / p} q ^ {1 / q}.$$

The number $$c$$ is defined as the inequality holds when we take a boundary as $$x to 1$$, I checked it for different values $$p$$ graphically and it seems to be true. It applies to equality in the case $$p = q = 2$$,

## Measure theory – Show that there is \$ delta> 0 \$ for the inequality to hold (Lebesgue integral)

To let $$(X, mathcal {A}, mu)$$ be a measure room, $$u in mathcal {L} ^ 1 ( mu)$$ and $$K_n = {| u | leq n }$$ for each $$n geq 1$$,

Show that there is $$Delta> 0$$ in order to:

$$forall E in mathcal {A}: mu (E) < delta Rightarrow bigg | int_Eu d mu bigg | < frac {1} {2019}$$

I can not fully understand these questions. Can someone explain or give a hint how to tackle this problem? It seems I have to show that for $$mu (E) < delta$$ there is a $$epsilon = frac {1} {2019}$$, then the inequality holds

Can I use the following for anything?
$$Bigg | int_Eu d mu Bigg | leq n cdot mu (E) + int_ {X backslash K_n} {| u |} d mu$$

## Number theory – An inequality inspired by the abc conjecture and two questions

To let $$text {rad} (x) = prod_ {p | x} p$$ be the radical (= product of the prime-number division $$x$$) from $$x$$ for a natural number $$x$$, To the $$x / y> 0$$. $$gcd (x, y) = 1$$ To let $$text {rad} (x / y) = text {rad} (x) cdot text {rad} (y)$$, If $$d$$ is a metric for nautical numbers, so that $$d (a, b) <1$$ and $$d (a, b)$$ is a rational number (I will call such a metric a "rational metric") and for $$epsilon> 0$$ define:

$$D_ {d, epsilon} (a, b) = 1 – frac {1} { text {rad} ( frac {1} {1 – d (a, b)}) ^ {1+ epsilon}}$$

By an argument similar to the answer of @ GregMartin in the related question, $$D_ {d, epsilon}$$ is a metric for natural numbers:

$$D (a, b) = 0$$ iff $$1 = text {rad} ( frac {1} {1-d (a, b)}) ^ {1+ epsilon}$$ iff $$frac {1} {1-d (a, b)} = 1$$ iff $$d (a, b) = 0$$ iff $$a = b$$,

The triangular inequality leads to the following inequality:

$$frac {1} { text {rad} ( frac {1} {1-d (a, c)}) ^ {1+ epsilon}} + frac {1} { text {rad} ( frac {1} {1-d (b, c)}) ^ {1+ epsilon}} = 1+ frac {1} { text {rad} ( frac {1} {1-d ( a, b)}) ^ {1+ epsilon}}$$

This argument comes from @GregMartin:

Both summands are on the left side $$<1/2$$ then the inequality is automatically true because of the $$1$$ On the right side.
Let the first summand be $$ge 1/2$$, Then

$$2 ge text {rad} ( frac {1} {1-d (a, c)}) ^ {1+ epsilon}$$

This can only happen if the right-hand side of the last inequality is $$= 1$$ therefore $$a = c$$, But then the inequality to be proved becomes a true equality.

That proves it $$D$$ is a metric for natural numbers.

Now consider two rational metrics $$d_1, d_2$$ and let it go $$d (a, b) = d_1 (a, b) + d_2 (a, b) -d_1 (a, b) d_2 (a, b)$$

My first question is:
is $$d$$ also a rational metric?

If so, you might consider:

The inequality inspired by the related question, and thus by the abc conjecture, is:

$$max (d_1 (a, b), d_2 (a, b))

My second question is:
Can you imagine rational metrics where the above inequality is "easy" to prove?

(Let for example $$d$$ be the trivial metric for natural numbers. To let $$d_1 (a, b) = d (a, b) / (d (a, 1) + d (b, 1) + d (a, b))$$ be a $$1/2$$ Turn stone house from $$d$$, which is then a rational metric, and similarly let $$d_2 (a, b) = d (a, b) / (d (a, 2) + d (b, 2) + d (a, b))$$ and define $$d_ {12} = d_1 + d_2-d_1 d_2$$, I wonder if $$d_ {12}$$ is a rational metric and if the above inequality is easy to prove or wrong?)

Connected:
A reinterpretation of the \$ abc \$ conjecture in metric spaces?

