Here is my problem.
Suppose $f:\mathbb{R}^n\rightarrow \mathbb{R}\cup\{+\infty\}$ is a positive function. Define $f_n(x)=\inf \{f(y)+nd(x,y):y\in \mathbb{R}^n\}$. Prove that $f_n(x)$ is Lipschitz continuous and $f_n(x)\rightarrow f(x)$.
I find $f_n(x)$ are increasing, $f_n(x)\leq f(x)$ and for each $n$, $f_n(x)$ is upper semi-continuous without using the condition $f\geq 0$. I wonder how this condition make sense. Thanks~
The condition $f \ge 0$ guarantees that $f_n(x)$ is finite for all $x$, and it would suffice to require that $f$ is bounded below.
The example $f(x) = - \Vert x \Vert^2$ shows that without a lower bound, $f_n(x)$ can be identically $-\infty$.
The Lipschitz continuity follows from the triangle inequality: For fixed $x_1, x_2 \in \Bbb R$ and all $y \in \Bbb R$ we have $$ f(y) + nd(x_1, y) \le f(y) + nd(x_2, y) + nd(x_1, x_2) $$ which implies $f_n(x_1) \le f_n(x_2) + nd(x_1, x_2)$. Now exchange $x_1$ and $x_2$ and conclude that $$ |f_n(x_1) - f_n(x_2) | \le n d(x_1, x_2) \, . $$
$f_n(x) \to f(x)$ holds if $f$ is lower semi-continuous at $x$, see for example Lower semicontinuous function as the limit of an increasing sequence of continuous functions.
Without lower semi-continuity it can be wrong. A counter example would the function defined by $f(0) = 1$ and $f(x) = 0$ otherwise, where $0 = f_n(0) \not \to f(0) = 1$.
