Approx. in $L^p$: Proving Existence of Subseq & $g$

[joypav]

Problem:
Assume $E$ is a measurable set, $1 \leq p < \infty$, and $f_n \rightarrow f$ in $L^p(E)$. Show that there is a subsequence $(f_{n_k})$ and a function $g \in L^p(E)$ for which $\left| f_{n_k} \right| \leq g$ a.e. on $E$ for all $k$.

Proof:

Maybe use?:

• $f_n \rightarrow f$ in $L^p(E)$ iff $lim_{n\rightarrow \infty}\int_{E}\left| f_n \right|^p = \int_{E}\left| f \right|^p $
• In class we have covered the proof for the Riesz-Fischer theorem. So applying that theorem gives the existence of a subsequence $(f_{n_k})$ in $L^p(E)$ so that $f_{n_k} \rightarrow f$ a.e. However, I'm thinking this isn't what I need because we don't necessarily need the subsequence to converge to $f$.

I'm not sure how to show existence of such a $g$... perhaps using Lusin's theorem? I am a bit confused by the relationship between $f$ and $g$. I haven't seen much on approximation in $L^p$ spaces.

Any help, hints, or helpful theorems would be much appreciated!

 

[joypav]

I think I may have gotten somewhere... if you check it out let me know if there are any issues!

Proof:
$f_n \rightarrow f$ in $L^p(E) \implies f$ is cauchy.
Then there exists a subsequence $(f_{n_k})$ such that
$\left| \left| f_{n_{k+1}} - f_{n_k} \right| \right|_p \leq \frac{1}{2^k}, \forall k \geq 1$

Define,
$s_j = \sum_{k=1}^{j} \left| f_{n_{k+1}} - f_{n_k} \right|, \forall j \geq 1$

Then $s_j > 0$ and $s_{j+1} \geq s_j, \forall j \geq 1$

$\left| \left| s_j \right| \right|_p = \left| \left| \sum_{k=1}^{j} \left| f_{n_{k+1}} - f_{n_k} \right| \right| \right|_p$
$\leq \sum_{k=1}^{j} \left| \left| f_{n_{k+1}} - f_{n_k} \right| \right|_p$
$\leq \sum_{k=1}^{\infty} \left| \left| f_{n_{k+1}} - f_{n_k} \right| \right|_p$
$\leq \sum_{k=1}^{\infty} \frac{1}{2^k} = 1$
$\implies \left| \left| s_j \right| \right|_p \leq 1, \forall j \geq 1$
$\implies s_j \in L^p$

$s = \sum_{j=1}^{\infty} s_j$
By the Monotone Convergence Theorem,
$\int_{E}\left| s \right|^p = \sum_{j=1}^{\infty}\int_{E}\left| s_j \right|^p$
$\int_{E}\left| s \right|^p = \lim_{{j}\to{\infty}} \int_{E}\left| s_j \right|^p \leq 1$
$\implies s \in L^p(E)$
$s$ is finite a.e. on $E$ and $s_j$ converges to $s$ in $E$.

Since $s_j$ converges in $E$, $s_j$ is Cauchy.
$\implies (f_{n_k})$ is Cauchy
$\implies (f_{n_k})$ converges to $f'$ in $E$

Moreover,
$\left| f_{n_l}(x) - f_{n_k}(X) \right| \leq s(x), \forall k, l \in \Bbb{N}$ a.e. $x$ in $E$
$s \in L^p$, so by Lebesgue Dominated Convergence, we have
$(f_{n_k}) \rightarrow f'$ in $L^p \implies f'=f$ a.e. $x$ in $E$

Define,
$g = s + f' = s + f$ (a.e.)
Then,
$g \in L^p(E)$ and $\left| f_n(x) \right| \leq g(x)$ a.e. $x$ in $E$.

 

[Euge]

Yes, your proof is correct. Great job!