Simple Inequality for Proving Equivalent Besov Seminorms

Question

For $f\in L^{p}(\mathbb{R}^{n})$, $1\leq p<\infty$, and $h\in\mathbb{R}^{n}$, define the quantity

$$I_{p}(h):=\left(\int_{\mathbb{R}^{n}}\left|f(x+h)-f(x)\right|^{p}dx\right)^{1/p}$$

and define the quantity

$$\omega_{p}(f,\delta):=\sup_{\left|h\right|\leq\delta}I_{p}(h), \qquad\forall 0<\delta<\infty$$

Now define

$$\overline{\omega}_{p}(f,\delta):=\left(\delta^{-n}\int_{\left|h\right|\leq\delta}I_{p}^{p}(h)dh\right)^{1/p}$$

As part a step towards proving two seminorms for Besov spaces are equivalent, I am trying to prove the following inequality:

$$v_{n}^{1/p}2^{-n-2}\omega_{p}(f,\delta)\leq\overline{\omega}_{p}(f,\delta)\leq v_{n}^{1/p}w_{p}(f,\delta), \qquad\forall \delta>0 \tag{*}$$

Can anyone give me a suggestion to get me on the right track?

Edit: So, Stephen Montgomery-Smith came up with a nice proof of an inequality with the sharper lower constant $v_{n}^{1/p}2^{-1-n/p}$. I am still curious if anyone sees a "cruder argument" that gives the lower estimate in the question.

Answer 1

The right hand inequality is easy. For the left hand inequality: $$ I_p(h_1 + h_2) \le I_p(h_1) + I_p(h_2) $$ So $$ I_p(h)^p \le (I_p(\tfrac12h + k) + I_p(\tfrac12h - k))^p \le 2^{p-1}(I_p^p(\tfrac12h + k) + I_p^p(\tfrac12h - k)) $$ So if $|h| \le \delta$, $$ I_p(h)^p \le 2^{p-1} v_n^{-1} (\delta/2)^{-n} \int_{|k| \le \delta/2}(I_p^p(\tfrac12h + k) + I_p^p(\tfrac12h - k)) \, dk \\ \le 2^{p+n}v_n^{-1} \delta^{-n} \int_{|k| \le \delta}I_p^p(k) \, dk $$ So $$ \omega_p(f,\delta) \le 2^{1+n/p} v_n^{-1/p} \bar\omega_p(f,\delta) $$ Seeing as $1+n/p \le 2+n$, this proves something slightly stronger, and so it probably isn't the intended argument. Or maybe I made an error somewhere. But I think the idea is valid.