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 October 17th, 2010, 11:24 AM #1 Newbie   Joined: Oct 2010 Posts: 1 Thanks: 0 Proving that a space is continuous, continuous at 0, and bdd My problem is that I have a space $T:E \to F \textrm{ that is a linear transformation from the normed linear space } E \textrm{ to the normed linear space } F.$ I need to prove the following are equivalent: A. T is continuous. B. T is continuous at 0. C. T is bounded. Thanks.
 October 18th, 2010, 01:14 PM #2 Senior Member   Joined: Feb 2009 Posts: 172 Thanks: 5 Re: Proving that a space is continuous, continuous at 0, and $A\Rightarrow B$ Obvious. To show $B\Rightarrow C$ we show that $B\Rightarrow A$ and then that $A\Rightarrow C$. $B\Rightarrow A$ Suppose $T$ is continuous at . Take $x\in E$ and $\epsilon>0$. Since $T$ is continuous at  there exists $\delta>0$ such that $\|T(y)-T(0)\|=\|T(y)\|<\epsilon=$ for all $x\in E$ such that $\|y\|<\delta$. Hence, if $\|x-y\|<\delta$ we have that $\|T(x-y)-T(0)\|=\|T(x-y)\|=\|T(x)-T(y)\|<\epsilon=$ wich gives us that $T$ is continuous at $x$. Thus $T$ is continuous. $A\Rightarrow C$ Suppose $T$ is continuous. Consider the set $A=\{x\in E:\|x\|=1\}$. We have that $A$ is a compact set since the norm is a continous function from $E$ to $\mathbb{R}$, $A=f^{-1}(\{1\})$ and $\{1\}$ is compact on $\mathbb{R}$. Hence since $T$ is continous $T(A)=\{T(x)\in A\}" /> is a compact set wich gives us that $B=\{\|T(x)|\in A\}" /> is a real bounded set. Take $M=\sup B$. We have that $\|T(x)\|\leq M$ for all $x\in A$ wich gives us that for all $x\in E\backslash\{0\}$, $\frac{\|T(x)\|}{\|x\|}\leq M$. Hence $\|T(x)\|\leq M\|x\|$ for all $x\in E$ since $T(0)=0=M\|0|$. Thus $T$ is bounded and then $B\Rightarrow C$. $C\Rightarrow A$ Suppose $T$ is bounded. Then there exists $M>0$ such that $\|T(x)\|\leq M\|x\|$ for all $x\in E$. Take $x\in E$ and $\epsilon>0$. Consider $\delta=\frac{\epsilon}{M}$. If $\|x-y\|<\delta$ we have that $\|T(x)-T(y)\|=\|T(x-y)\|\leq M\|x-y\|
October 18th, 2010, 05:51 PM   #3
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Re: Proving that a space is continuous, continuous at 0, and

Quote:
 Originally Posted by parasio Just making a few corrections $A\Rightarrow B$ Obvious. To show $B\Rightarrow C$ we show that $B\Rightarrow A$ and then that $A\Rightarrow C$. $B\Rightarrow A$ Suppose $T$ is continuous at . Take $x\in E$ and $\epsilon>0$. Since $T$ is continuous at  there exists $\delta>0$ such that $\|T(y)-T(0)\|=\|T(y)\|<\epsilon=$ for all $x\in E$ such that $\|y\|<\delta$. Hence, if $\|x-y\|<\delta$ we have that $\|T(x-y)-T(0)\|=\|T(x-y)\|=\|T(x)-T(y)\|<\epsilon=$ wich gives us that $T$ is continuous at $x$. Thus $T$ is continuous. $A\Rightarrow C$ Suppose $T$ is continuous. Consider the set $A=\{x\in E:\|x\|=1\}$. We have that $A$ is a compact set since the norm is a continous function from $E$ to $\mathbb{R}$, $A=f^{-1}(\{1\})$ and $\{1\}$ is compact on $\mathbb{R}$. Hence since $T$ is continous $T(A)=\{T(x)\in A\}" /> is a compact set wich gives us that $B=\{ |T(x)|\in A\}" /> is a real bounded set. Take $M=\sup B$. We have that $\|T(x)\|\leq M$ for all $x\in A$ wich gives us that for all $x\in E\backslash\{0\}$, $\frac{\|T(x)\|}{\|x\|}\leq M$. Hence $\|T(x)\|\leq M\|x\|$ for all $x\in E$ since $T(0)=0=M\|0|$. Thus $T$ is bounded and then $B\Rightarrow C$. $C\Rightarrow A$ Suppose $T$ is bounded. Then there exists $M>0$ such that $\|T(x)\|\leq M\|x\|$ for all $x\in E$. Take $x\in E$ and $\epsilon>0$. Consider $\delta=\frac{\epsilon}{M}$. If $\|x-y\|<\delta$ we have that $\|T(x)-T(y)\|=\|T(x-y)\|\leq M\|x-y\|

