In Michael Taylor's PDE - Nonlinear equations, there are two estimates (3.28) & (3.29), that he says can be easily seen. I have tried Fourier transform and fundamental theorem in calculus to express the $L^{\infty}$ norm, but I still cannot efficiently relate it to the $L^2$ norm in the inequalities. I will appreciate if anyone can offer a proof of them. The inequalities are listed below.
$\Vert{u}\Vert^2_{L^\infty}\le C\Vert D^{m+1}u\Vert^2_{L^2} + C\Vert D^{m-1}u\Vert^2_{L^2}$ , for $u\in C_0^{\infty}(\mathbb{R}^{2m})$,
and
$\Vert{u}\Vert^2_{L^\infty}\le C\Vert D^{m+1}u\Vert^2_{L^2} + C\Vert D^{m}u\Vert^2_{L^2}$ , for $u\in C_0^{\infty}(\mathbb{R}^{2m+1})$,
within which $\Vert D^ku\Vert_{L^p} = \sum_{|\alpha|=k}\Vert D^{\alpha}u\Vert_{L^p}$
My trying using Fourier transform:
\begin{align} u(x) &= C\int_{\mathbb{R}^{2m}}\ e^{ix\xi}\ \hat{u}(\xi)d\xi \\ &= C\int_{\mathbb{R}^{2m}}\ e^{ix\xi}\frac{1}{(1+|\xi|^2)^{(m+1)/2}}(1+|\xi|^2)^{(m+1)/2}\ \hat{u}(\xi)d\xi \\ \implies |u(x)| &\le C\int_{\mathbb{R}^{2m}}\ |\frac{1}{(1+|\xi|^2)^{(m+1)/2}}|\cdot|(1+|\xi|^2)^{(m+1)/2}\ \hat{u}(\xi)|d\xi\\ &\le C\ [\int_{\mathbb{R}^{2m}}\ \frac{1}{(1+|\xi|^2)^{m+1}}d\xi]^{1/2}\cdot [\int_{\mathbb{R}^{2m}}\ (1+|\xi|^2)^{m+1}\ |\hat{u}(\xi)|^2d\xi]^{1/2} \\ &= C\ [\int_{\mathbb{R}^{2m}}\ (1+|\xi|^2)^{m+1}\ |\hat{u}(\xi)|^2d\xi]^{1/2}. \end{align} I do know that there is an estimate of this form $$c_{1,k}(1+|\xi|^2)^k\le\sum_{|\alpha|\le k}\xi^{2\alpha}\le c_{2,k}(1+|\xi|^2)^k$$ which I think I can use to have the following estimate \begin{align} |u(x)| &\le C\ [\int_{\mathbb{R}^{2m}}\ (1+|\xi|^2)^{m+1}\ |\hat{u}(\xi)|^2d\xi]^{1/2} \\ &\le C\ [\sum_{|\alpha|\le m+1}\int_{\mathbb{R}^{2m}}\ \xi^{2\alpha}\ |\hat{u}(\xi)|^2d\xi]^{1/2} \end{align} That's where I was stuck. If I am summing over $|\alpha|\le m+1$, then it will lead me to the $H^{m+1}$ norm, but not just $\Vert D^{m+1}u\Vert_{L^2}$. I think the tricky point is that I cannot avoid the constant "1" or any positive constant $\epsilon$, or else the integral $[\int_{\mathbb{R}^{2m}}\ \frac{1}{(1+|\xi|^2)^{m+1}}\ d\xi]^{1/2}$ won't be finite.
My trying of fundamental theorem of calculus seems not to be applicable.