Spring 2012 - Problem 1

construction, Lp spaces

Some of the following statements about sequences of functions $f_n$ in $L^3\p{\br{0, 1}}$ . Indicate these and provide an appropriate counterexample.

If $f_n$ converges to $f$ almost everywhere then a subsequence converges to $f$ in $L^3$ .
If $f_n$ converges to $f$ in $L^3$ then a subsequence converges almost everywhere.
If $f_n$ converges to $f$ in measure then the sequence converges to $f$ in $L^3$ .
If $f_n$ converges to $f$ in $L^3$ then the sequence converges to $f$ in measure.

False. Consider $f_n = n^{1/3}\chi_{\br{0,\frac{1}{n}}}$ , which converges to $f = 0$ everywhere except at $x = 0$ . But
$\norm{f_n - f}_{L^3}^3 = \int_0^{1/n} n \,\diff{x} = 1,$
so no subsequence can possibly converge to $f$ in $L^3$ .
True.
False. The same example as in (1) works: for any $\epsilon > 0$ , there exists $N$ so that $N^{1/3} \geq \epsilon$ . Thus, for $n \geq N$ , $\abs{f_n - f} = n^{1/3} \geq N^{1/3} \geq \epsilon$ on $\br{0, \frac{1}{n}}$ and $0$ otherwise, so
$m\p{\set{\abs{f_n - f} \geq \epsilon}} = m\p{\br{0, \frac{1}{n}}} = \frac{1}{n} \xrightarrow{n\to\infty} 0.$
But as we saw, $f_n$ does not converge to $f$ in $L^3$ .
True.