Let $$f_n (x) := \frac{1}{n} \sum_{i=1}^n \exp(-a_i \cdot x^2),$$ with $x \in \mathbb{R}$, and $(a_i)_{i \in \mathbb{N}}$ monotone sequence with $a_i >0 \; \forall i \in \mathbb{N}$.

I have to determine whether $f_n (x)$ converges for $x \in \mathbb{R}$, and eventually if it converges uniformly.

My attempt:

as $f_n(x)$ is:

  • bounded for each $x \in \mathbb{R}$: we have $ 0 \leq f_n(x) \leq 1 \; \forall n \in \mathbb{N}$ and $\forall x \in \mathbb{R}$,
  • monotone: we have $f_n(x) \geq f_{n+1}(x) \; \forall n \in \mathbb{N}$ and $\forall x \in \mathbb{R}$,

then it converges pointwise.

To understand whether it converges uniformly, as I could not determine explicitely the limit f(x), I tried to use Cauchy criteria (Cauchy Criterion for Uniform Convergence of Functions).

In the trivial case in which $a_i=a>0$ (the $a_i$'s are all the same) then it is clearly uniformly convergent, as $f_n(x) = f(x) = \exp(-a \cdot x^2)$, but in the general case I could neither show uniform convergence nor an objection to it.

Any idea? Any thought would be greatly appreciated.

  • 1
    $\begingroup$ How did you get monotonicity? $\endgroup$ Sep 13 '21 at 9:50
  • $\begingroup$ Thank you for the question. It would be sufficient to assume that $(a_i)_{i \in \mathbb{N} }$ is a monotone sequence to state this, but in general I do not have this condition. So firstly I need to review my conclusion on pointwise convergence. Thank you! $\endgroup$
    – G.Rossi
    Sep 13 '21 at 10:06
  • $\begingroup$ Ok, as expected there was a missing requirement in the exercise. $(a_i)_{i \in \mathbb{N}}$ is a monotone sequence (otherwise the sequence of $f_n$ is in general not even converging pointwise). I modified the post. Thank you again for your question. $\endgroup$
    – G.Rossi
    Sep 13 '21 at 11:08

I suppose you are assuming that $a_i$'s are monotonically increasing. [The case when they are decreasing is easier and I will leave that to you].

Case 1): $(a_i)$ increases to $\infty$. In this case $e^{-a_ix^{2}} \to 0$ as $i \to \infty$ for each $x \neq 0$. Hence, $f_n(x)$ also tends to $0$ for $x \neq 0$. But $f_n(0)=1$ for all $n$. If $f_n$ converges uniformly then the limit function $f$ is continuous. But $f(x)=0$ for $ \neq 0$ and $f(0)=1$. This proves that the convergence is not uniform.

Suppose $a_i$ increases to a finite limit $a$. Then $f_n(x) \to e^{-ax^{2}}$ uniformly. Here are some hints for the proof: Split $f_n(x)$ into the sum of $\frac 1 n\sum\limits_{i=1}^{k}e^{-a_ix^{2}}$ and $\frac 1 n\sum\limits_{i=k+1}^{n}e^{-a_ix^{2}}$. It is easy to see that the first part tends to $0$ uniformly for fixed $k$. We can choose $k$ so large that $a-\epsilon <a_i <a+\epsilon$ for $i >k$. Now use Squeeze Theorem to finish the proof.

  • $\begingroup$ Thank you for your exhaustive answer. I try to complete it in case $a_i$'s are decreasing in the next answer. Could you please give me a quick feedback if it is correct? $\endgroup$
    – G.Rossi
    Sep 13 '21 at 12:43

Case in which $a_i$'s are decreasing.

Case 1): Decreasing to $0$. Then is $f_n(x) \rightarrow 1 \; \forall x \in \mathbb{R}$.

By definition of zero sequence for $a_i$ it holds that $\forall \varepsilon > 0 \; \exists N_\varepsilon : | a_n - 0 | < \varepsilon \; \forall n \geq N_\varepsilon$.

Then (definition of uniform convergence) $\forall \varepsilon >0, \exists N_\varepsilon : \forall x \in \mathbb{R} , \forall n \geq N_\varepsilon$ is

$\left| \frac{1}{n} \sum_{i=1}^n \exp(- a_i \cdot x^2) - 1 \right| \leq \left| \frac{1}{n} \sum_{i=1}^n \exp(- \varepsilon \cdot x^2) - 1 \right|. $

As $\varepsilon$ is arbitrary small the first term in the absolute value is discretionary close to $1$, $\forall n \geq N_\varepsilon$, and therefore we have uniform convergence.

Case 2): Decreasing to $a>0$. Similarly as shown by Kavi Rama Murthy in the answer above (https://math.stackexchange.com/q/4249197).


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