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 <q <1 $, 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.
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$ 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:

$$ max (d_1 (a, b), d_2 (a, b)) <D_ {d, epsilon} (a, b) $$

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?)

Thanks for your help!

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
$$ frac {q ^ 6 – 1} {q ^ 6 – q ^ 5} = I (q ^ 5) leq I (q ^ k) < frac {q} {q-1} leq frac {5} {4} $$
$$ frac {8} {5} leq frac {2 (q-1)} {q} <I (n ^ 2) = frac {2} {I (q ^ k)} leq frac {2} {I (q ^ 5)} = frac {2 (q ^ 6 – q ^ 5)} {q ^ 6 – 1}, $$
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 $?