real analysis – Prove there exists a sequence $sum_{n=1}^{infty} a_n$ = inf $A$

Let $A$ be a nonempty set of real numbers with a lower bound. Prove there exists a sequence $(a_n)_{n=1}^{infty}$ such that $a_n in A $ for all $n$ and $lim_{n to infty} a_n = inf(A)$.

I try to apply the definition of convergence but struggle to connect that to $lim_{n to infty} a_n = inf(A)$. Any helps would be appreciated!

navier stokes – inf sup condition

I am considering the discrete inf-sup condition for the discrete Stokes problem

$ inf_{q in Q_h} sup_{v in V_h} frac{(q, nabla cdot v)}{| q | | nabla v|} geq beta > 0$

This means that
$ inf_{q in mathbb{R}^p} sup_{v in mathbb{R}^n} frac{v^T B q}{| q | | v|} geq beta > 0$

Can a do the following argumentation:

$ inf_{q in mathbb{R}^p} sup_{v in mathbb{R}^n} frac{v^T B q}{| q | | v|} = inf_{q in mathbb{R}^p, |q | = 1 } sup_{v in mathbb{R}^n, | v | = 1} v^T B q = inf_{q in mathbb{R}^p, |q | = 1 } |B q|$

Is this correct? Or am I missing something?

I would be very grateful about any help.

Best,
frieder

$ S: = Big { left (m + n right) ^ { frac {1} {mn}}: ; m, n in mathbb N Big } inf S, sup S $

If available, you will find $ sup S, inf S $
$$ S: = Big { left (m + n right) ^ { frac {1} {mn}}: ; m, n in mathbb N Big } $$
$$ m = 1, n = 1: ; left (m + n right) ^ { frac {1} {mn}} = 2 $$

For a firm $ n in mathbb N, m in infty ; $:

$ displaystyle lim_ {m bis infty} (m + n) ^ { frac {1} {mn}} = lim_ {m bis infty} Bigm (m left (1+ frac { n} {m} right) Big) ^ { frac {1} {m} cdot frac {1} {n}} = lim_ {m to infty} left ( sqrt (m) {m} left (1+ frac {n} {m} right) ^ { frac {1} {m} right) ^ { frac {1} {n}} $$$= lim_ {m bis infty} sqrt (n) { sqrt (m) {m}} lim_ {m bis infty} left (1+ frac {n} {m} right) ^ { frac {1} {mn}} = e $$

$$ f (m): = left (1+ frac {1} {mn} right) ^ { frac {1} {mn}} $$

$$ m & # 39;> m implies f (m & # 39;)> f (m) $$
$$ m + n geq 2 (mn) ^ { frac {1} {2}} Big / ^ { frac {1} {mn}} $$
$$ (m + n) ^ { frac {1} {mn}} geq sqrt { sqrt (mn) {mn}} ; sqrt (mn) {2}> 1 $$
Then I concluded: $$ f (x) in langle 1, e rangle implies inf S = 1, sup S = e $$
Is that correct?

complex analysis – $ f $ is completely without zeros, then $ r> 0 $ m (r) = inf left { left | f (z) right |: left | z right | = r right } $, $ m $ does not increase

$ f $ is whole function without zeros, then for $ r> 0 $ $ m (r) = inf left { left | f (z) right |: left | z right | = r right } $. $ m $ does not increase the function.

I have an idea how to solve this problem, but I encountered a problem.

If $ f $ is constant, then it's trivial.

If $ f $ is not constant with $ f (z) neq 0 $ for each $ z in mathbb {C} $, and $ f $ is holomorphic, then we define
$$ g (r): = frac {1} {m (r)} $$ then
$$ = frac {1} { inf left { left | f (z) right |: left | z right | = r right }} $$
$$ = sup left { frac {1} {f (z)}: left | z right | = r right } $$

because $ left { frac {1} {f (z)}: left | z right | = r right } $ is closed and limited $ g $ takes its maximum values ​​at the limit.

The question is: I assumed that $ g $ increases with $ r $and of it $ g $ is an increasing function, but unfortunately I can't find any right arguments for this statement and I would like if someone could explain this?

Algorithms – Create a priority search tree to determine the number of points in the range [-inf, qx] X [qy, qy’] from a series of points sorted by y-coordinates

A priority search tree can be created for a set of points P in time O (n log (n)). However, if the points are sorted by the y-coordinates, it takes O (n). I find algorithms for constructing the tree if the points are not sorted.

I found a way to do this as follows:

  1. Construct a BST on the points.
    Since the points are sorted, it will take O (n) time

  2. Min-heap the BST on the x-coordinates
    This will take theta (n) time

The total time complexity will be O (n).

Is this a valid approach to creating a priority search tree in O (n) time?

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Functional Analysis – Calculate $ inf_ {a, b, c in mathbb {R}} int _ {- 1} ^ {1} left | x ^ {3} -ax ^ {2} -b xc right | ^ {2} dx $

calculation $$ inf limits_ {a, b, c in mathbb {R}} , int _ {- 1} ^ {1} left | x ^ {3} -ax ^ {2} -b xc right | ^ {2} mathrm dx $$

I'm new to Hilbert space, I see similar questions with the formula: $ langle f, g rangle = int f g , mathrm d mu $ (real $ L ^ {2} $), but I have problems transforming the problem into the formula: what is the systematic way to find it $ f $ and $ g $? What role does orthogonality play in the question?

nt.number theory – approximation for $ inf {r> 0, (n-r, n + r) in mathbb {P} ^ {2} } $ by minimizing a distance

Under Goldbach's guess $ r_ {0} (n): = inf {r> 0, (n-r, n + r) in mathbb {P} ^ {2} } $ and $ k_ {0} (n): = pi (n + r_ {0} (n)) – pi (n-r_ {0} (n)) $, The PNT implies that one can assume $ dfrac {2r_ {0} (n) -1} {k_ {0} (n)} sim log n $,

As for everyone $ n> 1 $ one has $ tau (n) geq $ 2The equality occurs then $ n $ is prime, a very good approximation (actually a narrow upper bound) of $ k_ {0} (n) $ is given by the function $ S_ {r_ {0} (n)} (n) $ from where $ S_ {r} (n): = left ( sum_ {m = nr} ^ {n + r} dfrac {2 ^ m} { tau (m) ^ m} right) – frac {1 } {2} $,

Can you prove that? $ r_ {0} (n) $ is the positive integer $ r $ that minimizes the amount $ vert S_ {r} (n) – dfrac {2r-1} { log n} vert $? If so, you can get an upper limit for $ r_ {0} (n) $ in terms of $ n $?