Let $L : \mathbb{R}^d \rightarrow \mathbb{R}^d$ be a linear operator and ${\| \cdot \|}_{\infty} $ be the maximum norm. Are the following conditions equivalent?
- For $x \in \mathbb{R}^d$ with $Lx = f$ it holds ${\| x \|}_{\infty} \leq C {\| f \|}_{\infty}$.
- For $x, y \in \mathbb{R}^d$, with $Lx = f_1, Ly = f_2$ it holds ${\| x - y \|}_{\infty} \leq C {\| f_1 - f_2 \|}_{\infty}$.
Clearly taking $y = 0$ in the second statement yields $f_2 = 0$, which leads to the first statement. Is the converse true? If not, how can the second condition be modified so that the two are equivalent? I have considered the following:
\begin{align} \| x - y \| = \| x - y + Lx - Lx +Ly - Ly\| &\leq \| Lx - x\| + \| Ly -y\| + \| Lx - Ly \| \\ &= | Lx - x\| + \| Ly -y\| + \| f_1 - f_2 \| \end{align}
(1) $\Longrightarrow$ (2):
Set
$z = x - y; \tag 1$
set
$g = f_1 - f_2; \tag 2$
then
$Lz = L(x - y) = Lx - Ly = f_1 - f_2 = g; \tag 3$
thus, via our OP Holden's stipulated item (1),
$\Vert z \Vert_\infty \le C \Vert g \Vert_\infty; \tag 4$
therefore,
$\Vert x - y \Vert_\infty = \Vert z \Vert_\infty \le C \Vert g \Vert_\infty = C\Vert f_1 - f_2 \Vert_\infty, \tag 5$
which is Holden's item (2).
(2) $\Longrightarrow$ (1):
As Holden him/herself points out, taking
$y = 0, \; f_1 = f; \tag 6$
yields
$f_2 = Ly = L0 = 0, \tag 7$
whence
$\Vert x \Vert_\infty = \Vert x - y \Vert_\infty \le C \Vert f_1 - f_2 \Vert_\infty = C \Vert f - 0 \Vert_\infty = C \Vert f \Vert_\infty, \tag 8$
which is Holden's item (1).
Thus the two stipulated hypotheses are, in fact, equivalent.