# Uniformly convergence of sequence of functions

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.

• How did you get monotonicity? Sep 13 '21 at 9:50
• 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! Sep 13 '21 at 10:06
• 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. 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 $$a-\epsilon for $$i >k$$. Now use Squeeze Theorem to finish the proof.

• 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? 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).