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| bio | website | |
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| age | ||
| visits | member for | 1 year, 5 months |
| seen | 3 mins ago | |
| stats | profile views | 119 |
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Apr 12 |
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Given $L(y) = \int_ {1}^{y}\frac{1}{t}dy$ , $y\in(0, \infty)$ $n$ starts at $2$ in the last sum... |
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Apr 12 |
answered | Does the closed graph theorem presuppose that the domain is closed? |
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Apr 12 |
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Given $L(y) = \int_ {1}^{y}\frac{1}{t}dy$ , $y\in(0, \infty)$ yes it does. it ensures differentiability which is stronger. |
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Apr 12 |
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Egorov's Theorem - Counterexample in Infinite Case Here your hypothesis of finite measure fails since $m(\mathbb{R}) = \infty$ and the conclusion fails as well because for any set of finite measure $F$, your function won't converge uniformly to the zero function. |
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Apr 12 |
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Egorov's Theorem - Counterexample in Infinite Case The definition states that if you have a sequence of functions in a set of finite measure $E$ which converge pointwise to $f$ almost everywhere. Then for every $\epsilon >0$ you can find a measurable set $F \subset E$ such that $m(F) < \epsilon$ and $f_n$ converges uniformly to $f$ on $E \setminus F$. |
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Apr 11 |
answered | Egorov's Theorem - Counterexample in Infinite Case |
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Apr 11 |
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Given $L(y) = \int_ {1}^{y}\frac{1}{t}dy$ , $y\in(0, \infty)$ you basically have it. the FTC tells you that your function is differentiable (and therefore continuous) at all $x \in (0,\infty)$. The increasing bit you get (as mentioned by Mark) by noting that the integral of a positive function over an interval gives you a positive number. Let $z > y$ be two numbers in $(0,\infty)$. Then $L(z) - L(y) = \int_y^z \frac{1}{t} dt > 0$ |
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Apr 11 |
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Given $L(y) = \int_ {1}^{y}\frac{1}{t}dy$ , $y\in(0, \infty)$ And forgot to mention, you also multiply by the derivative with respect to $y$ of the upper bound function, in your case this is $f(y)= zy$ so you get the $z$ in the numerator. It's basically chain rule. |
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Apr 11 |
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Given $L(y) = \int_ {1}^{y}\frac{1}{t}dy$ , $y\in(0, \infty)$ The first and the second term appear because you need to use the fundamental theorem of calculus for differentiating an integral. Basically you evaluate the integrand at the endpoints... |
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Apr 11 |
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Equivalence of measures and $L^1$ functions got that now. thanks! |
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Apr 11 |
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Let $p_1 , p_2$ be prime. Then prove that the only divisors of $p_1 p_2$ are $1 , p_1 , p_2 , p_1 p_2 $. No worries. Good luck! |
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Apr 10 |
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Let $f:\Bbb R^2→\Bbb R:(0,0)\mapsto 0 \quad (x,y)\mapsto\frac{x^2y^2}{x^4+y^4}$ Yeah, that is the right idea: $\lim_{x \to 0} f(x,x) = 1/2 \neq 0 = \lim_{x \to 0} f(x,0)$ |
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Apr 10 |
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Let $p_1 , p_2$ be prime. Then prove that the only divisors of $p_1 p_2$ are $1 , p_1 , p_2 , p_1 p_2 $. Also, I think you should be less rude to people who are trying to help you. And be humble when you are asking for help, advice, etcetera. |
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Apr 10 |
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Let $p_1 , p_2$ be prime. Then prove that the only divisors of $p_1 p_2$ are $1 , p_1 , p_2 , p_1 p_2 $. Oh well, whatever... That is such an elementary fact, it is sometimes taken as the definition of prime. Your original question didn't refer to this at all anyway. |
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Apr 10 |
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Let $p_1 , p_2$ be prime. Then prove that the only divisors of $p_1 p_2$ are $1 , p_1 , p_2 , p_1 p_2 $. Of course they are ALL possible divisors, this is easy to prove. Suppose you have your unique prime factorization of say $b$. And suppose another number $a$ divides $b$. Then we can apply the FTA to $a$ and get another prime factorization for $a$. If $p$ is a prime in the factorization of $a$, it must be in the factorization of $b$. Why? Well, you can use the property that if a prime number divides a product, it must divide one of the factors. So you have proven that every divisor of $b$ has to be made out of the primes in the factorization of $b$. Done! |
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Apr 8 |
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Let $p_1 , p_2$ be prime. Then prove that the only divisors of $p_1 p_2$ are $1 , p_1 , p_2 , p_1 p_2 $. What do you mean? Mathematics is not a subjective field. Read the Fundamental Theorem of Arithmetics carefully. It says that every number can be descomposed uniquely (up to order) into a product of primes. Your number is already in such a form. So if you take all combinations of the factors, you get ALL divisors. |
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Apr 8 |
answered | Let $p_1 , p_2$ be prime. Then prove that the only divisors of $p_1 p_2$ are $1 , p_1 , p_2 , p_1 p_2 $. |
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Apr 7 |
accepted | Equivalence of measures and $L^1$ functions |
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Apr 6 |
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Equivalence of measures and $L^1$ functions So for example in the case of $X=[0,1]$, the Lebesgue measure $m$ and the Gauss measure $\mu(B) = \frac{1}{\log 2}\int_B \frac{dx}{1+x}$ this is ok. But how do I prove it? |
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Apr 5 |
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Equivalence of measures and $L^1$ functions I see! And what about if the function $g$ is continuous. Your function has a jump at $0$. |
