This is very very similar to my other question asked three months ago. That time there was no Banach space because an integral in the norm definition allowed a counter-example.
Once again I have a normed space (I'll denote it $C^2[a;b]$) which consists of continuous real functions whose first and second derivatives are also continuous in interval $[a;b]$.
For all $x,y \in C^2[a;b]$ and for all $\lambda \in \mathbb{R}$ the norm is defined as follows
$$ \|x(t)\| = |x(a)| + |x(b)| + \max_{t \in [a;b]} | x''(t) |;$$
It is in fact a valid norm following the definition.
But is it a Banach space? I find it intuitively hard to believe because the first derivative is not taken into consideration but I'm still struggling with formal arguments.
It is a Banach space.
Consider a Cauchy sequence $x_n$. Clearly $x_n(a)$ and $x_n(b)$ converge, moreover $x_n''$ converges uniformly. Does $x_n'$ converge? Obviously $x_n'(t)-x_n'(a)$ converges uniformly since it is just an integral. But from the equality
$$x_n(b)-x_n(a)=\int_a^b x_n'(t)dt = \int_a^b x_n'(t) - x_n'(a) + x_n'(a) dt = \\ = \int_a^b x_n'(t)-x_n'(a)dt + \int_a^b x_n'(a)dt = \int_a^b x_n'(t)-x_n'(a)dt + (b-a)x_n'(a)$$
we easily conclude that $x_n'(a)$ converges and hence $x_n'(t)$. It follows that $x_n$ also converges.
EDIT: Regarding the definition: as we have uniform convergence of the second derivative, we are almost done - we know that $x'$ converges up to a constant $c_1$ and $x$ converges up to an affine function (constant $c_2$ + a linear term coming from integrating $c_1$). We just need some additional information to control these two constants. Instead of using $x(a)$ and $x(b)$ it is possible to use $x(a)$ (to control $c_2$) and $x'(a)$ (to control $c_1$) or even $x(d)$ and $x'(e)$ for any $d,e\in [a,b]$.
Note also that with the current definition we don't have to stick to $a,b$ - we can use any $f\neq g\in [a,b]$ instead.