I'm given the following dilation operator
$$D:L^{p}\rightarrow L^{p},\ (Df)(x):=2^{1/2}f(2x),$$
and I have to show that it is well-defined. Here's where I've gotten so far:
For an arbitrary function $f\in L^{p}$ I know that
$$||f||_{p}=\left(\int_{-\infty}^{\infty}|f(x)|^{p}\ \text{d}x\ \right)^{1/p}<\infty$$
So then I considered the norm of $(Df)(x)$:
$$||Df||_{p}=\left(\int_{-\infty}^{\infty}|2^{1/2}f(2x)|^{p}\ \text{d}x\ \right)^{1/p}=\left(2^{p/2}\int_{-\infty}^{\infty}|f(2x)|^{p}\ \text{d}x\ \right)^{1/p}$$
But now I'm kinda stuck. My aim is to somehow be able to construct an inequality involving the norm of $f$, $||f||_{p}$, since this would be an argument that the operator is well-defined.
Any help would be appreciated!
Like $f \in L^p$ so
$$||Df(x)||_{p}=||2^{\frac{1}{2}} f(2x)||_{p}=\left(\int_{-\infty}^{\infty}|2^{1/2}f(2x)|^{p}\ \text{d}x\ \right)^{1/p}=\left(\int_{-\infty}^{\infty}|2^{1/2}f(2x)|^{p}\ \frac{1}{2} \text{d}(2x)\ \right)^{1/p}=\left(\int_{-\infty}^{\infty}|2^{1/2}f(x)|^{p}\ \frac{1}{2} \text{d}x\ \right)^{1/p}= 2^{\frac{1}{2}-\frac{1}{p}}\left(\int_{-\infty}^{\infty}|f(x)|^{p}\ \text{d}x\ \right)^{1/p}<\infty$$
Does this help ?