Discontinuous linear mapping between infinite-dimension vector space

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SUMMARY

The discussion centers on the discontinuity of linear mappings between infinite-dimensional normed vector spaces, specifically using differentiation as an example. The operator T, defined as T:X→X:f→f′, where X is the set of all real polynomials on [0,1] equipped with the sup-norm, demonstrates this discontinuity. For the polynomial p_n(x)=x^n, while the sup-norm \|p_n\|_\infty equals 1, the image under T results in \|T(p_n)\|_\infty=n\|p_n\|_\infty, indicating that T is unbounded. This finding has significant implications in functional analysis.

PREREQUISITES
  • Understanding of linear mappings in vector spaces
  • Familiarity with normed vector spaces and sup-norm
  • Basic knowledge of differentiation as a linear operator
  • Concept of bounded and unbounded operators in functional analysis
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  • Study the properties of linear operators in infinite-dimensional spaces
  • Explore examples of bounded and unbounded operators in functional analysis
  • Learn about the implications of discontinuous linear mappings in mathematical analysis
  • Investigate the sup-norm and its applications in various mathematical contexts
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Mathematicians, students of functional analysis, and anyone interested in the properties of linear mappings in infinite-dimensional spaces will benefit from this discussion.

yifli
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It is known that any linear mapping between two finite dimensional normed vector space is continuous (bounded).

Can anyone give me an example of a linear mapping between two infinite dimensional normed vector space that is discontinuous?

Thanks
 
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Hi yifli! :smile:

yifli said:
It is known that any linear mapping between two finite dimensional normed vector space is continuous (bounded).

Can anyone give me an example of a linear mapping between two infinite dimensional normed vector space that is discontinuous?

Thanks

The standard example is a very familiar linear mapping: differentiation. Let X be the set of all real polynomials on [0,1]. Equip this with the sup-norm, i.e.

\|f\|_\infty=\sup_{t\in [0,1]}{|f(t)|}

Let

T:X\rightarrow X:f\rightarrow f^\prime

Let p_n(x)=x^n, then \|p_n\|_\infty=1, but

\|T(p_n)\|_\infty=n\|p_n\|_\infty

thus the operator T is not bounded.

This is a very tragic result and has a lot of bad consequences...
 

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