Let $f:[0;1]\rightarrow\mathbb{R}$ a continuous function. Let $Z$ denote the set of zeros of $f$.

If $Z$ is finite, it's easy to prove that $\lim\limits_{n\rightarrow+\infty}\left|\displaystyle\int_{0}^1\text{e}^{nx}f(x)\,\text{d}x\right|=+\infty$.

But is it also true if $f\ne0$ and $Z$ is infinite ?

  • $\begingroup$ I assume you exclude the case $\forall x,\, f(x)=0$ because in that the integral and the limit are zero $\endgroup$ – marwalix May 8 '19 at 15:28
  • $\begingroup$ @marwalix You can assume it. In fact, the OP says that (s)he assumes it. $\endgroup$ – ajotatxe May 8 '19 at 15:29
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    $\begingroup$ I guess the split is between $Z$ of (Lebesgue) measure zero, finite or infinite $\endgroup$ – marwalix May 8 '19 at 15:31
  • $\begingroup$ The difficulty comes when $f(1)=0$ (and hence, $1$ can be the limit of a subsequence of zeros). The maxima near $1$ should decrease at exponential speed. What about $f(x)=e^{1/(1-x)}\sin(x-1)$? $\endgroup$ – ajotatxe May 8 '19 at 15:48
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    $\begingroup$ That is false ; if $0$ is an accumulation point of $Z$ and $Z^c$, then we can construct a function $f$ with $Z= \{f=0\}$ and where these integral have very wild behaviour (I think we can have $\overline{\mathbb{R}}$ as the accumulation points of this sequence). The construction I see is a bit annoying (basically, construct recursively a subsequence $n_k$ going to infinity and the values of $f$ near $0$ such that the integral evaluated at $n=n_k$ take approximately whatever value you want). $\endgroup$ – D. Thomine May 10 '19 at 19:51

This starts with a description of an answer, and then drills down to details. This question is more interesting than many.

The original question's answer seems to be "no".

One can construct, inductively, a sequence of intervals $I_k=[a_k,a_{k+1}]$ (with $a_k$ increasing to $1$), and values for $f$ on $I_k$, and a sequence of $n_k\to\infty$, so that the value of $\int_0^1 \exp(n_kx)f(x)dx$ is pretty much determined by the values of $f$ on $[0,a_k]=I_0\cup I_1\cup \cdots\cup I_{k-1}$ and not affected much by subsequent tinkerings of $f$ on $[a_{k},1]=I_{k}\cup\cdots$. This can be done in such a way that $f$ is continuous and $|f(x)|\le1-x$ for all $x$. As EricWofsey comments, in effect, one can choose $a_{k+1}$ so large (that is, close to $1$) that $(1-a_{k+1})\exp(n_k)$ is as small as you want.

The upshot is a continuous $f$ such that $$\liminf_{k\to\infty}\quad (-1)^k \int_0^1 \exp(n_k x) f(x)dx = 1,$$ disproving the original conjecture.

Overall the construction I am describing is like the standard examples of bounded sequences without Cesaro averages. You know, 10 zeros followed by 100 ones followed by 1000 zeros followed by 10000 ones...

Now for some details. I apologize for the intricate notation. It is possible I have made a typo somewhere. It is more than possible that there is a simpler way to arrange the calculation or (better yet) a simpler way to see the result.

Let $\Lambda_\alpha(x)=\max(0,\alpha/2-|2x-(\alpha+1)|)$ be the piecewise linear function whose graph has an isosceles triangular spike centered at $(1+\alpha)/2$, of width $\alpha/2$ and height $\alpha/2$. Note that $\text{supp } \Lambda_\alpha = [(3\alpha+1)/4,(\alpha+3)/4]$, and that $|\Lambda_\alpha(x)|\le |1-x|$. The final form of $f$ will be $$f(x)=\sum_{k\ge0} \epsilon_k (-1)^k \Lambda_{a_k}(x),$$ where the $\epsilon_k$ are in $[0,1]$. The function $f$ satisfies $|f(x)|\le|1-x|$ on $[0,1]$.

We will construct $a_k, n_k, \epsilon_k$ inductively.

At stage $k$ we will have specified $f$ on the interval $[0,a_k]$, and the inductive step delivers a formula for $f$ on the interval $I_k=[a_k,a_{k+1}]$, thus extending the definition of $f$ to $[0,a_{k+1}]$.

Start with $k=0$ and $a_0=0$. Let $L>0$ be a constant, such as $L=1$.

At inductive stage $k$, chose $n$ so large that $0<\epsilon_k<1$, where $$\epsilon_k = \frac { L - (-1)^k\int_0^{a_k}f(x)e^{nx}dx}{\int_0^1\Lambda_{a_k}(x)e^{nx}dx},$$ and, if $k>0$, also $n\ge1+n_{k-1}$. This is possible because the integral in the denominator has larger exponential growth rate (at least $(3a_k+1)/4$) than that in the numerator, which is at most $a_k$. Denote the chosen $n$ by $n_k$. Finally, and this is the key point identified by EricWofsey in a comment, chose to be very close to $1$, as in $$a_{k+1} = \max((a_k+3)/4, 1-\exp(-2n_k)).$$ Note $a_{k+1}<1$.

The restriction of $f$ to $I_k=[a_k,a_{k+1}]$ is $(-1)^k\epsilon_k\Lambda_{a_k}$.

It is easy to see from this that $n_k\to\infty$ and that $a_k\to 1$. Checking that $f$ is continuous is routine.

By construction, $\int_0^{a_{k+1}}f(x)e^{n_kx}dx = (-1)^k L$ and this differs from $\int_0^1 f(x) e^{n_kx}dx$ by $\int_{a_{k+1}}^1 \exp(n_k)dx = O(\exp(-n_k))$.

  • $\begingroup$ I would add that the key point is, having chosen $n_1,\dots,n_k$ and the restriction of $f$ to $[0,a_{k+1}]$, you can then choose a bound for $f$ on $I_{k+1}$ which is small enough to have negligible effect on the integrals for $n_1,\dots,n_k$, and then choose $n_{k+1}$ large enough so that by varying $f$ within this bound on $I_{k+1}$ you can make the integral for $n_{k+1}$ small. $\endgroup$ – Eric Wofsey May 10 '19 at 22:42
  • $\begingroup$ @EricWofsey Exactly. Care to post a more complete answer? $\endgroup$ – kimchi lover May 10 '19 at 22:45
  • $\begingroup$ Why start with "this is not an answer", instead of just stating the answer (which appears to be simply "no") up front? $\endgroup$ – hmakholm left over Monica May 12 '19 at 12:34
  • $\begingroup$ @HenningMakholm Sorry about that. This was written in bits and pieces, and I seem to have finished it before I realized I was done. That's why it reads like a draft. $\endgroup$ – kimchi lover May 12 '19 at 12:42

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