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The last counterexample posted here led me to think a bit on the theme "What can be done then?". I came out with the following simple crank machine, that crumbles a positive $C^1_c$ function $f$ into a $C^1$ convergent series of squares, and that could also work in other ordered algebras (in fact I suspect it may be a known procedure). We may assume with no loss of generality $0\le f\le1/2$.

Consider the sequence of functions: $$\begin{cases} g_1=f\\ g_{n+1}=g_n-g_n^2 \end{cases}$$

It is immediate by the definition that $f=\sum_{k=1}^n g_k^2 +g_{n+1}$, so that we have $f=\sum_{k=1}^\infty g_k^2$ in various senses, provided that the remainder $g_{n}$ converges to zero in the same sense.

Uniform convergence of $g_n$: We have $0\le g_n\le1$ and ${1\over g_{n+1}}={1\over g_{n}}+{1\over 1-g_{n}}\ge {1\over g_{n}}+1,$ hence ${1\over g_{n}}\ge n$ and $g_n\le {1\over {n}}$.

Uniform convergence of $\partial_i g_n$: We have $\partial_i g_{n+1}=(1-2g_n)\partial_ig_n$. Since $0\le 1-2g_n\le1-g_n={g_{n+1}\over g_n}\le 1$ we also have $|\partial_i g_{n+1}|\le {g_{n+1}\over g_n}|\partial_ig_n|\le |\partial_ig_n|$. Iterating the latter inequalities we get

$$|\partial_i g_{n}|\le {g_{n}\over f}|\partial_if|\le|\partial_i f|$$ which immediately implies the uniform convergence of $\partial_i g_{n}$ to $0$ (indeed $\partial_i g_{n}$ converges uniformly to zero on any set $\{f\ge\epsilon\}$ and it is dominated by $|\partial_i f|$).

$$*$$

On the other hand, by David Speyer's example, we know that, in general, from third order on derivatives, can't converge uniformly to zero.

The last counterexample posted here led me to think a bit on the theme "What can be done then?". I came out with the following simple crank machine, that crumbles a positive $C^1_c$ function $f$ into a $C^1$ convergent series of squares, and that could also work in other ordered algebras (in fact I suspect it may be a known procedure). We may assume with no loss of generality $0\le f\le1/2$.

Consider the sequence of functions: $$\begin{cases} g_1=f\\ g_{n+1}=g_n-g_n^2 \end{cases}$$

It is immediate by the definition that $f=\sum_{k=1}^n g_k^2 +g_{n+1}$, so that we have $f=\sum_{k=1}^\infty g_k^2$ in various senses, provided that the remainder $g_{n}$ converges to zero in the same sense.

Uniform convergence of $g_n$: We have $0\le g_n\le1$ and ${1\over g_{n+1}}={1\over g_{n}}+{1\over 1-g_{n}}\ge {1\over g_{n}}+1,$ hence ${1\over g_{n}}\ge n$ and $g_n\le {1\over {n}}$.

Uniform convergence of $\partial_i g_n$: We have $\partial_i g_{n+1}=(1-2g_n)\partial_ig_n$. Since $0\le 1-2g_n\le1-g_n={g_{n+1}\over g_n}\le 1$ we also have $|\partial_i g_{n+1}|\le {g_{n+1}\over g_n}|\partial_ig_n|\le |\partial_ig_n|$. Iterating the latter inequalities we get

$$|\partial_i g_{n}|\le {g_{n}\over f}|\partial_if|\le|\partial_i f|$$ which immediately implies the uniform convergence of $\partial_i g_{n}$ to $0$ (indeed $\partial_i g_{n}$ converges uniformly to zero on any set $\{f\ge\epsilon\}$ and it is dominated by $|\partial_i f|$).

$$*$$

Rmk if for instance $f(x)=x^2+o(x^2)$ for $x\to0$ then also $g_n(x)=x^2+o(x^2)$ so that $g_n''(0)$ can't converge to zero, as illustrated in the last post.

On the other hand, by David Speyer's example, we know that, in general, for a smooth $f$, no series can't be convergent in $C^3$.

