Supremum of sequence of piecewise linear functions Suppose we have a sequence of continuous and piecewise linear functions $(f_n)_{n\in\mathbb{N}}$ mapping from $\mathbb{R}^d$ to $\mathbb{R}$. Assume this sequence of functions converge pointwise to some continuous function $f:\mathbb{R}^d \to \mathbb{R}$.
I want the show that there exists a continuous piecewise linear function $\bar f \neq f$ such that 
$$
\sup_{n\in\mathbb{N}} f_n(x) = \max \{f(x), \bar f(x)\}.
$$
Is the above true??? Is the below also true for some continuous piecewise linear function $\bar f \neq f$?
$$
|\sup_{n\in\mathbb{N}} f_n(x)| \leq \max \{|f(x)|, |\bar f(x)|\}.
$$
 A: Neither is true. For a function $g$ to be bounded above by a continuous function, it is necessary and sufficient that it is bounded above on compact intervals.
For a counterexample, consider the functions $f_n$ taking values $0,n,0$ on $0,1/n,2/n$ respectively and extending by linear interpolation to $\mathbb R\to\mathbb R.$ These converge pointwise to zero. But their pointwise supremum is unbounded on $[-1,1]$ so cannot be bounded by a continuous function.
If the function $g$ defined by $g(x)=\sup_nf_n(x),$ or $g(x)=|\sup_nf_n(x)|$ for the second question, is bounded above on compact intervals, and if you allow infinitely many domains in the definition of "piecewise", then it is easy to construct a piecewise upper bound. You can define
$$f_1(x)=\max(0,\sup_{r\in [-1,1]^d} g(x+r))\qquad\text{ for $x\in\mathbb Z^d$}$$
Let $\phi(x)=\max(0,2-\max(1,|x_1|,\dots,|x_d|)).$ Define $\overline{f}(x)=\sum_{y\in\mathbb Z^d}\phi(x-y)f_1(y).$ Note the only non-zero terms have $|y_i|<1,$ and since there is at least one integer point $y$ with $\max(|x_i-y_i|)\leq 1$ we get $\overline{f}(x)\geq f_1(y)\geq f(x).$
You do need infinitely many domains for $\overline{f}$ even in this case. Consider $f(x)=x^2,$ with $f_n$ linearly interpolating $f$ at multiples of $1/n$ in the range $[-n,n].$
