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

My answers:

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.