For $N\geq 1$, let $u\in\mathcal{C}_c^{\infty}(\mathbb{R}^N)$, $u$ radially symmetric and $r>0$. I am trying to justify this inequality (it appears in my lecture notes): $$\left(\int_{r}^{\infty} |u^{\prime}(s)|^p s^{N-1} ds\right)^{\frac1p} \leq \omega_N^{-\frac{1}{p}} \left(\int_{\mathbb{R}^{N}\setminus\{0\}}|\nabla u|^p dx\right)^{\frac1p},$$ for $1<p<N$.
I guess something related to the polar coordinates has been used, but how? Could anyone please explain me?
Moreover, why one could write $\mathbb{R}^{N}\setminus\{0\}$ instead of $\mathbb{R}^N$?
I hope someone could help. Thank you in advance!
In your equation $u'$ and $\nabla u$ stand for two different things. The first is a function of one variable and second is a function of $N$ variables so you must be missing a hypothesis.
Any point $x \in \mathbf R^N$ can be expressed as $x = s \omega$ where $s > 0$ and $\omega$ is a point on the unit sphere $\mathbf S^{N-1}$. The change of variable formula reads $$\int_{\mathbf R^N} F(x) \, dx = \int_{\mathbf S^{N-1}} \int_0^\infty F(s\omega) s^{N-1} \, ds d\omega$$ for suitable functions $F$.
If $F$ is radially symmetric there is a function $f$ satisfying $F(x) = f(|x|)$. Except at the point $x=0$ you have $$\frac{\partial |x|}{\partial x_i} = \frac{x_i}{|x|}$$ so that $$\frac{\partial F}{\partial x_i}(x) = f'(|x|) \frac{x_i}{|x|}.$$ Consequently $|\nabla F(x)| = |f'(|x|)|$ and thus $|\nabla F(s\omega)| = |f'(s)|$. Now integrate the $p$-power of these and apply the change of variable rule: \begin{align*} \int_{\mathbf R^N \setminus \{0\}} |\nabla F(x)|^p \, dx &= \int_{\mathbf S^{N-1}} \int_0^\infty |\nabla F(s\omega)|^p s^{N-1} \, ds d\omega \\ &= \int_{\mathbf S^{N-1}} \int_0^\infty |f'(s)|^p s^{N-1}\, ds d\omega \\ &= \omega_N \int_0^\infty |f'(s)|^p s^{N-1}\, ds \end{align*} where $\omega_N$ is the surface area of $\mathbf S^{N-1}$. From here the inequality is trivial.