The last counterexample posted here led me to think a bit on the theme "What can be done then?". I came out with the following simple crank machine, that crumbles a positive $C^1_c$ function $f$ into a $C^1$ convergent series of squares, and that could also work in other ordered algebras (in fact I suspect it may be a known procedure). We may assume with no loss of generality $0\le f\le1/2$.

Consider the sequence of functions: $$\begin{cases} u_0=f\\ u_{n+1}=u_n-u_n^2 \end{cases}$$

It is immediate by the definition that $f=\sum_{k=0}^n u_k^2 +u_{n+1}$, so that we have $f=\sum_{k=0}^\infty u_k^2$ in various senses, provided that the remainder $u_{n}$ converges to zero in the same sense.

Uniform convergence of $u_n$: We have $0\le u_n\le1$ and ${1\over u_{n+1}}={1\over u_{n}}+{1\over 1-u_{n}}\ge {1\over u_{n}}+1,$ hence ${1\over u_{n}}\ge n$ and $0\le u_n\le {1\over {n}}$.

Uniform convergence of $\partial_i u_n$: We have $\partial_i u_{n+1}=(1-2u_n)\partial_iu_n$. From the equalities and inequalities above we find $${1\over u_{n}}={1\over f}+\sum_{k=0}^{n-1}{1\over 1-u_{k}}\le {1\over f}+{1\over 1-f}+{n-1\over1-{1\over n}}={1\over f(1-f) } +n\le 2\big({1\over f} +n\big),$$ whence $1-2u_n\le1-{f\over1 +nf} .$ Therefore $$|\partial_i u_{n}|\le|\partial_i f|\prod_{k=0}^{n-1}\Big(1-{f\over1 +kf}\Big)={|\partial_i f| \over1 +(n-1)f}$$ which immediately implies the uniform convergence of $\partial_i u_{n}$ to $0$ (indeed $\partial_i u_{n}$ converges uniformly to zero on any set $\{f\ge\epsilon\}$ and it is dominated by $|\partial_i f|$).

I didn't try the challenging exercise of proving the uniform convergence of second order derivatives. On the other hand, by David Speyer's example, we know that, in general, from third order on derivatives can't converge uniformly to zero.

The last counterexample posted here led me to think a bit on the theme "What can be done then?". I came out with the following simple crank machine, that crumbles a positive $C^1_c$ function $f$ into a $C^1$ convergent series of squares, and that could also work in other ordered algebras (in fact I suspect it may be a known procedure). We may assume with no loss of generality $0\le f\le1/2$.

Consider the sequence of functions: $$\begin{cases} g_1=f\\ g_{n+1}=g_n-g_n^2 \end{cases}$$

It is immediate by the definition that $f=\sum_{k=1}^n g_k^2 +g_{n+1}$, so that we have $f=\sum_{k=1}^\infty g_k^2$ in various senses, provided that the remainder $g_{n}$ converges to zero in the same sense.

Uniform convergence of $g_n$: We have $0\le g_n\le1$ and ${1\over g_{n+1}}={1\over g_{n}}+{1\over 1-g_{n}}\ge {1\over g_{n}}+1,$ hence ${1\over g_{n}}\ge n$ and $g_n\le {1\over {n}}$.

Uniform convergence of $\partial_i g_n$: We have $\partial_i g_{n+1}=(1-2g_n)\partial_ig_n$. Since $0\le 1-2g_n\le1-g_n={g_{n+1}\over g_n}\le 1$ we also have $|\partial_i g_{n+1}|\le {g_{n+1}\over g_n}|\partial_ig_n|\le |\partial_ig_n|$. Iterating the latter inequalities we get

$$|\partial_i g_{n}|\le {g_{n}\over f}|\partial_if|\le|\partial_i f|$$ which immediately implies the uniform convergence of $\partial_i g_{n}$ to $0$ (indeed $\partial_i g_{n}$ converges uniformly to zero on any set $\{f\ge\epsilon\}$ and it is dominated by $|\partial_i f|$).

$$*$$

On the other hand, by David Speyer's example, we know that, in general, from third order on derivatives, can't converge uniformly to zero.

The last counterexample posted here led me to think a bit on the theme "What can be done then?". I came out with the following simple crank machine, that crumbles a positive $C^1_c$ function $f$ into a $C^1$ convergent series of squares, and that could also work in other ordered algebras (in fact I suspect it may be a known procedure). We may assume with no loss of generality $0\le f\le1/2$.