## Number Theory – A Bonse's inequality for semiprimes with a good mathematical content

A semiprime $$s$$ is a positive integer, which is the product of two primes, cf. Semiprine the encyclopedia Wikipedia. A well-known inequality with applications that are primes is Bonse's inequality, which I add as a reference from Wikipedia Bonse's inequality,

I was wondering if it is possible to generate a similar inequality $$s_k$$ refers to the $$k$$-th half (ie the sequence A001358 from the OEIS)

$$m (n) cdot (s_ {n + 1}) ^ {a} < left ( prod_ {k = 1} ^ n s_k right) ^ b$$
for a suitable arithmetic function (or sequence) $$m (n)$$ and constants $$a$$ and $$b$$ (I added this feature $$m (n)$$ and constant $$b$$ with the aim of providing flexibility in exploring our inequality).

Question. So you get a sharper inequality * including semiprimes $$s_k$$
$$(s_ {n + 1}) ^ {a} < frac {1} {m (n)} left ( prod_ {k = 1} ^ ns_k right) ^ b tag {1}$$
That holds $$forall n> N$$ for a suitable selection of $$N$$and for a suitable choice of a function $$m (n)> 0$$ and constants $$a$$ and $$b$$? Many thanks

* If you find a notable inequality or asymptotic inequality in your examinations, this does not fit exactly with the previous kind of inequality $$(1)$$ I think it reasonable for you to feel free to add it as an answer because I am asking for a good version of the inequality of a Bonse for Semiprimes.

With the words a sharper inequality I mean that your inequality is of the kind $$(1)$$ Have good mathematical content / meaning that it is a good Bonse's inequality for semiprimes. To emphasize, we take yours $$m (n)$$ as a positive calculation function $$m (n)> 0$$ for all $$n> N$$,

## Inequality – Bound truncated fold

To let $$f$$ to be a continuous stochastic process $$L ^ p ( omega)$$i.e. $$mathbb E | f (t) | ^ p leq C$$ for all $$t geq 0$$. $$g in L ^ infty ( mathbb R) cap C ^ infty ( mathbb R)$$ for all $$1 leq p leq infty$$ with support in $$(0, infty)$$, The convolution is defined as
$$F (t) = (f star g) (t) = int_0 ^ t f (s) g (t-s) dt.$$
I want to tie for all $$p geq 1$$, The quantity $$mathbb E | F (t) | ^ p$$, I've been dealing with Youngs inequality for turns that gives way $$p, q, r geq 1$$ so that $$p ^ {- 1} + q ^ {- 1} = 1 + r ^ {- 1}$$.
$$| F | _r leq | f | _p | g | _q.$$
Now choose $$r = infty$$We have for all $$t> 0$$ and a generic constant $$C$$
$$mathbb E | F (t) | ^ p leq int_0 ^ t mathbb E | f (s) | ^ p ds left ( int_0 ^ t | g (s) | ^ q ds right) ^ {p-1} leq Ct.$$
Since the folding should be a "liquid version" $$f$$I would have guessed to find a time-independent boundary, i. $$mathbb E | F (t) | ^ p leq C$$,

What do I miss here?

## Inequality – Prove \$ ( frac {a} {b ^ {2} +1} + frac {b} {c ^ {2} +1} + frac {c} {a ^ {2} + 1} ge frac {3} {2} \$

$$a, b, c> 0$$ and $$a + b + c = 3$$, to prove

$$frac {a} {b ^ {2} + 1} + frac {b} {c ^ {2} +1} + frac {c} {a ^ {2} +1} ge 3 / 2$$

Attempt:

Note that from AM-Gm

$$frac {a} {b ^ {2} + 1} + frac {b} {c ^ {2} +1} + frac {c} {a ^ {2} +1} ge 3 frac { sqrt (3) {abc}} { sqrt (3) {(a ^ {2} +1) (b ^ {2} +1) (c ^ {2} +1)}}$$

Now AM-GM again

$$a ^ {2} + b ^ {2} + c ^ {2} + 3 ge 3 sqrt (3) {(a ^ {2} +1) (b ^ {2} +1) (c ^ {2} +1)} … (1)$$

Then $$a + b + c = 3 ge 3 sqrt (3) {abc} implies 1 ge sqrt (3) {abc}$$, Also

$$a ^ {2} + b ^ {2} + c ^ {2} ge 3 sqrt (3) {(abc) ^ {2}}$$
times $$1 ge sqrt (3) {abc}$$ and will get

$$a ^ {2} + b ^ {2} + c ^ {2} ge 3 abc … (2)$$

subtract $$(1)$$ With $$(2)$$ and get

$$3 ge 3 sqrt (3) {(a ^ {2} +1) (b ^ {2} +1) (c ^ {2} +1)} – 3 abc$$
$$3 + 3 abc ge sqrt (3) {(a ^ {2} +1) (b ^ {2} +1) (c ^ {2} +1)}$$
$$frac {3abc} { sqrt (3) {(a ^ {2} +1) (b ^ {2} +1) (c ^ {2} +1)}} ge 1 – frac { 3} { sqrt (3) {(a ^ {2} +1) (b ^ {2} +1) (c ^ {2} +1)}}$$

How it goes on..?

