# 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 $\bar f \neq f$? $$|\sup_{n\in\mathbb{N}} f_n(x)| \leq \max \{|f(x)|, |\bar f(x)|\}.$$

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 $[-n,n].$

• Thanks. So if I replace pointwise convergence with uniform convergence, then the first equality should hold right? What about with compact convergence (uniform convergence on all compacts sets)? Oct 25, 2017 at 8:57
• @master_goon: if you have uniform convergence then $\|f_1-f\|_\infty\leq C$ for some $C$, which means $|\sup f_n(x)|$ is bounded by $|f_1(x)|+C,$ which is already piecewise linear (it might have infinite boundary due to the absolute value, though this can probably be avoided with a bit of care). The same argument applies for the restriction to a compact set.
– Dap
Oct 25, 2017 at 14:22
• Thanks again. I'm not familiar with your linear extension i.e. $\bar f(x + \theta)$ and I'm struggling to see its intuition. Can you explain this to me or point me to a reference, please? Oct 26, 2017 at 0:40
• @master_goon: sorry, that wasn't actually linear (it was the "Lovasz extension"). It should be pretty intuitive that you can get SOME piecewise linear function as long as the function is locally bounded. I've added another suggested construction but basically anything linear should work.
– Dap
Oct 26, 2017 at 8:53