Consider the sequence of functions: $$\begin{cases} u_0=f\\ u_{n+1}=u_n-u_n^2 \end{cases}$$

It is immediate by the definition that $f=\sum_{k=0}^n u_k^2 +u_{n+1}$, so that we have $f=\sum_{k=0}^\infty u_k^2$ in various senses, provided that the remainder $u_{n}$ converges to zero in the same sense.

Uniform convergence of $u_n$: We have $0\le u_n\le1$ and ${1\over u_{n+1}}={1\over u_{n}}+{1\over 1-u_{n}}\ge {1\over u_{n}}+1,$ hence ${1\over u_{n}}\ge n$ and $0\le u_n\le {1\over {n}}$.

Uniform convergence of $\partial_i u_n$: We have $\partial_i u_{n+1}=(1-2u_n)\partial_iu_n$. From the equalities and inequalities above we find $${1\over u_{n}}={1\over f}+\sum_{k=0}^{n-1}{1\over 1-u_{k}}\le {1\over f}+{1\over 1-f}+{n-1\over1-{1\over n}}={1\over f(1-f) } +n\le 2\big({1\over f} +n\big),$$ whence $1-2u_n\le1-{f\over1 +nf} .$ Therefore $$|\partial_i u_{n}|\le|\partial_i f|\prod_{k=0}^{n-1}\Big(1-{f\over1 +kf}\Big)={|\partial_i f| \over1 +(n-1)f}$$ which immediately implies the uniform convergence of $\partial_i u_{n}$ to $0$ (indeed $\partial_i u_{n}$ converges uniformly to zero on any set $\{f\ge\epsilon\}$ and it is dominated by $|\partial_i f|$).

I didn't try the challenging exercise of proving the uniform convergence of second order derivatives. On the other hand, by David Speyer's example, we know that from third order on, derivatives can't converge uniformly to zero.

The last counterexample posted here led me to think a bit on the theme "What can be done then?". I came out with the following simple crank machine, that crumbles a positive $C^1_c$ function $f$ into a $C^1$ convergent series of squares, and that could also work in other ordered algebras (in fact I suspect it may be a known procedure). We may assume with no loss of generality $0\le f\le1/2$.

Consider the sequence of functions: $$\begin{cases} u_0=f\\ u_{n+1}=u_n-u_n^2 \end{cases}$$

It is immediate by the definition that $f=\sum_{k=0}^n u_k^2 +u_{n+1}$, so that we have $f=\sum_{k=0}^\infty u_k^2$ in various senses, provided that the remainder $u_{n}$ converges to zero in the same sense.

Uniform convergence of $u_n$: We have $0\le u_n\le1$ and ${1\over u_{n+1}}={1\over u_{n}}+{1\over 1-u_{n}}\ge {1\over u_{n}}+1,$ hence ${1\over u_{n}}\ge n$ and $0\le u_n\le {1\over {n}}$.

Uniform convergence of $\partial_i u_n$: We have $\partial_i u_{n+1}=(1-2u_n)\partial_iu_n$. From the equalities and inequalities above we find $${1\over u_{n}}={1\over f}+\sum_{k=0}^{n-1}{1\over 1-u_{k}}\le {1\over f}+{1\over 1-f}+{n-1\over1-{1\over n}}={1\over f(1-f) } +n\le 2\big({1\over f} +n\big),$$ whence $1-2u_n\le1-{f\over1 +nf} .$ Therefore $$|\partial_i u_{n}|\le|\partial_i f|\prod_{k=0}^{n-1}\Big(1-{f\over1 +kf}\Big)={|\partial_i f| \over1 +(n-1)f}$$ which immediately implies the uniform convergence of $\partial_i u_{n}$ to $0$ (indeed $\partial_i u_{n}$ converges uniformly to zero on any set $\{f\ge\epsilon\}$ and it is dominated by $|\partial_i f|$).

I didn't try the challenging exercise of proving the uniform convergence of second order derivatives. On the other hand, by David Speyer's example, we know that, in general, from third order on derivatives can't converge uniformly to zero.