## Statistics – Why is this inequality of truncated random variables correct?

Given an order $$X_k$$ from $$k$$ independent random variables reading book says I can form "truncated" random variables, i.

$$X_k (n) = X_k cdot textbf {1} _ { { omega: | X_k ( omega) | leq n }}$$

Where $$omega$$ means a result from the sample space. Define $$S_n$$ and $$has {S} _n$$ as:

$$S_n = X_1 + X_2 + … + X_n$$

$$has {S} _n = X_1 (n) + X_2 (n) + … X_n (n)$$

So that means that $$has {S} _n$$ is the sum of $$n$$ truncated random variables. We also define $$m_n$$ as:

$$m_n = mathbb {E} (X_1 (n))$$

Since the variable distributions (of $$X_k$$) are the same ones we have $$m_n = mathbb {E} (X_k (n))$$ for all $$k geq 1$$,

The book reads that the following inequality is "evident" where given $$epsilon> 0$$, we have:

$$P bigg ( big | frac {S_n} {n} – m_n big | geq epsilon bigg) leq P bigg ( big | frac { has {S} _n} {n} – m_n big | geq epsilon bigg) + P bigg ( has {S} _n neq S_n bigg)$$

But I'm not clear at all and I got lost. How can we have the above inequality? (ie we could get some big ones $$X_k$$ in the $$S_n$$so that the left term is much larger than the truncated right term).

## Inequality – Can the following bounds be improved for \$ I (n ^ 2) – I (q ^ k) \$, where \$ q ^ k n ^ 2 \$ is an odd odd number with special primes \$ q \$?

The present question is tangentially related to my earlier question.

To let $$sigma (x)$$ denote the sum of dividers of the positive integer $$x$$, If $$sigma (y) = 2y$$, then $$y$$ is called a perfect number. Name the frequency index of $$z$$ by $$I (z) = sigma (z) / z$$,

Euler proved that, though $$N$$ is a odd perfect number, then it must necessarily have the form $$N = q ^ kn ^ 2$$ from where $$q$$ the special / euler prime is satisfactory $$q equiv k equiv 1 pmod 4$$ and $$gcd (q, n) = 1$$,

Then since then $$q$$ is a satisfying prime $$q equiv 1 pmod 4$$We have the lower limit $$q geq 5$$; That's why we have the limits
$$1 < frac {q + 1} {q} = I (q ^ k) leq frac {6} {5} implies frac {5} {3} leq I (n ^ 2) = frac {2} {I (q ^ k)} = frac {2q} {q + 1} <2$$
when $$k = 1$$, and
$$frac {q ^ 6 – 1} {q ^ 6 – q ^ 5} = I (q ^ 5) leq I (q ^ k) < frac {q} {q-1} leq frac {5} {4}$$
$$frac {8} {5} leq frac {2 (q-1)} {q}
when $$k> 1$$, (Note that $$5/4 <8/5$$.)

We consider two cases:

case 1 $$k = 1$$

$$I (n ^ 2) – I (q ^ k) = frac {2q} {q + 1} – frac {q + 1} {q} geq frac {5} {3} – frac {6} {5} = frac {7} {15}.$$

WolframAlpha yields $$q geq 5$$,

Case 2 $$k> 1$$

$$frac {2 (q ^ 6 – q ^ 5)} {q ^ 6 – 1} – frac {q ^ 6 – 1} {2 (q ^ 6 – q ^ 5)} geq I (n ^ 2) – I (q ^ k)> frac {8} {5} – frac {5} {4} = frac {7} {20}.$$

WolframAlpha yields $$q> 4.99872$$,

Here is my question:

Can these limits continue? $$I (n ^ 2) – I (q ^ k)$$ to be improved, where $$q ^ k n ^ 2$$ is an odd, perfect number with a special prim $$q$$, especially in the case when $$k> 1$$?