We have function $f:(\mathbb{R^2}, d_{x}) \rightarrow (\mathbb{R^2}, d_{y})$ ($d_{x}$ - maximum metric space, $d_{y}$ - British rail metric space). $f(x,y) = (\frac{x+y}{2}, \frac{y-x}{2})$.
I need to check if $f$ and $f^{-1}$ are continuous.
$\mathit{My\, work}:$ for example, I took point $(2,2)$, and $f(2,2) = (2,0)$. Here I used definition: $\forall \varepsilon > 0 \, \exists \delta >0: \, \max{\{|x-2|,|y-2|\}} < \delta, \, d_{y}(f(x,y), (2,0)) <\varepsilon$. Let it be, $x=y$ and $\varepsilon = 1$, so $|x-2| < \delta$. Let $\delta + 2 = x$. So, $d_{y}(f(\delta+2, \delta + 2), (2,0)) < \varepsilon \Rightarrow \sqrt{(\frac{4 + \delta}{2} -2)^2 +(\frac{2 + \delta}{2})^2} = \sqrt{\frac{2\delta^2 + 4\delta + 4}{4}} > \frac{\delta + 2}{2} > 1$. So, $f$ couldnt be continuous. Is it right solution?
And here for $f^{-1}$: $\left\{\begin{array}{l}a = \frac{x+y}{2}\\b= \frac{y-x}{2}\end{array}\right. \Rightarrow \left\{\begin{array}{l} y= b+ a\\x = a-b\end{array}\right. \Rightarrow f^{-1} =(a-b, b+a)$ - is continuous?? I dont know how to show it.
In $\mathbb{R}^2$ in the $d_y$ metric, all singleton sets except $\{0,0)\}$ are open, and $(0,0)$ has essentially the same neighbourhoods as in the usual metric on the plane. The max metric $d_x$ is just equivalent to the usual Euclidean metric.
$f$ is a bijection, sending $(0,0)$ to $(0,0)$ and as $f$ is usual-usual continuous, it is usual-$d_y$ continuous at $(0,0)$ as well. It is not continuous at any other point, because the usual topology (or the max metric) has no isolated points at all. The inverse function $f^{-1}$ is continuous at all points unequal to $(0,0)$ because those points are isolated in $d_y$, and similar reasoning as above will give that $f^{-1}$ is continuous at $(0,0)$ as well.
So $f^{-1}$ is everywhere continuous and $f$ only at the origin.