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Im trying to prove that the series

\begin{eqnarray} f_n(x) & = & 1\quad \mbox{for } n < x \leq n+1 \\ {} & = & 0 \quad \mbox{otherwise} \end{eqnarray}

converges weakly in $L_2[0,\inf]$

I did the following :

$<f_n(x),g> = |\int_{n}^{n+1} g(t)dt| \leq \int_{n}^{n+1} |g(t)|dt \leq \sqrt{\int_{n}^{n+1} |1|^2dt }*\sqrt{\int_{n}^{n+1} |g(t)|^2dt } = \sqrt{\int_{n}^{n+1} |g(t)|^2dt }$

so I need to prove that the last statement goes to zero when $n\to\infty $

I would really appreciate some hint :)

B. Mehta
  • 12,774
Bar Dubovski
  • 89

1 Answers1

1

Try to prove $\lim_{n \rightarrow \infty}\int_n^{n+1} g(x) dx = 0$ for all $g(x) \in L_2[0, \infty)$

Mudream
  • 534
  • Ty, I also got here but not quite sure how to prove this point, if we know that $g(x) \in L_2[0,∞)$ it means that $ |g|_{2}\equiv \left(\int _{[n.n+1]}|g|^{2};\mathrm {d} \mu \right)^{1/2}<\infty $

    what else can we learn from it to help solve the problem?

     – Bar Dubovski May 31 '17 at 09:28
  • let $<f_n, g>^2 = a_n^2 = \int_n^{n+1}g(x)^2dx$, then $\sum^n a_i^2 \leq |g|^2 < \infty$, then $\lim_{n \rightarrow \infty} a_n^2 = 0$ – Mudream May 31 '17 at 13:13
  • errrrrrrrr $f_n$ not converge !!?? – Mudream May 31 '17 at 13:15
  • could you please explain this claim $∑^{n}a_{i}^{2}≤|g|^2$? – Bar Dubovski May 31 '17 at 15:48
  • Thank you very much, you helped me figure it out :) – Bar Dubovski Jun 01 '17 at 06:42