Determine the completeness of $C^2[0,1]$ with respect to a non-standart norm Is $C^2[0,1]$ complete with respect to the norm $$\left\lVert{f}\right\rVert_A=\max_{x\in[0,1]}\left\lvert{f''(x)}\right\rvert+(\left\lvert{f(0)}\right\rvert^2+\left\lvert{f(1)}\right\rvert^2)^\frac{1}{2}?$$
It is evident, that $$\left\lVert{f}\right\rVert_A\le\sqrt{2}\left\lVert{f}\right\rVert_B,$$where $$\left\lVert{f}\right\rVert_B=\max_{x\in[0,1]}\left\lvert{f(x)}\right\rvert+\max_{x\in[0,1]}\left\lvert{f'(x)}\right\rvert+\max_{x\in[0,1]}\left\lvert{f''(x)}\right\rvert$$is standart norm in $C^2[0,1]$ and with respect to this norm space is complete. But is it enough to prove that $C^2[0,1]$ is complete with respect to $\left\lVert{.}\right\rVert_A$?
 A: If you have two norms $\|\cdot\|_A$ and $\|\cdot \|_B$ in a space $X$, such that $\|x\|_A\leq C\|x\|_B$ for all $x\in X$ , and the space $(X,\|\cdot\|_B)$ is complete, then it is not necessary that $(X,\|\cdot\|_A)$ is complete. One example is given by $$X=C[0,1],\quad \|f\|_B=\|f\|_{\infty},\quad\|f\|_A=\|f\|_2.$$ In your case, you can actually show that the two norms $\|\cdot \|_A$ and $\|\cdot\|_B$ are equivalent, and this will imply that $(C^2[0,1],\|\cdot\|_A)$ is complete. To do this, let $f\in C^2[0,1]$. Then there exists $c\in(0,1)$ such that $$f'(c)=f(1)-f(0).$$ Therefore, for all $x\in [c,1]$, $$\begin{align*}
|f'(x)|&\leq|f'(x)-f'(c)|+|f'(c)|\\
&\leq \int_c^x|f''(t)|\,dt+|f'(c)|\\
&\leq\max_{x\in[0,1]}|f''(x)|+|f(1)|+|f(0)|\\
&\leq \max_{x\in[0,1]}|f''(x)|+2(|f(1)|^2+|f(0)|^2)^{1/2}\\
&\leq 2\|f\|_A,
\end{align*}$$
and similarly, for $x\in[0,c]$, you obtain the same inequality. So, $$\max_{x\in[0,1]}|f'(x)|\leq 2\|f\|_A.$$ You also obtain a similar inequality for the maximum of $|f|$, which shows that $$\|f\|_B\leq c\|f\|_A$$ for all $f\in C^2[0,1]$. Therefore, the two norms $\|\cdot \|_A$ and $\|\cdot\|_B$ are equivalent.
