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by S.F.Papadopoulos
Sat Sep 28, 2019 6:34 pm
Forum: Linear Algebra
Topic: Linear Projection
Replies: 2
Views: 729

Re: Linear Projection

$f=I$
by S.F.Papadopoulos
Thu Mar 30, 2017 7:53 pm
Forum: Real Analysis
Topic: Series and continuous functions
Replies: 3
Views: 879

Re: Series and continuous functions

The first function is not continuous.(no Lebesgue integrable)

The condition 2 is not true.
The sequence must be decreasing
by S.F.Papadopoulos
Thu Dec 15, 2016 11:21 pm
Forum: Number theory
Topic: Subset of natural numbers
Replies: 2
Views: 1571

Re: Subset of natural numbers

Why there exist $a,b\in S$ with $(a,b)=1$?
by S.F.Papadopoulos
Thu Dec 15, 2016 8:23 am
Forum: Real Analysis
Topic: Existence of $x$
Replies: 1
Views: 640

Re: Existence of $x$

What if $ \int _{A}g=0$ and $ \int _{A}fg\neq 0 $ ?
by S.F.Papadopoulos
Thu Oct 27, 2016 2:14 pm
Forum: Real Analysis
Topic: Bounded sequence
Replies: 6
Views: 2721

Re: Bounded sequence

It is not bounded near zero.
by S.F.Papadopoulos
Thu Sep 08, 2016 2:05 pm
Forum: Algebra
Topic: $\mathbb{R}^5$ over $\mathbb{R}$
Replies: 2
Views: 1723

Re: $\mathbb{R}^5$ over $\mathbb{R}$

We have
1)Every finite extension is algebraic
2)The algebraic closure of real numbers are
the complex numbers
3)Every finite extension of real numbers are
the complex numbers.
by S.F.Papadopoulos
Mon Sep 05, 2016 10:50 pm
Forum: Algebra
Topic: An Interesting Exercise
Replies: 4
Views: 1903

Re: An Interesting Exercise

1) subspace of a vector space for me( Rudin Functional Analysis and other) mean linear subspace.
2)This is a simple version of Tietze Extension Theorem.










0
by S.F.Papadopoulos
Thu Sep 01, 2016 3:05 pm
Forum: Complex Analysis
Topic: Non existence of complex functions
Replies: 3
Views: 1244

Re: Non existence of complex functions

If we restrict the function to the half-plane where ${\rm Re}(z) \geq 0$ , then $g$ is bounded which by a known fact of complex analysis means it is constant. The above is false. $g(z)=\exp(-z)$.
by S.F.Papadopoulos
Thu Sep 01, 2016 2:24 pm
Forum: Complex Analysis
Topic: The function $f$ is constant
Replies: 2
Views: 1188

Re: The function $f$ is constant

Is easy to prove:
The range of entire no constant function is dense.
by S.F.Papadopoulos
Tue Aug 30, 2016 3:02 pm
Forum: Algebra
Topic: An Interesting Exercise
Replies: 4
Views: 1903

Re: An Interesting Exercise

1) If X is compact then is not subspace
2)The functions in Hahn-Banach theorem are linear