Search found 13 matches
- Sat Sep 28, 2019 6:34 pm
- Forum: Linear Algebra
- Topic: Linear Projection
- Replies: 2
- Views: 7354
Re: Linear Projection
$f=I$
- Thu Mar 30, 2017 7:53 pm
- Forum: Real Analysis
- Topic: Series and continuous functions
- Replies: 3
- Views: 4008
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
The condition 2 is not true.
The sequence must be decreasing
- Thu Dec 15, 2016 11:21 pm
- Forum: Number theory
- Topic: Subset of natural numbers
- Replies: 2
- Views: 6817
Re: Subset of natural numbers
Why there exist $a,b\in S$ with $(a,b)=1$?
- Thu Dec 15, 2016 8:23 am
- Forum: Real Analysis
- Topic: Existence of $x$
- Replies: 1
- Views: 2670
Re: Existence of $x$
What if $ \int _{A}g=0$ and $ \int _{A}fg\neq 0 $ ?
- Thu Oct 27, 2016 2:14 pm
- Forum: Real Analysis
- Topic: Bounded sequence
- Replies: 6
- Views: 8905
Re: Bounded sequence
It is not bounded near zero.
- Thu Sep 08, 2016 2:05 pm
- Forum: Algebra
- Topic: $\mathbb{R}^5$ over $\mathbb{R}$
- Replies: 2
- Views: 6615
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.
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.
- Mon Sep 05, 2016 10:50 pm
- Forum: Algebra
- Topic: An Interesting Exercise
- Replies: 4
- Views: 8431
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
2)This is a simple version of Tietze Extension Theorem.
0
- Thu Sep 01, 2016 3:05 pm
- Forum: Complex Analysis
- Topic: Non existence of complex functions
- Replies: 3
- Views: 5221
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. $g(z)=\exp(-z)$.
- Thu Sep 01, 2016 2:24 pm
- Forum: Complex Analysis
- Topic: The function $f$ is constant
- Replies: 2
- Views: 4617
Re: The function $f$ is constant
Is easy to prove:
The range of entire no constant function is dense.
The range of entire no constant function is dense.
- Tue Aug 30, 2016 3:02 pm
- Forum: Algebra
- Topic: An Interesting Exercise
- Replies: 4
- Views: 8431
Re: An Interesting Exercise
1) If X is compact then is not subspace
2)The functions in Hahn-Banach theorem are linear
2)The functions in Hahn-Banach theorem are linear