# Problem of the Week #226 - Sep 27, 2016

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#### Euge

##### MHB Global Moderator
Staff member
Here is this week's POTW:

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Let $(X,\mu)$ be a measure space, $f\in \mathcal{L}^1(\mu)$, and $\phi_n\in \mathcal{L}^1(\mu)$ such that $\sup_{n,t}\lvert \phi_n(t)\rvert \le 1$ and $\|\phi_n\|_1 \to 0$ as $n\to \infty$. Show that $\|f\phi_n\|_1 \to 0$ as $n\to \infty$.

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Remember to read the POTW submission guidelines to find out how to submit your answers!

#### Euge

##### MHB Global Moderator
Staff member
This week's problem was solved by Opalg . You can read his solution below.

Suppose that $\|f\phi_n\|_1 \not\to0$. Then there exist $\varepsilon>0$ and a subsequence $(\|f\phi_{n_k}\|_1)$ such that $\|f\phi_{n_k}\|_1 \geqslant \varepsilon$ for all $k.$

A sequence that converges in the $L^1$-norm has to have a subsequence that converges pointwise almost everywhere. So the subsequence $(\phi_{n_k})$ has a sub-subsequence that converges almost everywhere to $0.$ Replacing $(\phi_{n_k})$ by this sub-subsequence, we may assume that (1) $\|f\phi_{n_k}\|_1 \geqslant \varepsilon$ for all $k$, and (2) for almost all $t$, $\phi_{n_k}(t) \to0$ as $k\to\infty.$

Since $|\phi_n(t)| \leqslant1$ (for all $n$ and $t$), it follows that $|f\phi_{n_k}(t)| \leqslant |f(t)|$ (for all $k$ and $t$). Thus the functions $|f\phi_{n_k}|$ are dominated by the integrable function $|f|.$ From (2), $|f\phi_{n_k}| \to0$ pointwise almost everywhere as $k\to\infty.$ Hence by the dominated convergence theorem $$\displaystyle \|f\phi_{n_k}\|_1 = \int |f\phi_{n_k}| \to0$$ as $k\to\infty$, which contradicts (1).

The contradiction shows that $\|f\phi_n\| \to0$ as $n\to\infty.$

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