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On a problem sheet we were asked to find a sequence of functions $(f_n)_{n \geqslant 0} \in L^p [0,1]$ such that $\lim_{n \to \infty} ||f_n||_p = 0$ but $\lim_{n \to \infty} f_n (x)$ doesn't exist $\forall x \in (0,1)$ and we were then asked to find a subsequence $(f_{n_k})$ such that $\lim_{k \to \infty} f_{n_k}(x)= 0$ for almost all $x \in (0,1)$.

Basically my question is does such a subsequence always exist? It did in my example, but I couldn't think of a reason why this would always be true.

CameronJWhitehead
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2 Answers2

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In a general context, note that $L^p$ convergence $\implies$ convergence in probability $\implies$ almost sure convergence (up to an extraction!).

edit: for a simple example, consider $$ f(x) = 1_{nx<1} \text{ if $n$ is even}\\ f(x) = 1_{n(1-x)<1} \text{ if $n$ is odd} $$and the subsequence $\phi(n) = 2n$.

mookid
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We are given that $\lVert f_n\rVert_p$ converges to $0$. Good. But it can converge be arbitrarily slowly. However, given a sequence $(c_k)_k$ of real numbers, we can find a subsequence $(f_{n_k})_k$ such that $\lVert f_{n_k}\rVert_p\leqslant c_k$ for each $k$. In particular, we obtain for each $k$, $$\mu\{x,|f_{n_k}(x)|\geqslant k^{-1}\}\leqslant kc_k.$$

Choosing $c_k$ such that the series $\sum_k kc_k$ converges, we obtain that $\mu\left(\limsup_k\{x,|f_{n_k}(x)|\geqslant k^{-1}\} \right)=0$, hence for almost every $x\in [0,1]$, $|f_{n_k}(x)|\lt k^{-1}$ if $k\geqslant K(x)$.

Davide Giraudo
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