real analysis – Test of the motto du Bois-Reymond "by approximation"

I am curious about the following argument in Morrey ("Multiple integrals in the calculation of variations", Lemma 2.3.1). Suppose $ f in L ^ 1 (0,1) $ Y $$ int_0 ^ 1 fg , dx = 0 $$ for each test function $ g en C ^ infty_c (0,1) $. So, "by approximations", the same is true if $ g $ It is simply limited and measurable, in which case we can take $ g = mathrm {sign} (f) f, $ which it involves $ f = 0 $ Almost anywhere.

How exactly is one supposed to approach? Certainly not in the $ L ^ infty $ norm because the closure of $ C ^ infty_c (0,1) $ in this standard is not $ L ^ infty $ itself. Further, $ mathrm {sign} (f) f $ you don't need to be in $ L ^ infty $So how does this argument work?

real analysis: why is the total differential defined for functions in an open set $ U in mathbb {R ^ {n}} $

This is a follow-up question to my previous question.

Definition: a function $ f: D to mathbb {R} $ where $ D subset
> mathbb {R} $
is differentiable at a limit point $ z in D $ Yes

$ lim limits_ {h to 0} frac {f (z + h) -f (z)} {h}, h neq 0, z + h in D $
there then we say
$ f & # 39; (z) $ is the derivative of $ f $ to $ x $.

Note that the conditions $ h neq 0, z + h in D $ are necessary to ensure that the difference ratio is defined and $ z $ it needs to be a limit point of $ D $ such is possible to find a sequence that approaches $ z $ other than the constant sequence (for which the difference quotient is not defined). If this were not the case, the limit would not be unique.

It is also easy to show from this definition that this is equal to the existence of $ c = f & # 39; (z) in mathbb {R} $ such that $ forall h: x + h in D $:

$ f (z + h) = f (z) + ch + r_ {z} (h) $ with $ lim limits_ {h to 0} frac {r_ {z} (h)} {h} = 0 $.

Note that $ f & # 39; (z) h $ it is a linear map defined for all $ h: z + h in D $. Conditions again $ h neq 0, z + h in D $ are necessary for the expression to make sense and $ z $ it must be a limit point for $ c $ be unqiue

The definition of limit is also easy to generalize to functions with vector values, but it makes no sense for functions of more than one variable. So, the idea is to define the differentiability in terms of the existence of a linear map.

Definition: Let $ U subset mathbb {R ^ {n}} $ be open. A function $ f: U to mathbb {R ^ {}} $ is differentiable in
$ z in U $ if there is a linear map $ L: mathbb {R ^ {n}} to
> mathbb {R m} $
such that $ forall h: x + h in U $

$ f (z + h) = f (z) + L_ {z} (h) + r_ {z} (h) $ with $ lim limits_ {h a 0}
> frac {r_ {z} (h)} { | h |} = 0 $
.

This definition is taken from some class notes on real analysis and the linear map. $ L $ can be considered as the best approximation of the change in the value of the function $ f $ close $ z $.

It seems clear to me why we come up with the idea of ​​differentiability as linear approximation. However, what I really don't understand is why we require $ U $ be open. Of course, we need to be able to approach a point so that it is differentiable, but I feel that the condition is stronger than necessary. In particular, we did not need it for the case of functions of a variable. Can we not reuse the same conditions as in the case of a variable?

Another thing that bothers us is that we require $ L $ to be defined in $ mathbb {R ^ n} $. This also seems to be unnecessary and it would be sufficient to require it to be defined for $ h: x + h in U $.

Am I missing something? Any clue is appreciated. Thank you!

real analysis – Gaussian integral convergence

How do we show that the Gaussian integral$ displaystyle { large {{ int _ {{- infty}} ^ {{ infty}}} {{e}} ^ {{- {{x}} ^ {{2}}}} {d } {x}}} $ converges?

I can discover the convergence for $ displaystyle { large {{ int _ {{0}} ^ {{ infty}}} {{e}} ^ {{- {{x}} ^ {{2}}}} {d} { x PS as follows:

We know $ e ^ {- x ^ 2} leq e ^ {x} $ when $ x geq 1 $ and after this we get

$ displaystyle { large {{ int _ {{0}} ^ {{ infty}}} {{e}} ^ {{- {{x}} ^ {{2}}}} {d} { x} = { int _ {{0}} ^ {{1}}} {{e}} ^ {{- {{x}} ^ {{2}}}} {d} {x} + { int_ {{1}} ^ {{ infty}}} {{e}} ^ {{- {{x}} ^ {{2}}}} {d} {x}}} $

and to $ displaystyle { large {{ int _ {{1}} ^ {{ infty}}} {{e}} ^ {{- {x}}} {d} {x} = lim _ {{ {s} to infty}} { int _ {{1}} ^ {{s}}} {{e}} ^ {{- {x}}} {d} {x} = lim _ { {{s} to infty}} { left (- {{e}} ^ {{- {x}}} {{ mid} _ {{1}} ^ {{s}}} right) } = lim_ {{{s} to infty}} { left (- {{e}} ^ {{- {s}}} + {{e}} ^ {{- {1}}} right)} = frac {{1}} {{e}}}} $

so from the comparison test, this shows that

$ displaystyle { large {{ int _ {{0}} ^ {{ infty}}} {{e}} ^ {{- {{x}} ^ {{2}}}} {d} { x PS it also converges, but I am confused with the one where the lower limit is $ – infty $.

