Convergence in distribution (weak convergence)

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SUMMARY

The discussion centers on the convergence of the distribution defined by the function f_n(x) = n·I[|x|<1/(2n)] in D'(\mathbb{R}) to the delta distribution δ_0. The key conclusion is that f_n^2 - nδ_0 converges in D'(\mathbb{R}), which can be shown using the second-order Taylor expansion of the test function φ. The calculations reveal that (f_n^2, φ) = n(f_n - δ_0, φ), leading to the conclusion that the limit approaches zero as n approaches infinity.

PREREQUISITES
  • Understanding of distributions in D'(\mathbb{R})
  • Knowledge of the delta distribution δ_0
  • Familiarity with indicator functions and their properties
  • Proficiency in Taylor series expansions
NEXT STEPS
  • Study the properties of distributions in D'(\mathbb{R})
  • Learn about the delta distribution and its applications
  • Explore the use of Taylor series in functional analysis
  • Investigate convergence concepts in distribution theory
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Mathematicians, students of functional analysis, and anyone studying the properties of distributions and convergence in mathematical analysis.

christoff
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Homework Statement


Consider the function [tex]f_n(x)=n\cdot I\left[|x|<\frac{1}{2n}\right][/tex] considered as a distribution in [itex]D'(\mathbb{R})[/itex], where [itex]I[/itex] denotes the indicator function. Recall that [itex]f_n[/itex] converges to [itex]\delta_0[/itex], the delta distribution, in [itex]D'(\mathbb{R})[/itex]. Show that [itex]f_n^2-n\delta_0[/itex] converges in [itex]D'(\mathbb{R})[/itex].

The Attempt at a Solution



By a few calculations, we can show that for any test function [itex]\phi[/itex] and natural [itex]n>0[/itex], that
[tex](f_n^2,\phi)=n(f_n-\delta_0,\phi).[/tex]
At this point, I would be tempted to say that this converges to zero (as a sequence of real numbers) because [itex](f_n-\delta_0,\phi)[/itex] "must" dominate the convergence of [itex]n[/itex] to infinity. However, this isn't rigorous, and is more of a hunch than anything. However, any attempts I've made to prove it have not gone anywhere. I feel as though it must have something to do with the smoothness of the test functions, but I can't see where to incorporate this.

At this point, I would appreciate a nudge in the right direction.
Thanks in advance!
 
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christoff said:

Homework Statement


Consider the function [tex]f_n(x)=n\cdot I\left[|x|<\frac{1}{2n}\right][/tex] considered as a distribution in [itex]D'(\mathbb{R})[/itex], where [itex]I[/itex] denotes the indicator function. Recall that [itex]f_n[/itex] converges to [itex]\delta_0[/itex], the delta distribution, in [itex]D'(\mathbb{R})[/itex]. Show that [itex]f_n^2-n\delta_0[/itex] converges in [itex]D'(\mathbb{R})[/itex].

The Attempt at a Solution



By a few calculations, we can show that for any test function [itex]\phi[/itex] and natural [itex]n>0[/itex], that
[tex](f_n^2,\phi)=n(f_n-\delta_0,\phi).[/tex]
At this point, I would be tempted to say that this converges to zero (as a sequence of real numbers) because [itex](f_n-\delta_0,\phi)[/itex] "must" dominate the convergence of [itex]n[/itex] to infinity. However, this isn't rigorous, and is more of a hunch than anything. However, any attempts I've made to prove it have not gone anywhere. I feel as though it must have something to do with the smoothness of the test functions, but I can't see where to incorporate this.

At this point, I would appreciate a nudge in the right direction.
Thanks in advance!

Try using the second-order Taylor expansion of ##\phi(x)## in the evaluation of ##(f_n^2,\phi)##.
 
Yep, that worked. Thanks! It's actually very easy once you apply the Taylor expansion :)
 

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