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Let $(X,F,u)$ be a measure space and suppose $\{ f_{n} \}$is a sequence of measurable functions with the property that for all $ n>1 $,

$$ u \{ x\in X:|f_{n}(x)|\geq \lambda \} \leq C\exp(-\lambda^{2}/n) $$ for all $\lambda >0$. (Here C is a constant independent of $n$.) Let $n_{k}=2^{k}$. Prove that $$ \limsup_{k \to \infty} \frac{|f_{n_k}|}{\sqrt{n_{k}\log (\log(n_{k}))}}\leq 1\text{ a.e.} $$

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no, there is no other information about the space. – user53800 Dec 20 '12 at 12:40
This isn't true in general, e.g., $f_n = 0$ – mike Dec 20 '12 at 12:40
what do you mean by $f_{n}$independent? – user53800 Dec 20 '12 at 12:42
sorry, you caught me in the middle of editing that, I think the inequality goes the wrong way, because you want the $f_i$ to be fairly big , – mike Dec 20 '12 at 12:44
I am sorrry about that, thanks a lot, you are right, I have already correct it. – user53800 Dec 20 '12 at 12:47
up vote 1 down vote accepted

Let $\varepsilon>0$ and

$$A_k := \left[ \frac{|f_{n_k}|}{\sqrt{n_k \cdot \log \log n_k}} \geq 1+ \varepsilon \right] $$

We want to show that $$\sum_{k \geq 0} u(A_k)< \infty \tag{1} $$ From this we conclude by the Borel-cantelli theorem that

$$\limsup_{k \to \infty} \frac{|f_{n_k}|}{\sqrt{n_k \cdot \log \log n_k}} \leq 1+ \varepsilon$$

(this part of Borel-cantelli holds also for measure spaces, not only probability spaces.)

So remaining is to show (1). Using the decay property we obtain

$$u(A_k) = u \left(\left[ |f_{n_k}| \geq (1+\varepsilon) \cdot \sqrt{n_k \cdot \log \log n_k} \right] \right) \leq C \cdot \exp \left( -(1+\varepsilon)^2 \cdot \frac{n_k \cdot \log \log n_k}{n_k} \right) \\ = C \cdot \exp \left(\log \left(\log n_k \right)^{-(1+\varepsilon)^2} \right) = C \cdot \frac{1}{(k \cdot \log 2)^{(1+\varepsilon)^2}}$$


$$\sum_{k \geq 0} u(A_k) \leq \frac{C}{(\log 2)^{(1+\varepsilon)^2}} \cdot \sum_{k \geq 0}\frac{1}{k^{(1+\varepsilon)^2}} < \infty$$

notation $[f_{n_k} \in B] := \{x \in X; f_{n_k}(x) \in B\}$ denotes the pre-image of $B$.

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I am in a rush so I'll give a hint to the new version. I'll flesh it out if you don't figure it out yourself by tomorrow.

Hint: Let $B_\epsilon$ be the set $\{ \limsup \frac{|f_{n_k}|}{\sqrt{n_k \log \log n_k}} > 1 + \epsilon$. We can write $B_0$ as a countable union of sets of the form $B_\epsilon$ for $\epsilon > 0$. Next, we can show that each $B_\epsilon$ for $\epsilon > 0$ has measure 0, using the decay property you assumed.

Edit: the below addresses a previous version of the question

As stated it is false.

Let $X = (0,1)$, $F$ be the Borel $\sigma$-algebra, and $u$ the usual Lebesgue measure.

Let $f_n(x) \equiv n$. Then we have that

$$ u\{ |f_n| \geq \lambda\} = \begin{cases} 1 & n \geq \lambda \\ 0 & n < \lambda \end{cases} $$

Observe that if $n \geq \lambda$, $e^{-\lambda/n} \geq e^{-n/n} = e^{-1}$. So the required condition is satisfied for any $C > e$.

Now, we have

$$ \frac{|f_{n_k}|}{\sqrt{n_k \log \log n_k}} = \sqrt{ \frac{n_k}{\log\log n_k}} \nearrow \infty $$

so we have that precisely everywhere

$$ \limsup \frac{|f_{n_k}|}{\sqrt{n_k \log \log n_k}} > 1$$

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Thanks for your effort, however I have a typo and the change the direction of the inequality – user53800 Dec 20 '12 at 13:35
I am sorry again, there should be $\lambda ^{2}$there, and I really have no idea with the $n_{k}=2^{k}$ – user53800 Dec 20 '12 at 13:39

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