2
$\begingroup$

If $\{f_n\}\subset L^+$, $f\in L^+$ such that $\{f_n\}$ is dominated by $f$ where $\int f < \infty$ then $$\limsup\int f_n\leq \int \limsup f_n$$

Attempted proof - Consider the sequence $\{f - f_n\}\subset L^+$ then applying Fatou's lemma we have $$\int (\liminf(f-f_n))\leq \liminf \int (f - f_n)$$ We know that $$\int(\liminf (f - f_n)) = \int \lim_{k\rightarrow \infty} \inf_{n\geq k}(f - f_n)$$ Then from the Monotone Convergence theorem, $$\int \lim_{k\rightarrow \infty}\inf_{n\geq k}(f-f_n) = \lim_{k\rightarrow \infty}\int \inf_{n\geq k}(f - f_n)$$ Therefore we have, $$\lim_{k\rightarrow \infty}\int \inf_{n\geq k}(f - f_n) = \lim_{k\rightarrow \infty}\int \inf_{n\geq k} f - \lim_{k\rightarrow \infty}\int \inf_{n\geq k}f_n = \int f - \lim_{k\rightarrow \infty}\int \sup_{n\geq k}f_n$$

I am not sure if I am on the right track or where to go from here. Just some hints should help me enough.

Attempted proof #2 - Consider the sequence $\{f - f_n\}\subset L^+$ then applying Fatou's lemma we have \begin{align*} &\int (\liminf(f-f_n))\leq \liminf \int (f - f_n)\\ \Leftrightarrow &\int\liminf f - \int \liminf f_n \leq \liminf\int f - \liminf\int f_n\\ \Leftrightarrow &\int f - \int \limsup f_n \leq \int f - \limsup\int f_n \end{align*} Rearranging and cancelling out $\int f$ (which is possible since $\int f <\infty$) then we have $$\limsup\int f_n\leq \int \limsup f_n$$

$\endgroup$
  • $\begingroup$ $\sup_{n\geqslant k}f-f_n$ is decreasing with $k$, so monotone convergence would not apply. $\endgroup$ – Math1000 Jun 24 '16 at 23:33
  • $\begingroup$ bummer, how should I solve this then? $\endgroup$ – Wolfy Jun 24 '16 at 23:34
  • 1
    $\begingroup$ Apply Fatou's lemma to the sequence $\{f-f_n\}$. $\endgroup$ – carmichael561 Jun 24 '16 at 23:39
  • $\begingroup$ @carmichael561 ok I re-edited it could you provide another hint? $\endgroup$ – Wolfy Jun 24 '16 at 23:45
  • $\begingroup$ For any sequence $\{f_n\}$, $\liminf(-f_n)=-\limsup f_n$. So use this and the linearity of the integral, then cancel $\int f$ and rearrange. $\endgroup$ – carmichael561 Jun 25 '16 at 0:01
1
$\begingroup$

Congratulations! Your proof 2 is more direct, and it is correct.

If $\{f_n\}\subset L^+$, $f\in L^+$ such that $\{f_n\}$ is dominated by $f$ where $\int f < \infty$ then $$\limsup\int f_n\leq \int \limsup f_n$$

Proof 2 - Consider the sequence $\{f - f_n\}_n\subset L^+$ then applying Fatou's lemma we have \begin{align*} &\int (\liminf(f-f_n))\leq \liminf \int (f - f_n)\\ \Leftrightarrow &\int\liminf f - \int \liminf f_n \leq \liminf\int f - \liminf\int f_n\\ \Leftrightarrow &\int f - \int \limsup f_n \leq \int f - \limsup\int f_n \end{align*} Rearranging and cancelling out $\int f$ (which is possible since $\int f <\infty$) then we have $$\limsup\int f_n\leq \int \limsup f_n$$

Now let us look to your proof 1. The appproach you took here is viable, but there are some adjustments need.

Proof 1 - Consider the sequence $\{f - f_n\}\subset L^+$ then applying Fatou's lemma we have $$\int \liminf(f-f_n)\leq \liminf \int (f - f_n)$$ We know that $$\int\liminf (f - f_n) = \int \lim_{k\rightarrow \infty} \inf_{n\geq k}(f - f_n)$$ Since $\{\inf_{n\geq k}(f - f_n)\}_k$ is a non_decreasing sequence of non-negative functions in $L^+$ and $\{\inf_{n\geq k}(f - f_n)\}_k \to \lim_{k\rightarrow \infty} \inf_{n\geq k}(f - f_n)$, we have, from the Monotone Convergence theorem, $$\int \lim_{k\rightarrow \infty}\inf_{n\geq k}(f-f_n) = \lim_{k\rightarrow \infty}\int \inf_{n\geq k}(f - f_n) \tag{1}$$

Now, since for all $k$ and $n\geq k$, $ \inf_{n\geq k}(f - f_n)\leq f_n$, we have that, for all $k$ and $n\geq k$,
$$ \int \inf_{n\geq k}(f - f_n)\leq \int (f-f_n)$$ So we have, for all $k$,
$$ \int \inf_{n\geq k}(f - f_n)\leq \inf_{n\geq k}\int(f- f_n) $$ So we have $$ \lim_{k\rightarrow \infty}\int \inf_{n\geq k}(f - f_n)\leq \lim_{k\rightarrow \infty}\inf_{n\geq k}\int(f- f_n) \tag {2}$$ From $(1)$ and $(2)$, we have $$\int \lim_{k\rightarrow \infty}\inf_{n\geq k}(f-f_n) \leq \lim_{k\rightarrow \infty}\inf_{n\geq k}\int(f- f_n) \tag{3}$$

Therefore we have, \begin{align*} \int f- \int\lim_{k\rightarrow \infty}\sup_{n\geq k} f_n&=\int (f- \lim_{k\rightarrow \infty}\sup_{n\geq k}f_n)=\int \lim_{k\rightarrow \infty}\inf_{n\geq k}(f-f_n)\leq \lim_{k\rightarrow \infty}\inf_{n\geq k} \int (f - f_n)= \\ &=\lim_{k\rightarrow \infty}\inf_{n\geq k} \left ( \int f - \int f_n \right)= \int f - \lim_{k\rightarrow \infty}\sup_{n\geq k}\int f_n\\ \end{align*}

Since $\int f<+\infty$, we can conclude that $$\lim_{k\rightarrow \infty}\sup_{n\geq k}\int f_n \leq \int\lim_{k\rightarrow \infty}\sup_{n\geq k} f_n$$ that is to say: $$\limsup \int f_n \leq \int\limsup f_n$$

Remark: Please note that your attempted proof 1, using Monotone convergence theorem is viable, but when we write down the details, we can see that we proving again Fatou's Lemma as part of proof 1. Please note that $(3)$ is essently Fatou's Lemma applied to $\{f-f_n\}_n$. And the rest of proof 1 after $(3)$ is essentially your proof 2. That is why Proof 2 is more concise, more direct and more elegant.

$\endgroup$
2
$\begingroup$

You may try proving and using the following dual to the monotone convergence theorem:

Suppose $\{g_n\}\subseteq L^+$, $g_n$ decreases pointwise to $g$ , and $g_1$ is integrable. Then, $\int g=\lim\int g_n$.

This is Exercise 2.15 in Folland (1999, p.52).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.