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A problem from functional analysisHi 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 , we denote by the space of bounded linear operators . (The problem) Let be a complex vector space with an inner product and assume that . 1) Show that for all if and only if is the zero operator. 2) Show next that for all if and only if is the zero operator. 3) Are these results true if the vector space is a real vector space? |

Re: A problem from functional analysis1. 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. |

Re: A problem from functional analysisCorrection: 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. |

Re: A problem from functional analysisSuppose V is complex, and for all Then for all and Expand these out, and you should be able to deduce that for all and therefore (2) reduces to (1). |

Re: A problem from functional analysisThanks 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 be any operater in but not the zero operator. Then there exists such that and so we have . Choose and then we have found such that . Thus must be the zero operator. 2) Again the if part is trivial. The only if part I show by choosing again as any operator in but not the zero operator. Then we know there exists such that . Assume now that . Then we have Thus must be the zero operator. 3) These results are not true if is a real vector space. This is showed by an example. Choose and let be given by for . Then is linear, bounded and not the zero operator. Equip with the standard inner product and then we have for all . |

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