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Let $ ~~\cup_{k=-1}^{\infty} U_k = \mathbb{R} $ be an open covering of $\mathbb{R}$. It is a well known fact that partitions of unity subbordinate to the cover exists, i.e. there exists smooth functions $ \varphi_{k} : U_k \rightarrow \mathbb{R} $ with compact support such that $$ \sum_{k=-1}^{\infty} \varphi_k(x)^2 \equiv 1.$$

My first question is vague: Do there exist partitions of unity subbordinate to the cover if we impose some additional conditions on the derivatives (and the nature of the conditions are in terms of equalities, not inequalities.) ?

A more precise question is as follows: Given a smooth function $f: \mathbb{R} \rightarrow \mathbb{R}$, do there exist functions
$ \varphi_{k} : U_k \rightarrow \mathbb{R} $ with compact support such that in addition to being a partition of unity subbordinate to the cover, it also satisfies $$ \sum_{k=-1}^{\infty} \varphi_{k}^{\prime}(x)^2 \equiv f(x)$$ ? Here prime denotes derivative with respect to $x$. I assume the answer should depend on what $f(x)$ is. Ideally $\varphi_k$ should be smooth functions, but at the very least they ought to be $C^1$.

I would prefer if we do not assume $f$ is nowhere vanishing, but if there is an answer assuming that, I would still like to see it.

Remark: The answer probably also depends on the open covering.

$\textbf{Very specific question:}$ Choose some number $\tau \in (\sqrt{2}, 2) $. Say $\tau = 1.5$. Now define $$ U_k = \{ x \in \mathbb{R}: \frac{2^k}{\tau} < |x| < 2^k \tau \} \qquad k=0,1,2, \ldots $$

$$ U_{-1} = (-1,1).$$

The collection $\{U_k\}_{k=-1}^{\infty} $ is an open covering of $\mathbb{R}$. I want smooth functions $\varphi_k: U_k \rightarrow \mathbb{R}$ and numbers $n_k$ such that $$ \sum_{k= -1}^{\infty} \varphi_k(x)^2 \equiv 1 $$ and $$ \sum_{k=-1}^{\infty} n_k^2 (\varphi_k(x)^2 + \varphi_k^{\prime}(x)^2) \equiv e^x$$

It is easy to see that if I set $n_k =1$ and $f(x) = e^x-1$, then it is the previous question I had asked. This question is "easier", because one is allowed to choose the numbers $n_k$ (in other words there is a better chance that the answer here might be yes, because of the freedom in choosing $n_k$).

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  • $\begingroup$ @Pietro I am very sorry, I made a mistake. I have corrected the question now. Now if you derive the equation you get $\sum_{k} 2 \varphi_k(x) \varphi^{\prime}(x) \equiv 0.$ $\endgroup$
    – Ritwik
    Commented Sep 5, 2013 at 11:28
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    $\begingroup$ Yet I think some assumption on the cover is in order. What if all the $U_k$ but one are empty, or more generally, what if all the $U_k$ but one are disjoint from the interval $[0,1]$? $\endgroup$
    – Pietro Majer
    Commented Sep 5, 2013 at 11:30
  • $\begingroup$ I agree, the answer probably depends on the open covering. In my specific case, the second condition you have suggested is actually true, i.e. all but one of the $U_k$ is disjoint from $[0,1]$. In fact any point in $\mathbb{R}$ belongs to at most two $U_k$ in the specif case I have. $\endgroup$
    – Ritwik
    Commented Sep 5, 2013 at 11:45
  • $\begingroup$ It is pretty hard for a function supported on a short interval and having not too large derivative to rise to $1$ and, as you said itself, in your case you have only two shots at each point. Since I see no point in obtaining the full description to cover just one particular case, I suggest you just tell what exactly your "specific case" is. $\endgroup$
    – fedja
    Commented Sep 5, 2013 at 12:03
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    $\begingroup$ @Ritwik As posed, the answer is trivially "No" because on $(-0.1,0.1)$, the function $\varphi=\varphi_{-1}$ should satisfy $\varphi^2=1$ and $n(\varphi^2+(\varphi')^2)=e^x$ simultaneously. However it makes more or less clear what you might really be after. It beats me how you could hope to get a sum of squares equal to $e^x-1$ for $x<0$ though ;). $\endgroup$
    – fedja
    Commented Sep 5, 2013 at 23:43

1 Answer 1

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I hope the following construction will give you what you really need. If not, you'll have to explain why.

Take any nice locally finite covering $\mathbb R\subset\cup_j U_j$ and take any smooth partition of unity $1=\sum_j\psi_j^2$ subordinated to this covering. Take any smooth positive function $F$ on $\mathbb R$. Choose the numbers $n_j>0$ so that $\sum_j n_j^2(\psi_j^2+(\psi_j')^2)\le F/2$. Define the smooth function $\theta$ by $$ (\theta^{\,\prime})^2\sum_j n_j^2\psi_j^2=F-\sum_j n_j^2(\psi_j^2+(\psi_j')^2) $$
Finally, associate with each $U_j$ two functions $\varphi_{j,0}=\psi_j\cos\theta$ and $\varphi_{j,1}=\psi_j\sin\theta$ and enjoy the identities $$ \sum_{j,k}\varphi_{j,k}^2=1, \quad \sum_{j,k}n_j^2(\varphi_{j,k}^2+(\varphi_{j,k}')^2)=F\,. $$

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  • $\begingroup$ When you are defining $\theta^{\prime}$, did you intend to put the $\psi_j^2$ on the rhs? If the $\psi_j^2$ is not there, then every thing works out as you have said, unless I am making some calculation mistake. $\endgroup$
    – Ritwik
    Commented Sep 6, 2013 at 12:20
  • $\begingroup$ @Ritwik Let's see: $\varphi_0^2+\varphi_1^2=\psi^2$, $(\varphi_0')^2+(\varphi_1'^2)=\psi'^2+\psi^2(\theta')^2$, so, yes, $\psi_j^2$ should be there (unless the condition is that the pure sum of squared derivatives with coefficients $n_j^2$ is prescribed instead, in which case it shouldn't). $\endgroup$
    – fedja
    Commented Sep 6, 2013 at 12:30
  • $\begingroup$ I am sorry, you were absolutely right; there is no mistake in what you said. I was making a simple error. $\endgroup$
    – Ritwik
    Commented Sep 6, 2013 at 13:13
  • $\begingroup$ In any case, this is a great answer. This is exactly what I was after. $\endgroup$
    – Ritwik
    Commented Sep 6, 2013 at 13:14
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    $\begingroup$ @Ritwik Glad to hear it :-). The moral is: ask exactly what you need and let other people generalize or consider partial cases. Unless you can solve the problem yourself, you never know what the first step or the first reduction should be. $\endgroup$
    – fedja
    Commented Sep 6, 2013 at 13:43

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