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So I have this problem that I want to see if I have solved right.

Let ${f_n}$ be a function sequence in $L_2(\mathbb{R})$ for which the following holds:

$\sum_{n=1}^{\infty}\sqrt{\int_{- \infty}^{\infty}|f_n(x)|^2dx} < ∞ $

and we define for $n ∈ \mathbb N∪\{∞\}$: $φ_n(x) = \sum_{k=1}^{n}|f_k(x)|$ .

Show that for almost all $x ∈ \mathbb{R}$ $φ_∞(x) < ∞$ and $\sum_{n=1}^{\infty} f_n(x)$ converges.

Hint: Since this statement was part of the proof of the completeness of $L_p (µ)$, you may not use this statement here, but you may apply the Minkowski inequality.

So what I have done is:

With Minkowski follows:

$\sum_{n=1}^{\infty}(\int_{- \infty}^{\infty}|f_n(x)|^2dx)^{\frac{1}{2}}>(\int_{- \infty}^{\infty}\sum_{n=1}^{\infty}|f_n(x)|^2dx)^{\frac{1}{2}} >(\int_{- \infty}^{\infty}|\sum_{n=1}^{\infty}f_n(x)|^2dx)^{\frac{1}{2}} $ If $\sum_{n=1}^{\infty} f_n(x)$ does not converge we would have $(\int_{- \infty}^{\infty}|\sum_{n=1}^{\infty}f_n(x)|^2dx)^{\frac{1}{2}}= \infty$ and this would be a contraposition to our assumpitions.

For the same idea if $\phi_{infty}=\infty$ than $(\int_{- \infty}^{\infty}\sum_{n=1}^{\infty}|f_n(x)|^2dx)^{\frac{1}{2}}= \infty$ and this would be as before a contraposition.

Have I done any mistake?

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You do not explain how your two $>$ would follow from Minkowski, they should be $\ge,$ and anyway, the second one $$\left(\int_{- \infty}^{\infty}\sum_{n=1}^{\infty}|f_n(x)|^2dx\right)^{\frac{1}{2}} \ge\left(\int_{- \infty}^{\infty}|\sum_{n=1}^{\infty}f_n(x)|^2dx\right)^{\frac{1}{2}} $$ would be wrong.

E.g. for $f_1=f_2=2\,{\bf1}_{[0,1]}$ and $f_n=0\quad\forall n>2$, the LHS is $\sqrt8$ and the RHS is $\sqrt{16}.$

Instead, write: by Minkowski, $$\forall n\in\Bbb N\quad\|\varphi_n\|_2\le\sum_{k=1}^n\|f_n\|_2\le M:=\sum_{k=1}^n\|f_n\|_2$$ hence by monotone convergence, $$\|\varphi_\infty\|_2=\sup_{n\in\Bbb N}\|\varphi_n\|_2\le M<\infty.$$ Therefore, $\varphi_\infty$ is a.e. finite.

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  • $\begingroup$ Can you please give me a corrected way to prove my statements? $\endgroup$ Feb 12, 2023 at 14:06

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