# Small exercise on Minkowski inequality

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?

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.