If ## f\in L_{p}^{\rm loc}(\mathbb{R}^{n}) ## and ## 1\leq p<\infty ##, then a stronger version of Lebesgue differentiation theorem holds: $$\lim\limits_{r\rightarrow 0}\dfrac{\|f\chi_{B(x,r)}\|_{L_{p}(\mathbb{R}^{n})}}{\|\chi_{B(x,r)}\|_{L_{p}(\mathbb{R}^{n})}}=|f(x)|$$ for almost all ##...
\sum\limits_{n=1}^{\infty}|x_{n}|=0\Rightarrow\lim \limits_{n\rightarrow\infty}\sum\limits_{k=1}^{n}|x_{k}|=0. Since s_{n}= \sum\limits_{k=1}^{n}|x_{k}| is an increasing sequence it can converge only its supremum. So \sup\limits_{n}s_{n}=0 . Thus \forall n,\, 0\leq s_{n} =...
Thank you for the answer i also think like you. This is an error in the wiki. But i saw several functional analysis book which write the proof of proposition same as in wiki. So,
Who is wrong?
I read the proof of the proposition "every cauchy sequence in a metric spaces is bounded" from
http://www.proofwiki.org/wiki/Every_Cauchy_Sequence_is_Bounded
I don't understand that how we can take m=N_{1} while m>N_{1} ?
In fact i mean that in a metric space (A,d) can we say that...
Mark44
The proposition which i want to prove is exactly the proposition which you write. I meant natural numbers by IN. Thanks for your suggestions.
I am waiting for your answers.
Homework Statement
Is the proposition
\sum^{\infty}_{n=1}|x_{n}|=0 ⇔\foralln\inIN x_{n}=0
true? If it is true how can we prove that ?
Homework Equations
The Attempt at a Solution
I proved the \Leftarrow side of proposition but i could not prove the \Rightarrow side of proposition.
"In his studies on Fourier Series, W.H.Young has analyzed certain convex functions \Phi:IR\rightarrow\bar{IR}^{+} which satisfy the conditions : \Phi(-x)=\Phi(x), \Phi(0)=0, and lim_{x\rightarrow\infty}\Phi(x)=+\infty. Then \Phi is called a Young function.
Several interesting nontrivial...
we know that \int_{|y-z|}^{y+z}D(y,z,w)\frac{w^{2m+1}}{2^m\Gamma(m+1)}dw=1
how can we calculate the integral
\int_{|y-z|}^{y+z}(1+2^kw)^aD(y,z,w)\frac{w^{2m+1}}{2^m\Gamma(m+1)}dw
find the general solution of yux+xuy=yu+xex ( the solution is in the form of u(x,y)=yex+f(y2-x2)ex )
if at first the value of u(x,y) on the upper half of hyperbola (that is y>=1) has been given as φ,show that if φ has not been given as a special form there is no solution.find that special form...