Results 1 to 6 of 6

Math Help - mapping

  1. #1
    Member
    Joined
    Feb 2010
    Posts
    133

    mapping

    Hello,

    I am considering the following map:

    T:L^{1}(0,2) \to T:L^{1}(0,2), (Tf)(x)=\int_{0}^{x}tf(t)

    I wan't to show that L^{1}(0,2) maps to L^{1}(0,2). I have done the following (but get stuck):

    \int_{0}^{2}|(Tf)(x)|dx=\int_{0}^{2}\vert \int_{0}^{x}tf(t)dt \vert dx \leq \int_{0}^{2} \int_{0}^{x}|tf(t)|dt dx

    How should I continue so that I and up with an expression which shows that \int_{0}^{2}|(Tf)(x)|dx<\infty? Swich of integral?

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by surjective View Post
    Hello,

    I am considering the following map:

    T:L^{1}(0,2) \to T:L^{1}(0,2), (Tf)(x)=\int_{0}^{x}tf(t)

    I wan't to show that L^{1}(0,2) maps to L^{1}(0,2). I have done the following (but get stuck):

    \int_{0}^{2}|(Tf)(x)|dx=\int_{0}^{2}\vert \int_{0}^{x}tf(t)dt \vert dx \leq \int_{0}^{2} \int_{0}^{x}|tf(t)|dt dx

    How should I continue so that I and up with an expression which shows that \int_{0}^{2}|(Tf)(x)|dx<\infty? Swich of integral?

    Thanks.
    Am I missing something or is \displaystyle \int_0^2\left|T_f(x)\right|\text{ }dx\leqslant \int_0^2\int_0^x |tf(t)|\text{ }dx\leqslant \int_0^2\int_0^2 |tf(t)|\text{ }d\text{ }dx\leqslant 2\int_0^2\int_0^2|f(t)|\text{ }dx\text{ }dt<\infty?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Feb 2010
    Posts
    133
    Hey,

    How did you get:

    \int_0^2\int_0^x |tf(t)|\text{ }dx\leqslant \int_0^2\int_0^2 |tf(t)|\text{ }d\text{ }dx\leqslant 2\int_0^2\int_0^2|f(t)|\text{ }dx\text{ }dt

    Sorry, but it's not quite clear. Where does the upper limit 2 on the second integral come from and where does the constant 2 in the last expression come from??
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by surjective View Post
    Hey,

    How did you get:

    \int_0^2\int_0^x |tf(t)|\text{ }dx\leqslant \int_0^2\int_0^2 |tf(t)|\text{ }d\text{ }dx\leqslant 2\int_0^2\int_0^2|f(t)|\text{ }dx\text{ }dt

    Sorry, but it's not quite clear. Where does the upper limit 2 on the second integral come from and where does the constant 2 in the last expression come from??
    Use the fact that for a Lebesgue integral \displaystyle \left|\int f\;dm\right|\leqslant \int |f|\;dm, to get the first step, the second clearly comes since for each x\in[0,2] it's evidently true that \displaystyle \int_0^x |tf(t)|\text{ }dt\leqslant \int_0^x|tf(t)|\text{ }dt+\int_x^2 |tf(t)|\text{ }dt=\int_0^2 |tf(t)|\text{ }dt then the last step comes from the fact that |t|\leqslant 2 on [0,2]
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Feb 2010
    Posts
    133
    Hello,

    To show that the operator mentioned in the beginning is bounded, I have done the following using the above calculations p=1:

    \Vert (Tf)(x) \Vert = \int_0^2\int_0^x |tf(t)|dtdx\leq 2\int_0^2\int_0^2|f(t)|dtdx

    If I could somehow get "rid" of one of the integrals then I would be home free?!!! Any suggestions?

    Thanks.
    Last edited by surjective; April 5th 2011 at 05:06 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by surjective View Post
    Hello,

    To show that the operator mentioned in the beginning is bounded, I have done the following using the above calculations p=1:

    \Vert (Tf)(x) \Vert = \int_0^2\int_0^x |tf(t)|dtdx\leq 2\int_0^2\int_0^2|f(t)|dtdx

    If I could somehow get "rid" of one of the integrals then I would be home free?!!! Any suggestions?

    Thanks.
    I think you may be overthinking this.. \displaystyle \int_0^2\int_0^2 |f(t)|\text{ }dt\text{ }dx=4\int_0^2|f(t)|\text{ }dt
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. mapping
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: April 10th 2011, 12:14 AM
  2. Mapping
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: March 10th 2011, 01:52 AM
  3. Complex open mapping & conformal mapping problems.
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: February 22nd 2011, 07:26 AM
  4. mapping
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: November 13th 2008, 08:24 PM
  5. Mapping: one-to-one and onto
    Posted in the Discrete Math Forum
    Replies: 7
    Last Post: November 6th 2008, 07:41 PM

Search Tags


/mathhelpforum @mathhelpforum