$\mathbf {The \ Problem \ is}:$ Let $f:\mathbb{R^2}\to \mathbb{R}$ be defined by $(x,y)\mapsto \frac{1}{\sqrt{2}}(x+y).$ Find a sequence $(p_i)_{i\in \mathbb {N}}$ of points in $\mathbb{R}^2$ so that $f(.)=\operatorname{lim}_{i\to \infty} (d(.,p_i) - d(0,p_i)).$
Here $d$ is the distance function in $\mathbb{R}^2$ with respect to the standard metric and convergence equals convergence on compact sets .
$\mathbf {My \ approach}:$ Here $f$ is indeed a distance function because $||\nabla{f}||=1$ identically on $\mathbb{R}^2.$ But I can't exactly approach to find a sequence to satisfy this property . I was thinking to define $g: \mathbb{R}^2\to \mathbb{R}$ by $g(a)=d(a,p_i)$ and then the right hand side of the question looks like $\operatorname{lim}_{i\to \infty} \operatorname{\int}_0^l D(g\circ \gamma)(r) dr $ on a compact set where $\gamma:[0,l]\to \mathbb{R}^2$ is a smooth curve with $\gamma(0)=0$ and $\gamma(l)=a.$ But I still don't know how to find the sequence $(p_i)_{i\in \mathbb{N}}.$ Thanks in advance for any help .
Let $p_i = (-i,-i)$. For any $a=(x,y) \in \mathbb R^2$, if you plot the triangle with vertices $0,a,p_i$ for large $i$ you will see that the line segment $\overline{0 p_i}$ has slope $+1$ and that the line segment $\overline{a p_i}$ has slope closer and closer to $1$ as $i \to \infty$. From this one can intuit that the limit of $d(a,p_i) - d(0,p_i)$ equals the desired quantity, which (if $x+y>0$) is equal to the length of the slope $+1$ line segment that connects the point $a$ to the slope $-1$ line given by the equation $x+y=0$, i.e. the line $f(x,y)=0$.
To prove that it holds, \begin{align*} d(a,p_i) &= \sqrt{(x+i)^2 + (y+i)^2} \\ &= \sqrt{x^2+y^2 + 2(x+y)i + 2i^2} \\ d(0,p_i) &= \sqrt{2} \, i \\ d(a,p_i) - d(0,p_i) &= \sqrt{x^2+y^2 + 2(x+y)i + 2i^2} - \sqrt{2} \, i \\ &= \frac{\sqrt{x^2+y^2 + 2(x+y)i + 2i^2} - \sqrt{2} \, i }{1} \end{align*} and now you can finish computing the limit by first rationalizing the numerator, and then multiplying the resulting numerator and denominator $\frac{1}{i}$.
Once one has that limit calculation in hand, one can easily use it to observe that the convergence is uniform as $a=(x,y)$ varies over any compact subset of $\mathbb R^2$.