mapping, linear and bounded

**surjective** · March 6th 2011, 02:31 PM

Hey,

For $p \in[1,\infty[$ I have the following dilation operator: $D:L^{p}(\mathbb{R}) \to L^{p}(\mathbb{R}), (Df)(x)=2^{\frac{1}{2}}f(2x)$

I want to show that:

1) D maps $L^{p}(\mathbb{R})$ into $L^{p}(\mathbb{R})$
2) D is linear and bounded

My answers:

1)

Firstly: $L^{p}(\mathbb{R}) = \{f:\mathbb{R}\to \mathbb{C} | \int_{-\infty}^{\infty}|f(x)|^{p}dx < \infty \}$

So to show that D maps $L^{p}(\mathbb{R})$ into $L^{p}(\mathbb{R})$ we must have that:

$\int_{-\infty}^{\infty}|2^{\frac{1}{2}}f(2x)|^{p}dx < \infty$

Correct so far? If yes, how to continue?

2)

Would it be sufficient first to show that an operator $T:V_{1} \to V_{2}$ is bounded if there exists a $K\geq 0$ such that:

$\Vert Tv \Vert_{V_{2}} \leq K \Vert v \Vert_{V_{1}}$

Then I would show that the mapping $T:V_{1} \to V_{2}$ is linear if:

$T(\alpha v+\beta w)=\alpha T(v)+\beta T(w)$

Appreciate the effort.

**Drexel28** · March 6th 2011, 06:48 PM

Originally Posted by surjective

Hey,

For $p \in[1,\infty[$ I have the following dilation operator: $D:L^{p}(\mathbb{R}) \to L^{p}(\mathbb{R}), (Df)(x)=2^{\frac{1}{2}}f(2x)$

I want to show that:

1) D maps $L^{p}(\mathbb{R})$ into $L^{p}(\mathbb{R})$
2) D is linear and bounded

My answers:

1)

Firstly: $L^{p}(\mathbb{R}) = \{f:\mathbb{R}\to \mathbb{C} | \int_{-\infty}^{\infty}|f(x)|^{p}dx < \infty \}$

So to show that D maps $L^{p}(\mathbb{R})$ into $L^{p}(\mathbb{R})$ we must have that:

$\int_{-\infty}^{\infty}|2^{\frac{1}{2}}f(2x)|^{p}dx < \infty$

Correct so far? If yes, how to continue?

2)

Would it be sufficient first to show that an operator $T:V_{1} \to V_{2}$ is bounded if there exists a $K\geq 0$ such that:

$\Vert Tv \Vert_{V_{2}} \leq K \Vert v \Vert_{V_{1}}$

Then I would show that the mapping $T:V_{1} \to V_{2}$ is linear if:

$T(\alpha v+\beta w)=\alpha T(v)+\beta T(w)$

Appreciate the effort.

The first part looks fine to me. Moreover, you may be overthinking part two. Linearity is immediate and boundedness is almost as easy.

**surjective** · March 6th 2011, 07:31 PM

Hey,

For part 1) I was hoping that you could help me do "something" with the indefinite integral so that it clearly stands out that it in fact is < $\infty$ .

For part 2) : So checking the two criterias, i wrote down, would be sufficient?

**tonio** · March 7th 2011, 06:17 PM

Originally Posted by surjective

Hey,

I have the following operator: $D: L^{p}(\mathbb{R}) \to L^{p}(\mathbb{R}), (DF)(x)=2^{\frac{1}{2}}f(2x)$

How do I show that $\int_{-\infty}^{\infty}|2^{\frac{1}{2}}f(2x)|^{p}dx< \infty$

How do I show that the operator is linear?

I posted this question before along with my suggestions but i got stuck.

Greatly appreciate the assistance.

Please do define $L^p(\mathbb{R})$ . I don't think this is a standard notation, though I could be wrong.

Tonio

**Drexel28** · March 7th 2011, 08:21 PM

Originally Posted by tonio

Please do define $L^p(\mathbb{R})$ . I don't think this is a standard notation, though I could be wrong.

Tonio

He already did define it. $L^p(\mathbb{R})$ is the space of all functions whose $p^{\text{th}}$ power is Lebesgue integrable over $\mathbb{R}$ .

**tonio** · March 8th 2011, 03:11 AM

Originally Posted by Drexel28

He already did define it. $L^p(\mathbb{R})$ is the space of all functions whose $p^{\text{th}}$ power is Lebesgue integrable over $\mathbb{R}$ .

He did not define it in the post I was addressing to, as you can see in my message. Apaprently in another post, which

now appears at the top of the list, the definition is there, but that didn't show when I answered.

Tonio

**Opalg** · March 8th 2011, 03:38 AM

Originally Posted by surjective

Hey,

For part 1) I was hoping that you could help me do "something" with the indefinite integral so that it clearly stands out that it in fact is < $\infty$ .

Make the substitution $u=2x$ in the integral.

**surjective** · March 8th 2011, 12:34 PM

Hello,

Thanks for response. I have done the following to show that the operator is bounded:

$\Vert (Df)(x) \Vert = \left( \int_{-\infty}^{\infty}|2^{\frac{1}{2}}f(2x)|^{p}dx \right)^{\frac{1}{p}}= \left( \int_{-\infty}^{\infty}|(2^{\frac{1}{2}}f(2x))^{p}|dx \right)^{\frac{1}{p}}\leq \left( \int_{-\infty}^{\infty}|2^{\frac{p}{2}}||f(2x)|^{p}dx \right)^{\frac{1}{p}} = 2^{\frac{1}{2}} \left(\int_{-\infty}^{\infty}|f(2x)|^{p}dx\right)^{\frac{1}{p}} $

Is the above correct? I suspect that there is an error since I should end up with $\Vert (Df)(x) \Vert\leq K \Vert f \Vert$

Thanks.

Thread: mapping, linear and bounded

LinkBack

Thread Tools

Display

mapping, linear and bounded

bounded operator

Similar Math Help Forum Discussions

Linear Mapping..

linear mapping

Continuous mapping - closed, open, compact, bounded

linear mapping

linear mapping

Search Tags