notation: What is the topology $ epsilon ^ {00} $ -?

"Yes $ F $ is a locally convex space, then the topology in $ F ^ * $ (the topological dual of F) will be the strong topology, and the topology in $ F ^ {**} $ it will be $ epsilon ^ {00} $-topology "This is the first time I see this notation, it's described as the uniform convergence topology in the polar of the p-unit balls, but I'd like to know what exactly is this $ epsilon ^ {00} $-refers to the topology, is also mentioned in some other articles without any other explanation, is it always the same as explained here?

The article I am reading at this time was published in 1974, "Linear operators and vector measurements" by Brooks and Lewis.

general topology – How is $ overline {c _ { epsilon}} $ convex and $ c: = bigcap _ { epsilon> 0} overline {c _ { epsilon}} neq emptyset $ in a CAT $ (0) $ space?

Leave $ (X, d) $ be a cat$ (0) $ space, $ {x_n } subset X $ be limited and $ K subset X $ be closed and convex Define $ varphi: , X longrightarrow mathbb {R}, $ by $ varphi (x) = limsup limits_ {n to infty} d (x, x_n) $ for each $ x in X. $ Then, there is a single point. $ u in K $ such that $$ varphi (u) = inf limits_ {x in K} varphi (x). $$

Test

Leave $ r = inf limits_ {x in K} varphi (x) $ Y $ epsilon> 0. $ Then, there is $ x_0 in K $ such that $ varphi (x_0) <r + epsilon. $ This implies that there is $ N in mathbb {N} $ such that
$$ x_0 in bigcup_ {n geq N} left ( bigcap_ {k = n} B (x_k, r + epsilon) cap K right) subset bigcup_ {n geq 1} left ( bigcap_ {k = n} B (x_k, r + epsilon) cap K right). $$
Define $ c _ { epsilon}: = bigcup_ {n geq 1} left ( bigcap_ {k = n} B (x_k, r + epsilon) cap K right). $ It is clear that $ c _ { epsilon} $ It is convex

Question: How is it $ overline {c _ { epsilon}} $ convex and $ c: = bigcap _ { epsilon> 0} overline {c _ { epsilon}} neq emptyset? $

Details: The current article that I am reviewing is Dhompongsa et al. To show that $ overline {c _ { epsilon}} $ it is convex, Dhompongsa et al., referring to Proposition 1.4 (1). I just have a hard time understanding it. Any help, please?

finite automata – ε-NFA to DFA – initial state with only epsilon transitions

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logic – kPDA handling multiple epsilon transitions

I am assigned to build a kPDA with 2 batteries that handle {w # w, where w is a string of (0,1) *}. I understand that the # delineates the two chains, but I'm not sure of the logic when starting the batteries with epsilon. I asked my teacher, verbally, if this works and he said no, but I can not think otherwise.

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How can I prove that I use a test $ epsilon – delta $ that $ lim_ {x rightarrow frac {1} {e}} (e ^ {x ^ {x ^ x}}) <2 $?

It is not a homework question. I just wanted to refresh my tests of the epsilon delta and this happened to me: I struggled for an hour, I have no idea where to start.

limits: What values ​​of $ alpha $, if any, produce $ f = mathcal {O} ( epsilon ^ alpha) $ as $ epsilon downarrow 0 $?

I need to solve the following exercise:

Exercise: What values ​​of $ alpha $, if applicable, performance $ f = mathcal {O} ( epsilon ^ alpha) $ as $ epsilon downarrow 0 $ Yes
$$
f = dfrac {1} {1 – e ^ epsilon}
$$

What I have tried:

I know I can use the following theorem:

Yes
$$
lim_ {e downarrow e_o} dfrac {f ( epsilon)} { phi ( epsilon)} = L,
$$

where $ – infty <L < infty $, so $ f = mathcal {O} ( phi) $ as $ epsilon downarrow epsilon_0 $.

So I think I need to evaluate the limit
$$
lim _ { epsilon downarrow 0} dfrac {1} { epsilon ^ alpha (1 – e ^ epsilon)}
$$

and find out for what values ​​of $ alpha $ exists. The limit exists if $ epsilon ^ alpha (1 – e ^ epsilon) $ converges to infinity or any other number that is not $ 0 $. I know that $ 1 – e ^ epsilon a 1 – 1 = 0 $ for $ epsilon downarrow 0 $. Then the limit exists if $ epsilon ^ alpha to infty $ faster than $ 1 – e ^ epsilon a 0 $because in that case $ epsilon ^ alpha (1 – e ^ epsilon) a infty $ Y

$$ lim _ { epsilon to 0} dfrac {1} { epsilon ^ alpha (1-e ^ epsilon)} = 0 $$
Yes $ alpha> 0 $ so $ lim _ { epsilon downarrow 0} epsilon ^ alpha = 0 $ so we know that at least $ alpha leq 0 $ necessary. Yes $ alpha = 0 $ so $ lim _ { epsilon downarrow 0} epsilon ^ alpha = 1 $. Yes $ alpha <0 $ so $ lim _ { epsilon downarrow 0} epsilon ^ alpha = infty $. That's why I think that for all values ​​of $ alpha <0 $ we have that $ f = mathcal {O} ( epsilon ^ alpha) $.

Question: If my reasoning about when the limit exists is correct, I think the weak point of my solution so far lies in the fact that I do not really show / know whether it is true or not. $ epsilon ^ alpha to infty $ faster than $ 1 – e ^ epsilon a 0 $. Is this true? And if so, how can I prove what it is?

Probability: anti-concentration: upper limit for $ P ( sup_ {a in mathbb S_ {n-1}} sum_ {i = 1} ^ na_i ^ 2Z_i ^ 2 ge epsilon) $

Leave $ mathbb S_ {n-1} $ be the unitary sphere in $ mathbb R ^ n $ Y $ z_1, ldots, z_n $ be a sample i.i.d of $ mathcal N (0, 1) $.

Dice $ epsilon> 0 $ (It can be assumed that it is very small), which is reasonable upper limit for the tail probability $ P ( sup_ {a in mathbb S_ {n-1}} sum_ {i = 1} ^ na_i ^ 2z_i ^ 2 ge epsilon) $ ?

  • Using ideas from this other answer (MO link), you can set the not uniform Limit of anti-concentration: $ P ( sum_ {i = 1} ^ na_i ^ 2z_i ^ 2 le epsilon) le sqrt {e epsilon} $ for all $ a in mathbb S_ {n-1} $.

  • The uniform analog is another story. Can coverage numbers be used?

Linear algebra: if $ B $ is a small perturbation of the positive-defined matrix $ A $, do we have $ B> epsilon A $?

Suppose $ A = (a_ {ij}) $ is a positive-defined (symmetric) matrix, and $ B $ It's another symmetric matrix.

Question: Yes $ B $ It's in a small neighborhood $ U $ of $ A $, then it seems that $ B $ It must also be positive-defined. Also why value of $ epsilon> 0 $ we can find a neighborhood $ U $ so that $ B> epsilon A $?

Yes, both $ A $ Y $ B $ They are diagonal matrices, so this is trivial. But in general, since we can only diagonalize one of them and I'm afraid there will be some problem.