Do I have a Banach space given the following norm?

Question

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.

Answer 1

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.