Results 1 to 6 of 6

Math Help - Norm of an Operator

  1. #1
    Junior Member
    Joined
    Jan 2010
    Posts
    32

    Norm of an Operator

    I need to show that the operator T:L^2([0,1],\mathbb{R}) \rightarrow L^2([0,1],\mathbb{R}) defined by (Tf)(x)=xf(x) has norm 1

    There is a hint on the question saying show it must be at least 1-\frac{1}{n} for all n

    The norm for L^2 is \|f\|_2=\left(\int_{0}^{1}|f(x)|^2dx\right)^\frac{  1}{2}

    Any help would be great
    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Feb 2010
    Posts
    133
    Hello,

    Try showing that the operator is linear. Then apply the definition of a bounded linear operator and the norm of the operator will naturally follow.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2010
    Posts
    32
    Okay, so I have shown T is linear, pretty straighforward. I am stuggling with the bounded part.

    I think I somehow need to show \|Tf\|\leq R\|f\| for some R>0. Please correct me if I'm wrong.

    I went along the lines of the cauchy schwarz inequality but I'm a bit stuck.

    Thanks in advance
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by ejgmath View Post
    I need to show that the operator T:L^2([0,1],\mathbb{R}) \rightarrow L^2([0,1],\mathbb{R}) defined by (Tf)(x)=xf(x) has norm 1

    There is a hint on the question saying show it must be at least 1-\frac{1}{n} for all n

    The norm for L^2 is \|f\|_2=\left(\int_{0}^{1}|f(x)|^2dx\right)^\frac{  1}{2}
    You know that \|f\|_2^2 = \int_{0}^{1}|f(x)|^2dx, and \|Tf\|_2^2 = \int_{0}^{1}x^2|f(x)|^2dx. Multiplying the integrand by x^2 has the effect of making it smaller, because x^2<1 throughout the interval except at the right-hand endpoint. That tells you that  \|Tf\|_2 < \|f\|_2, and the only hope of replacing that inequality by something approaching equality would by if the "mass" of the function f is concentrated close to 1. So try taking f(x) = x^n. For large values of n, that function will be very small through most of the interval and then shoot up to the value 1 when x=1.

    If you caculate the integrals for \|f\|_2^2 and \|Tf\|_2^2 for the function f(x)=x^n, you should find that \frac{\|Tf\|}{\|f\|} = \sqrt{\frac{2n+3}{2n+5}}, which may not be "at least 1-\tfrac1n" as the hint would like, but it serves the same purpose of being close to 1.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Feb 2010
    Posts
    133

    bounded operator

    Hello,

    Would the following be correct:

    \Vert (Tf)(x) \Vert = \Vert xf(x) \Vert \leq \vert x \vert \cdot \Vert f \Vert .

    Since

    \max_{x\in [0,1]} x=1

    we see that:

    \Vert (Tf)(x) \Vert \leq 1 \cdot \Vert f \Vert
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by surjective View Post
    Hello,

    Would the following be correct:

    \Vert (Tf)(x) \Vert = {\color{red}\Vert xf(x) \Vert \leq \vert x \vert \cdot \Vert f \Vert} .

    Since

    \max_{x\in [0,1]} x=1

    we see that:

    \Vert (Tf)(x) \Vert \leq 1 \cdot \Vert f \Vert
    No, the inequality in red is not justified. (The left side is a number, the right side is a function of x, which is 0 when x=0. But the left side is not 0.)

    The reason that \|Tf\|\leqslant\|f\| is that if f, g are functions satisfying 0\leqslant g(x)\leqslant f(x), for all x in the interval [a,b], then \int_a^b\!\!\!g(x)\,dx\leqslant\int_a^b\!\!\!f(x)\  ,dx. In this case, since x^2|f(x)|^2\leqslant|f(x)|^2 for 0\leqslant x\leqslant1, it follows that \int_0^1\!\!\!x^2|f(x)|^2dx\leqslant\int_0^1\!\!\!  |f(x)|^2dx. Therefore \|Tf\|^2\leqslant\|f\|^2.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Differential Operator Norm
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: November 29th 2011, 06:13 AM
  2. [SOLVED] Norm of powers of the Volterra Operator
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: June 18th 2011, 02:49 AM
  3. Norm Of The Convolution Operator
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: August 15th 2010, 08:33 AM
  4. Operator norm
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 16th 2010, 07:32 PM
  5. Different Forms of Operator Norm
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: September 21st 2009, 02:48 PM

Search Tags


/mathhelpforum @mathhelpforum