Approximate simple function $f$ by a sequence of continuous functions on $\mathbb{R}^d$ such that $\|f_n\|_\infty\leq \|f\|_\infty$

Question

Let $f=\sum_{i=1}^n c_i 1_{\Delta_i}$ be a simple function on $\mathbb{R}^d$, where $c_i\in\mathbb{C}$. Then we can find sequnces of continuous functions $\{f_k^{(i)}\}$ for each $i=1,\ldots,n$ such that $f_k^{(i)}\to 1_{\Delta_i}$ pointwise. Then $\sum_{i=1}^n c_i f_k^{(i)}\to f$ pointwise. But here I can't make $$\Big\|\sum_{i=1}^n c_i f_k^{(i)}\Big\|\leq \|f\|_\infty.$$

Is this possible?

Answer 1

Another possibility is an approximation of the identity via convolution; by linearity it smoothes the $1_{\Delta_i}$, and of course $\|f*\phi_\epsilon\|_\infty\le \|f \|_\infty$, because the convolution at any point is an integral mean.

Answer 2

I will show that it is possible. Moreover, the functions $f_k^m$ I will construct will be independent of the numbers $c_l$ (I'm not sure if this is needed for you, but clearly this extra condition can only make our life harder).

For a subset $L\subset \{1, \ldots, n\}$, $L\neq \varnothing$ let $S_L$ denote the intersection of $\Delta_l$ over $l\in L$ and $\mathbb{R}^d \backslash \Delta_l$ over $l\notin L$. These sets are disjoint, have finite measure and any $\Delta_l$ is a finite union of some of them. Let us also instantly discard all $S_L$ with measure zero since they clearly will not impact us in any way and let's call the remaining sets $S_1, \ldots , S_N$.

For each $S_k$ there exists a sequence of compact sets $K_{k, m}$ such that $K_{k, 1}\subset K_{k, 2}\subset\ldots \subset S_k$ and the measure of $K_{k, m}$ tends to the measure of $S_k$ as $m$ tends to $\infty$. On the other hand, there are open sets $U_{k, m}$ such that $U_{k, 1}\supset U_{k, 2}\supset\ldots \supset S_k$ and the measure of $U_{k, m}$ tends to the measure of $S_k$ as $m$ tends to $\infty$.

Let us fix $m$ for now. Since $K_{k, m}\subset S_k$, these compacts are clearly disjoint. Thus, there exists $\varepsilon_m > 0$ such that they are a distance at least $\varepsilon_m$ from each other. By $V_{k, m}$ let us denote the intersection of $U_{k, m}$ with $\varepsilon_m/2$-neighborhood of $K_{k, m}$. Then $V_{k, m}$ are still open, they are now disjoint, they still contain $K_{k, m}$ and they are contained in $U_{k, m}$.

By the Urysohn's lemma there exists a continuous function $g_k^m$ such that it is equal to $1$ on $K_{k, m}$, and it is equal to $0$ outside of $V_{k, m}$. Moreover, we can assume that $0 \le g_k^m \le 1$ everywhere.

Finally, for a set $\Delta_l$ let $f_l^m$ to be the sum of $g_k^m$ over all $S_k$ which intersect $\Delta_l$ (that is, such that the corresponding set $L$ contains $l$). I claim that this works.

First of all, $g_k^m$ tend to $1_{S_k}$ almost everywhere (they tend to $1$ on the union of $K_{k, m}$ and tend to $0$ outside of the intersection of $U_{k, m}$). Hence, since $\Delta_l$ is the disjoint union of the corresponding sets, $f_l^m$ tends to $1_{\Delta_l}$ almost everywhere.

As for the norm estimate, we have $$\|f\|_{\infty} = \sup_{L\subset\{1, \ldots, n\}\,:\,|S_L|>0}\Big|\sum_{l\in L} c_l\Big|$$ since on each $S_L$ the function is equal to this exact constant. When we are computing $\|\sum c_l f_l^m\|$ by our construction they live on different $V_{k, m}$'s which are disjoint, and on each one of them we have $g_k^m$ (which is at most $1$ in absolute value, and does not depend on $l$, so there can be no cancellations) times the same sum of $c_l$'s, hence it is at most the norm.