Hilbert Space Help: Showing Norm Does Not Satisfy Parallelogram Law

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SUMMARY

The discussion focuses on the space of continuously differentiable functions on the interval [a,b], denoted as W[a,b], and its inner product defined by the formula \(\langle f,g \rangle = \int_a^b (f(x) \cdot \overline{g(x)} + f'(x) \cdot \overline{g'(x)}) \, dx\). Participants aim to demonstrate that the norm derived from this inner product does not satisfy the parallelogram law, which is a critical property in the context of normed spaces. Additionally, there is a suggestion to explore whether W[a,b] is a complete metric space.

PREREQUISITES
  • Understanding of inner product spaces and their properties
  • Familiarity with the concept of norms in functional analysis
  • Knowledge of the parallelogram law in the context of normed spaces
  • Basic calculus, particularly integration of functions
NEXT STEPS
  • Investigate the properties of inner products in functional spaces
  • Learn about the parallelogram law and its implications in normed spaces
  • Study the completeness of metric spaces, particularly in relation to W[a,b]
  • Explore examples of spaces that do and do not satisfy the parallelogram law
USEFUL FOR

Mathematicians, students of functional analysis, and anyone interested in the properties of inner product spaces and their implications in advanced calculus.

gravenewworld
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How can I show that the space of all continuously differentiable functions on [a,b] denoted W[a,b] with inner product (f,g)=Integral from a to b of (f(x)*conjugate of g(x)+f'(x)*conjugate of g'(x)).

Should I show that the norm does not satisfy the parallelogram law?
 
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What do you want to show ?

And just to clarify, you have the inner product defined as :

[tex]\langle f,g \rangle = \int _a ^b (fg^* + f'g'^*)dx[/tex] ?
 
I think I need to show that W[a,b] is not a complete metric space?

And yes that is right inner product.
 

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