Variables cv.complex: Bromwich integral transformed into integral in the real axis

I am new to complex integration and reverse transformations of Laplace. I already asked this question in math.se but I didn't get an answer.

The author of a textbook states that the inverse transformation of Laplace has expression
$$
f (t) = frac {2 exp (bt)} { pi} int_0 ^ infty Re bigl ( hat {f} (b + iu) bigr) cos (ut) , math {d} u.
$$

Get this formula by replacing $ s = b + iu $ in the integral of Bromwich
$$
f (t) = frac {1} {2 pi i} int_ {bi infty} ^ {b + i infty} exp (st) hat {f} (s) , mathrm {d } s.
$$

However, I have numerically verified this formula and it does not seem to be true:

fhat <- function(s) 1/(s+3) # Laplace transform of exp(-3x)
b <- 5 
integrand <- function(u, x){
  Re(fhat(b+1i*u))*cos(x*u)
}
x <- 2 
2*exp(b*x)/pi * integrate(integrand, 0, Inf, x = x, subdivisions = 10000)$value
# -0.1124648
exp(-3*x)
# 0.002478752

by $ b = -2 $ the result is close to the expected value $ exp (-3x) $:

b <- -2
2*exp(b*x)/pi * integrate(integrand, 0, Inf, x = x, subdivisions = 10000)$value
# 0.002479138

I understood $ b $ must be chosen anywhere to the right of the singularities of $ hat {f} $ (here $ -3 $) I'm wrong? Here the result depends on the choice of $ b $. Do I misunderstand something or is there something wrong in this textbook?

Problem with importing large land from the real world to Unity. The sea is higher!

Sample Image

I am trying to import some real-world land to Unity with the help of Global Mapper. The data I am using is downloaded from CGIAR SRTM (http://srtm.csi.cgiar.org/,http://srtm.csi.cgiar.org/srtmdata/). The land part works well but I am having problems with the sea part.

Weird thing

This is a bigger view of that strange thing in the first image. This stranger is really the sea. This is happening because I am using the gradient shader. Lokks like:

nothing

The sea is represented with white, but when Unity reads the .raw file, white means the highest part. Does anyone know how to deal with problems like this? Thank you very much.

real analysis – Question about compactness in metric spaces

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real analysis: shows that a sub sequence is a subset of the original sequence

My main question is the fourth point, but I hope you can clarify some things along the way.

The definition of a sequence says that a function $ a: mathbb {N} a S $ it's a sequence in a set $ S $denoted $ (a_n) $.

  1. Can I freely restrict the domain of the function? $ to $ and still call it sequence? In particular, a) is it valid to define $ a_n $ in a finite subset of $ mathbb {N} $ b) in an infinite subset of $ mathbb {N} $?

Let's say that in each term of the sequence $ (a_n) _ {n in mathbb {N}} $, I need to define a new sequence from there. First I define a sequence $ (m_k) $ what maps $ {k in mathbb {N}: k geq n } to mathbb {N}, forall n in mathbb {N} $ with $ m_k <m_ {k + 1} $ (assuming affirmative in question 1b because the domain is an infinite subset of $ mathbb {N} $) Then how $ (a_ {m_k}) $ is the sequence composition $ (a_n) $ and the growing sequence $ (m_k) $, by definition $ (a_ {m_k}) $ it is a sequence of $ (a_n) $.

  1. Is it a correct way to show that these new sequences $ (a_ {m_k}) $ Are they subsequences?

A set of defined points for the sequence. $ (a_n) $ it is $ left {a_n: n in mathbb {N} right } $.

  1. How do you define a set of points for a sub sequence? $ (a_ {m_k}) $? It is $ forall n in mathbb {N}, left {a_ {m_k}: k geq n right } $ penalty fee?

Assuming affirmative in question 3, the main question is:

  1. How to prove that $ forall n in mathbb {N}, left {a_ {m_k}: k geq n right } subseteq left {a_n: n in mathbb {N} right } $? In other words, that the set of points defined for a sub-sequence is a subset of the set of points of the original sequence. I think it should take a term of $ left {a_ {m_k}: k geq n right } $ and somehow I deduce that it is also on the set $ left {a_n: n in mathbb {N} right } $, but I'm not sure how to do it rigorously.

To give context to my questions, I want to show that $ sup { left {a_n: n in mathbb {N} right }} geq sup { left {a_ {k}: k geq n right }} $. Given the $ (a_n) $ is bounded, set $ left {a_n: n in mathbb {N} right } $ Y $ left {a_ {k}: k geq n right } $ They are delimited. Having $ left {a_ {k}: k geq n right } subseteq left {a_n: n in mathbb {N} right } $ I would test the supremums as in the question Test the supremum of a subset is smaller than the supremum of the whole.

CPanel VPS license for $ 5, is it real?

Hi

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Are these & # 39; The difference between Elizabeth Warren and Bernie Sanders: Bernie is real and she is another Fake Flake?

Bernie being "real" is no better. The guy is a pink rag that praised Castro, said the bread lines are good and said killing rich people is a noble practice.

He is also the guy who wrote a short novel about how women really like to be raped.

How the hell can anyone vote for him? Ok, he is & # 39; real & # 39; about his ideology, but guess what happened to Dahmer