See the title.
If I am not mistaken, this is true for $L^2$-spaces but is false for $L^1$-spaces. Say, should the space be reflexive?
Asked
Active
Viewed 118 times
0
-
For $1<p<\infty$.${}{}$ – David Mitra Jan 22 '14 at 09:47
-
See this post for an outline of a proof that you'll have weak convergence in $L_p$, $1<p<\infty$ with your conditions. In $L_1$, $f_n=n\chi_{[0,1/n]}$ furnishes a counterexample. – David Mitra Jan 22 '14 at 09:56
-
Yes, thanks. My aim is an abstract property for much more general spaces of functions. – limanac Jan 22 '14 at 10:00
1 Answers
0
If you only speak about the spaces $L^p(0,1)$, $1 \le p \le \infty$, this is always false.
Just consider the sequence \begin{equation*} u_n(t) = \begin{cases} n^{1/p} & \text{for }t \le 1/n \\ 0 & \text{else}, \end{cases} \end{equation*} with $n^{1/\infty} = 1$. Then you have $\|u\|_p^p = \int_0^1 u_n(t)^p \, \mathrm{d}t = 1$ and $u_n(t) \to 0$ for all $t > 0$.
Hence, $u_n$ does not converge.
gerw
- 31,359