October 18th, 2010, 06:33 PM   #4
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Re: Proving that a space is continuous, continuous at 0, and

Quote:
 Originally Posted by parasio $A\Rightarrow B$ Obvious. To show $B\Rightarrow C$ we show that $B\Rightarrow A$ and then that $A\Rightarrow C$. $B\Rightarrow A$ Suppose $T$ is continuous at . Take $x\in E$ and $\epsilon>0$. Since $T$ is continuous at  there exists $\delta>0$ such that $\|T(y)-T(0)\|=\|T(y)\|<\epsilon=$ for all $x\in E$ such that $\|y\|<\delta$. Hence, if $\|x-y\|<\delta$ we have that $\|T(x-y)-T(0)\|=\|T(x-y)\|=\|T(x)-T(y)\|<\epsilon=$ wich gives us that $T$ is continuous at $x$. Thus $T$ is continuous. $A\Rightarrow C$ Suppose $T$ is continuous. Consider the set $A=\{x\in E:\|x\|=1\}$. We have that $A$ is a compact set since the norm is a continous function from $E$ to $\mathbb{R}$, $A=f^{-1}(\{1\})$ and $\{1\}$ is compact on $\mathbb{R}$. Hence since $T$ is continous $T(A)=\{T(x)\in A\}" /> is a compact set wich gives us that $B=\{\|T(x)|\in A\}" /> is a real bounded set. Take $M=\sup B$. We have that $\|T(x)\|\leq M$ for all $x\in A$ wich gives us that for all $x\in E\backslash\{0\}$, $\frac{\|T(x)\|}{\|x\|}\leq M$. Hence $\|T(x)\|\leq M\|x\|$ for all $x\in E$ since $T(0)=0=M\|0|$. Thus $T$ is bounded and then $B\Rightarrow C$. $C\Rightarrow A$ Suppose $T$ is bounded. Then there exists $M>0$ such that $\|T(x)\|\leq M\|x\|$ for all $x\in E$. Take $x\in E$ and $\epsilon>0$. Consider $\delta=\frac{\epsilon}{M}$. If $\|x-y\|<\delta$ we have that $\|T(x)-T(y)\|=\|T(x-y)\|\leq M\|x-y\|

How can we know that your proof :B=>A is correct??

 October 18th, 2010, 07:30 PM #5 Senior Member   Joined: Feb 2009 Posts: 172 Thanks: 5 Re: Proving that a space is continuous, continuous at 0, and $B\Rightarrow A$ Suppose $T$ is continuous at . To show that $T$ is continuous you must show that $T$ is continuous at $x$ for all $x\in E$. Then you take an arbitrary $x\in E$ and show that for all $\epsilon>0$ there exists $\delta>0$ such that $\|T(x)-T(y)\|<\epsilon$ for all $y\in E$ such that $\|x-y\|<\delta$. This is what I did here. Take $x\in E$ arbitrary. Now take $\epsilon>0$ arbitrary. Since $T$ is continuous at  there exists $\delta>0$ such that $\|T(y)-T(0)\|=\|T(y)\|<\epsilon=$ for all $y\in E$ such that $\|y\|<\delta$. Hence, if $\|x-y\|<\delta$ we have that $\|T(x-y)-T(0)\|=\|T(x-y)\|=\|T(x)-T(y)\|<\epsilon=$ wich gives us that $T$ is continuous at $x$. Thus $T$ is continuous. There was a little mistake here, instead of "such that $\|T(y)-T(0)\|=\|T(y)\|<\epsilon=$ for all $x\in E$ such that $\|y\|<\delta$" it should be written such that "$\|T(y)-T(0)\|=\|T(y)\|<\epsilon=$ for all $y\in E$ such that $\|y\|<\delta$". In the demonstration above it's already corrected. If you have any problem just write me.
October 19th, 2010, 02:57 PM   #6
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Re: Proving that a space is continuous, continuous at 0, and

