1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Convergence in measure of the product of two functions

  1. Oct 19, 2013 #1
    [itex]f_{k} \overset{m}{\rightarrow} f[/itex] and [itex]g_{k} \overset{m}{\rightarrow} g[/itex] over [itex]E[/itex].
    Then:
    a)
    [itex]f_{k} + g_{k} \overset{m}{\rightarrow} f+g[/itex] over [itex]E[/itex]

    b)
    If [itex]| E | < + \infty[/itex], then [itex]f_{k} g_{k} \overset{m}{\rightarrow} fg[/itex] over [itex]E[/itex]. Show that the hiphotesis [itex]| E | < + \infty[/itex] is neccesary

    c) Let [itex]\{ \frac{f_{k}}{g_{k}} \}_{k \in \mathbb{N}}[/itex] a sequence of functions defined almost everywhere over [itex]E[/itex].
    If [itex]| E | \rightarrow g[/itex] over [itex]E[/itex] and [itex]g \neq 0[/itex] almost everywhere, then [itex] \frac{f_{k}}{g_{k}} \overset{m}{\rightarrow} \frac{f}{g} [/itex] over [itex]E[/itex].

    a)
    [itex]g_{k} \overset{m}{\rightarrow} g[/itex] means
    [itex]\forall \delta > 0 , \forall \varepsilon > 0 , \exists k_{0} / \forall k \geq k_{0}:[/itex]
    [itex] | \{ | g - g_{k} | \geq \delta \} | < \varepsilon [/itex]
    So for every [itex]\frac{\delta}{2} > 0, \frac{\varepsilon}{2} > 0[/itex] we have [itex]k^{'}_{0}, k^{''}_{0}[/itex] such that:

    [itex]\forall k \geq k^{'}_{0} : | \{ | g - g_{k} | \geq \frac{\delta}{2} \} | < \frac{\varepsilon}{2} [/itex]
    [itex]\forall k \geq k^{''}_{0} : | \{ | f - f_{k} | \geq \frac{\delta}{2} \} | < \frac{\varepsilon}{2} [/itex]
    [itex]k_{0} = max \{ k^{'}_{0}, k^{''}_{0} \}[/itex]
    Now this holds for all [itex]k \geq k_{0}[/itex]
    Then
    [itex]\forall k \geq k_{0} : | \{ | g - g_{k} | \geq \frac{\delta}{2} \} \cup \{ | f - f_{k} | \geq \frac{\delta}{2} \} | < \varepsilon [/itex]
    (The measure of the union is equal or lesser than sum of measures).
    [itex]\{ | (f - f_{k}) + (g - g_{k}) \geq \delta \} \subset \{ | f - f_{k} | + | g - g_{k} | \geq \delta \} \subset \{ | g - g_{k} | \geq \frac{\delta}{2} \} \cup \{ | f - f_{k} | \geq \frac{\delta}{2} \}[/itex]
    Then for monotony of the measure:
    [itex]\forall k \geq k_{0} | \{ | (f - f_{k}) + (g - g_{k}) \geq \delta \} | < \varepsilon[/itex]

    b) I guess I can prove:
    [itex]f_{k} \overset{m}{\rightarrow} f \Longrightarrow | f_{k} | \overset{m}{\rightarrow} |f |[/itex]
    [itex]f=f^{+}-f^{-}[/itex] and [itex]| f | = f^{+} + f^{-}[/itex] (so it's easy, we only need to prove that if [itex]f_{k} \overset{m}{\rightarrow} f[/itex] then [itex]f^{\mp}_{k} \overset{m}{\rightarrow} f^{\mp}[/itex]
    Then using that prove that if [itex]f_{k} \overset{m}{\rightarrow} f[/itex] then [itex]f^{2}_{k} \overset{m}{\rightarrow} f^{2}[/itex] (*)
    And finally notice that [itex]fg = \frac{1}{4} [ (f+g)^{2} - (f-g)^{2} ][/itex] (Using also that if [itex]f_{k} \overset{m}{\rightarrow} f[/itex] then [itex]-f_{k} \overset{m}{\rightarrow} -f[/itex])

    (*)
    [itex]| f_{k} | \overset{m}{\rightarrow} | f |[/itex] means
    [itex]\forall \delta > 0 , \forall \varepsilon > 0 , \exists k_{0} / \forall k \geq k_{0}:[/itex]
    [itex] | \{ | | f | - | f_{k} | | \geq \delta \} | < \varepsilon [/itex]

    What I thought was:
    [itex]f^{2} - f^{2}_{k} = (\underbrace{f+f_{k}}_{\text{if }f <+\infty \text{ it tends to }2f})(\underbrace{f-f_{k}}_{ \to 0}) [/itex]
    And later do something like this:
    [itex] | \{ | f - f_{k} | \geq \frac{\delta}{f+f_{k}} \} | < \varepsilon [/itex]
    But I realized that this doesn't make any sense at all because [itex]f+f_{k}[/itex] depends of the value of [itex]x[/itex].

    Any hint?
     
    Last edited: Oct 19, 2013
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?



Similar Discussions: Convergence in measure of the product of two functions
  1. Product measures (Replies: 0)

Loading...