My Math Forum (http://mymathforum.com/math-forums.php)
-   Real Analysis (http://mymathforum.com/real-analysis/)
-   -   A problem from functional analysis (http://mymathforum.com/real-analysis/15100-problem-functional-analysis.html)

 tach October 13th, 2010 09:27 AM

A problem from functional analysis

Hi

I'm currently following a first course in functional analysis, and a problem from my weekly assignment is bothering me quite a bit. It would be greatly appreciated if anyone could give me a hint or maybe even go through the steps in the solution. The problem is given below.

For a normed vector space $V$, we denote by $B(V)$ the space of bounded linear operators $T \, : \, V \rightarrow V$.

(The problem)
Let $V$ be a complex vector space with an inner product $(\cdot \, , \, \cdot)$ and assume that $T \in B(V)$.
1) Show that $(Tx,y)=0$ for all $x,y \in V$ if and only if $T$ is the zero operator.
2) Show next that $(Tx,x)=0$ for all $x \in V$ if and only if $T$ is the zero operator.
3) Are these results true if the vector space is a real vector space?

 mathman October 13th, 2010 04:00 PM

Re: A problem from functional analysis

1. and 2. The if parts are trivially obvious.

For 1. If Tx = y ? 0 for some x, the (Tx,y) ? 0. Therefore only if.
For 2. Simple example: 2 dimensional space with base vectors a and b, with (a,b)=0. Let Ta=b and Tb=-a, then (Tx,x)=0 for all x, therefore only if is false.

I don't see any distinction between real and complex spaces. However you may have to use complex conjugates for complex spaces in the above analysis.

 mathman October 15th, 2010 01:42 PM

Re: A problem from functional analysis

Correction: The analysis for 1. holds for both both real and complex vector spaces. The analysis for 2. is only for real vector spaces. I believe that 2. is true for complex vector spaces, but I haven't been able to work it out completely.

 mattpi October 16th, 2010 06:00 AM

Re: A problem from functional analysis

Suppose V is complex, and $\langle Tx,x\rangle=0$ for all $x\in V.$

Then for all $x,y\in V:$

$\langle T(x+ y),x+ y\rangle=0,$ and $\langle T(x+ iy),x+ iy\rangle=0.$

Expand these out, and you should be able to deduce that $\langle Tx,y\rangle=0$ for all $x,y\in V,$ and therefore (2) reduces to (1).

 tach October 20th, 2010 05:53 AM

Re: A problem from functional analysis

Thanks for responding both of you. I solved the problem in the following way (hopefully no typo's)

1)
The if part is trivial. The only if part I show by contraposition. Let $T$ be any operater in $B(V)$ but not the zero operator. Then there exists $x \in V$ such that $Tx \neq 0$ and so we have $(Tx,Tx)= llTxll^2 \neq 0$. Choose $y=Tx \in V$ and then we have found $x,y \in V$ such that $(Tx,y) \neq 0$. Thus $T$ must be the zero operator.

2)
Again the if part is trivial. The only if part I show by choosing again $T$ as any operator in $B(V)$ but not the zero operator. Then we know there exists $x,y \in V$ such that $(Tx,y) \neq 0$. Assume now that $(Tx,x)= 0 \, \forall x \in V$. Then we have



$0 = (T(x+iy),x+iy) = (Tx+iTy,x+iy) = (Tx,x) + (iTy,iy) + (Tx,iy) + (iTy,x) = \bar{i}(Tx,y) + i(Ty,x) =
Thus $T$ must be the zero operator.
These results are not true if $V$ is a real vector space. This is showed by an example. Choose $V=\mathbb{R}^2$ and let $T : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be given by $T((a,b))= (-b,a)$ for $a,b \in \mathbb{R}$. Then $T$ is linear, bounded and not the zero operator. Equip $V$ with the standard inner product and then we have $(Tx,x)= 0$ for all $x \in \mathbb{R}^2$.