Recent content by fderingoz

\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

i found the general solution of the equation. thanks for your helps but i can't understand anything rest of the question.i am waiting for your helps.

thanks for your advice i tried the method of characteristics,but i can not find the solution :blushing:

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...