How to prove that $H_p$ norm $(5.89)$ is equivalent to the commonly used norm $\|f\|_{H^p}=\|f\|_{L^{2}}+\|f'\|_{L^{2}}\ldots +\|f^{(p)}\|_{L^{2}}$?
The snapshot is from Page $170$ in "Nisio M. Stochastic control theory [J]. ISI Lecture Notes, $2015$, $9$"
In that definition, $p$ may be any real number, so it is meant to generalize your definition when $p$ is an integer. I'm not sure what $(1 - \Delta)^{p/2}$ means, but by Fourier transformation, $\mathcal{F}((1 - \Delta)\phi) = (1 + |\xi|^2)\hat{\phi}$, and I will conjecture that $$\mathcal{F}((1 - \Delta)^{p/2}\phi) := (1 + |\xi|^2)^{p/2}\hat{\phi} = \langle \xi \rangle^{p}\hat{\phi},$$ where $$\langle \xi \rangle := (1 + |\xi|^2)^{1/2}.$$ Thus since the Fourier transform is unitary on $L^2$, $$(\phi, \psi)_{H^p} = (\langle \xi \rangle^p\hat{\phi},\langle \xi \rangle^p \hat{\psi})_{L^2}.$$ Hence $$\|\phi\|_{H^p} = \|\langle \xi \rangle^p\hat{\phi}\|_{L^2}.$$ In contrast, your familiar norm on $H^p$ when $p = k$ is a positive integer is $$\|\phi\|=\sum_{|\alpha| \leq k}\|\partial^{\alpha}\phi\|_{L^2}.$$ Since the Fourier transform is unitary on $L^2$ and $\mathcal{F}(\partial^{\alpha}\phi) = (i\xi)^{\alpha}\hat{\phi}$, $$\|\phi\| = \sum_{|\alpha| \leq k}\|\partial^{\alpha}\phi\|_{L^2} = \sum_{|\alpha| \leq k}\|\xi^{\alpha}\hat{\phi}\|_{L^2}.$$ Now all that's left is to do simple estimation to obtain $\|\phi\|_{H^k} \leq C\|\phi\|$ and $\|\phi\| \leq C\|\phi\|_{H^k}$. The key inequalities here are $|\xi^{\alpha}| \leq \langle \xi \rangle^{|\alpha|}$ and $$\langle \xi \rangle^k = (1 + \xi_1^2 + \dots + \xi_n^2)^{k/2} \leq (1 + |\xi_1| + \dots + |\xi_n|)^k \leq C\sum_{|\alpha| \leq k}|\xi^{\alpha}|.$$