Quote:
 Originally Posted by parasio $B\Rightarrow A$ Suppose $T$ is continuous at . To show that $T$ is continuous you must show that $T$ is continuous at $x$ for all $x\in E$. Then you take an arbitrary $x\in E$ and show that for all $\epsilon>0$ there exists $\delta>0$ such that $\|T(x)-T(y)\|<\epsilon$ for all $y\in E$ such that $\|x-y\|<\delta$. This is what I did here. Take $x\in E$ arbitrary. Now take $\epsilon>0$ arbitrary. Since $T$ is continuous at  there exists $\delta>0$ such that $\|T(y)-T(0)\|=\|T(y)\|<\epsilon=$ for all $y\in E$ such that $\|y\|<\delta$. Hence, if $\|x-y\|<\delta$ we have that $\|T(x-y)-T(0)\|=\|T(x-y)\|=\|T(x)-T(y)\|<\epsilon=$ wich gives us that $T$ is continuous at $x$. Thus $T$ is continuous. There was a little mistake here, instead of "such that $\|T(y)-T(0)\|=\|T(y)\|<\epsilon=$ for all $x\in E$ such that $\|y\|<\delta$" it should be written such that "$\|T(y)-T(0)\|=\|T(y)\|<\epsilon=$ for all $y\in E$ such that $\|y\|<\delta$". In the demonstration above it's already corrected. If you have any problem just write me.

To get |y-x|<? you have to use the substitution : y= y-x , infering that x=0 and thus proving continuity not for all ,x?E ,BUT FOR x=0 only.

Correct??

 October 19th, 2010, 06:19 PM #7 Senior Member   Joined: Feb 2009 Posts: 172 Thanks: 5 Re: Proving that a space is continuous, continuous at 0, and You have that $\|T(y)\|<\epsilon$ for all $y\in E$ such that $\|y\|<\delta$. Now here's what I did. $x$ is a fixed point. For all $z\in E$ we have that $x-z\in E$ since $E$ is a linear space. Hence if you take $z\in E$ such that $\|x-z\|<\delta$ you have from the above sentence that $\|T(x-z)\|=\|T(x)-T(z)\|<\epsilon=$. Any problem write again.
October 20th, 2010, 03:12 PM   #8
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Re: Proving that a space is continuous, continuous at 0, and

Quote:
 Originally Posted by paraiso You have that $\|T(y)\|<\epsilon$ for all $y\in E$ such that $\|y\|<\delta$. Now here's what I did. $x$ is a fixed point. For all $z\in E$ we have that $x-z\in E$ since $E$ is a linear space. Hence if you take $z\in E$ such that $\|x-z\|<\delta$ you have from the above sentence that $\|T(x-z)\|=\|T(x)-T(z)\|<\epsilon=$. Any problem write again.
That substitution works.

Now i have a question concerning the laws of logic involved in the above proof.

Since we know that in every mathematical proof we have the laws of logic applied on the theorems,definitions, or axioms involved in the proof and hence giving us the statements of the proof,which would you concur, are the laws of logic taking place in the above proof??

 October 20th, 2010, 04:36 PM #9 Senior Member   Joined: Feb 2009 Posts: 172 Thanks: 5 Re: Proving that a space is continuous, continuous at 0, and Logic is present in everything you do in mathematics. I don't know the names of the laws applied here it's been a long time since I studied logic for the last time.
October 20th, 2010, 04:50 PM   #10
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Re: Proving that a space is continuous, continuous at 0, and

Quote:
 Originally Posted by paraiso Logic is present in everything you do in mathematics. I don't know the names of the laws applied here it's been a long time since I studied logic for the last time.

You mean long time ago you could recognise the laws of logic involved in a mathematical analysis proof and now due to the lapse of time you cannot??

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