I do not understand some steps of the proof of the following theorem
Theorem (Moreau-Yosida regularization). Let $X$ be a metric space, $f:X\longrightarrow\mathbb{R}\cup\{+\infty\}$ be a function bounded from below (i.e. there exists $m\in\mathbb{R}$ s.t. $f(x)\geq m$, $\forall x\in X$). Then $f$ is lower semicontinuous (l.s.c.) iff there exists a sequence $\{f_k\}_{k\in\mathbb{N}}$ of bounded and continuous functions $f_k$ from $X$ to $\mathbb{R}$ such that
- $f_k$ is Lipschitz continuous with constant $k$ for all $k\in\mathbb{N}\setminus\{0\}$
- $\{f_k(x)\}_{k\in\mathbb{N}}$ is monotonically increasing $\forall x\in X$
- $\lim_{k\rightarrow\infty}f_k(x)=f(x)$
Proof. The sufficiency of the condition is trivial since $f$ is the pointwise supremum of continuous (hence l.s.c.) functions $f_k$.
Conversely, given $f$ l.s.c. and bounded from below, define $$ h_k(x)=\inf\{f(z)+k\operatorname{dist}(x, z)\}. $$ Since the functions $x\longmapsto f(z)+k\operatorname{dist}(x, z)$ are Lipschitz continuous with constant $k$ for all $x\in X$, then $h_k$ is Lipschitz continuous with constant $k$. Moreover the sequence $\{h_k(x)\}_{k\in\mathbb{N}}$ is monotonically increasing for every fixed $x\in X$ and we have $\inf f\leq h_k(x)\leq f(x)$ (choosing $z=x$, right?). Now, assume by contradiction that $l(x):=\lim_{k\rightarrow\infty}h_k(x)<f(x)$. Then $l(x)<+\infty$. For every $k\in\mathbb{N}\setminus\{0\}$, by definition of $h_k$, we can choose $z_k\in X$ such that $$ h_k(z_k)+\frac{1}{k}\geq f(z_k)+k\operatorname{dist}(x, z_k), $$ from which $$ \operatorname{dist}(x, z_k)\leq\frac{h_k(z_k)+\frac{1}{k}-f(x_k)}{k}\leq\frac{l(x)+1-m}{k} $$ (here I do not understand why $h_k(z_k)+\frac{1}{k}\leq l(x)+1$).
Now, letting $k\rightarrow\infty$, we get that $z_k\rightarrow x$ and $$ l=\lim_{k\rightarrow\infty}h_k(x)\geq\liminf_{k\rightarrow\infty}f(z_k)\geq l(x) $$ where the last inequality follows from the lower semicontinuity of $f$. So we get a contradiction. Finally, setting $f_k(x)=\min\{h_k(x), k\}$ we get the boundedness property. $\square$
My question. I do not understand how we get the boundedness property. If $\min\{h_k(x), k\}=h_k(x)$, we get properties $1.$, $2.$ and $3.$ and the continuity of the $f_k$ (by point $1.$, right?) but $h_k$, in my opinion, can be unbounded. On the other hand, if $\min\{h_k(x), k\}=k$, how can I get point $3.$ and the fact that $f_k$ is bounded?
Thank You
$m \leq f_k(x) \leq k$ so each $f_k$ is a bounded function. (It is not asserted that the functions are uniformly bounded).