mapping

**surjective** · April 5th 2011, 01:11 PM

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.

**Drexel28** · April 5th 2011, 02:31 PM

Originally Posted by surjective

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 $\displaystyle \int_0^2\left|T_f(x)\right|\text{ }dx\leqslant \int_0^2\int_0^x |tf(t)|\text{ }dx\leqslant \int_0^2\int_0^2 |tf(t)|\text{ }d\text{ }dx\leqslant 2\int_0^2\int_0^2|f(t)|\text{ }dx\text{ }dt<\infty$ ?

**surjective** · April 5th 2011, 02:54 PM

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??

**Drexel28** · April 5th 2011, 03:13 PM

Originally Posted by surjective

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 $[0,2]$

**surjective** · April 5th 2011, 04:55 PM

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.

**Drexel28** · April 5th 2011, 06:16 PM

Originally Posted by surjective

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.. $\displaystyle \int_0^2\int_0^2 |f(t)|\text{ }dt\text{ }dx=4\int_0^2|f(t)|\text{ }dt$

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