Suppose there exists a sequence [itex]f_n[/itex] of square-integrable functions on [itex]\mathbb R[/itex] such that [itex]f_n(x) \to f(x)[/itex] in the L^2-norm with [itex]x \ f_n(x) \to g(x)[/itex], also in the L^2-norm. We know from basic measure theory that there's a subsequence [itex]f_{n_k}[/itex] with [itex]f_{n_k}(x) \to f(x)[/itex] for a.e. x. But my professor seems to be claiming that this somehow implies [itex]x \ f_{n_k}(x) \to g(x)[/itex] for a.e. x. I don't see why this is. Obviously, we know that [itex]x \ f_{m_k} \to g[/itex] a.e. for SOME subsequence of [itex]f_n[/itex]...but how do we know it works for the SAME subsequence? Can someone offer some guidance? Thanks!(adsbygoogle = window.adsbygoogle || []).push({});

**Physics Forums - The Fusion of Science and Community**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Convergence in the L^2 norm

Loading...

Similar Threads for Convergence norm | Date |
---|---|

I Weak Convergence of a Certain Sequence of Functions | Mar 27, 2018 |

I If A is an algebra, then its uniform closure is an algebra. | Dec 26, 2017 |

I Simple convergence proof | Sep 30, 2017 |

Sequence is norm convergent implies it's strongly convergent | Nov 8, 2015 |

Does L^2 Convergance Imply Convergance of L^2 norms? | Aug 16, 2013 |

**Physics Forums - The Fusion of Science and Community**