Equip $W_n^{p}[0,1]$ with the norm
$$\left\|f\right\|_{W_n^{p}} = \sum_{k=0}^{n}\left\|f^{(k)}\right\|_{L^p}.$$
I want to prove that this norm is equivalent to the norm $$\left\|f\right\|_2 = \left\|f\right\|_{L^p}+\left\|f^{(n)}\right\|_{L^p}.$$
It is obvious that $\left\|f\right\|_2\leq \left\|f\right\|_{W_n^p}$ so I want to find a constant $C>0$ such that $\left\|f\right\|_{W_n^p}\leq C\left\|f\right\|_2$. I've managed to solve this in the case $n=2$ but I don't know how to generalize this to the case $n>2$.
Solution in the case $n=2$: Let $f\in W^{2}_p[0,1]$. Let $\xi_{\mathrm{min}}$ and $\xi_{\mathrm{min}}'$ be points which minimizes $|f|$ and $|f'|$ respectively. Note that these functions are continuous so these points exists. Now Hölder's inequality implies that \begin{align*} |f'(x)|& = \left|f'(\xi_{\mathrm{min}}')+\int_{\xi_{\mathrm{min}}'}^{x}f''(t)\, dt\right| \\ & \leq |f'(\xi_{\mathrm{min}}')|+\int_{0}^{1}|f''(t)|\, dt \\ & \leq |f'(\xi_{\mathrm{min}}')|+\left(\int_{0}^{1}|f''(t)|^{p}\,dt\right)^{1/p} \end{align*} and by the mean-value theorem \begin{align*} 2|f(x)|\geq |f(x)-f(\xi_{\mathrm{min}})|=|f'(\xi)||x-\xi_{\mathrm{min}}|\geq |f'(\xi_{\mathrm{min}})||x-\xi_{\mathrm{min}}| \end{align*} for some $\xi$ between $x$ and $\xi_{\mathrm{min}}$. Integrating this inequality yields \begin{equation*} \frac{|f'(\xi_{\mathrm{min}}')|}{2} = \int_{0}^{1}|f'(\xi_{\mathrm{min}}')||x-\xi_{\mathrm{min}}|\, dx\leq 2\int_{0}^{1}|f(x)|\, dx \leq 2\|f\|_{L^p} \end{equation*} why \begin{equation*} |f'(x)|\leq 4\|f\|_{L^p}+\|f''\|_{L^p} \Rightarrow \|f'\|_{L^p}\leq 4\|f\|_{L^p}+\|f''\|_{L^p} \leq 4\|f\|_2 \end{equation*}
Now I don't see a way to generalise this method to the case $n>2$ using induction. For instance if we consider the case $n=3$ then applying the method to $f'$ we obtain $$\|f'\|_{L^p} \leq 4\|f\|_{L^p}+\|f''\|_{L^p}$$ but when we do the same estimate on $f''$ we again get a term containing $f'$ so this reasoning becomes circular.
This question has been previously asked in Sobolev space and equivalence of norms. The answer there however seems to be incomplete as it assumes that $f^{(k)}(0) = 0$ for every $k$ and the above calculation in the case that $n=2$ can be seen as a way to get around this fact using calculus.