Thread: mapping, linear and bounded

1. mapping, linear and bounded

Hey,

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

I want to show that:

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

1)

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

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

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

Correct so far? If yes, how to continue?

2)

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

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

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

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

Appreciate the effort.

2. Originally Posted by surjective
Hey,

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

I want to show that:

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

1)

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

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

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

Correct so far? If yes, how to continue?

2)

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

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

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

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

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.

3. 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 < $\displaystyle \infty$.

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

4. Originally Posted by surjective
Hey,

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

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

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

Tonio

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

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

6. Originally Posted by Drexel28
He already did define it. $\displaystyle L^p(\mathbb{R})$ is the space of all functions whose $\displaystyle p^{\text{th}}$ power is Lebesgue integrable over $\displaystyle \mathbb{R}$.

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

7. 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 < $\displaystyle \infty$.
Make the substitution $\displaystyle u=2x$ in the integral.

8. bounded operator

Hello,

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

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

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

Thanks.