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I am trying to show that the sequence of functions $f_n=2^{2n}\chi_{[0,2^{-3n})}$ converges in $\lVert \cdot \rVert_1$ but not $\lVert \cdot \rVert_2$. So in $\lVert \cdot \rVert_1$ we have:

$$\lVert f_n \rVert_1=2^{2n}\int_0^1 \chi_{[0,2^{-3n})}=2^{2n}2^{-3n}=2^{-n}\rightarrow 0\text{ as }n\rightarrow \infty$$

but in the $2$-norm I am getting:

$$\lVert f_n\rVert _2=2^{2n}\left ( \int_0^1 \lvert \chi_{[0,2^{-3n})}\rvert^2 \right )^{\frac{1}{2}}=2^{2n}\left ( \int_0^1\lvert 2^{-3n}\rvert^2 \right )^{\frac{1}{2}}=2^{2n}\times (2^{-6n})^{\frac{1}{2}}=2^{-n}$$ which I don't think is right.

Have I made a stupid mistake or am I using the wrong function to show that convergence in the $1$-norm does not imply convergence in the $2$-norm?

cheers

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    $\begingroup$ $|\chi_{[0,2^{-3n})}|^2 = \chi_{[0,2^{-3n})}$ and the second equation in the computation of the $2$-norm is incorrect. $\endgroup$
    – Martin
    Apr 26, 2013 at 9:22
  • $\begingroup$ @Martin oh dear yes so $||f_n||_2=2^{2n}\left ( \int_0^1|\chi_{[0,2^{-3n})}|^2 \right )^{\frac{1}{2}}=2^{2n}\times 2^{\frac{-3n}{2}}=2^{\frac{n}{2}}\rightarrow\infty$ as$n\rightarrow \infty$, thanks very much $\endgroup$
    – user73957
    Apr 26, 2013 at 9:25

2 Answers 2

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user73957 figured out that the correct computation of the $2$-norm should have been $$\lVert f_n\rVert_2=2^{2n}\left ( \int_0^1|\chi_{[0,2^{-3n})}|^2 \right )^{\frac{1}{2}}=2^{2n}\left ( \int_0^1\chi_{[0,2^{-3n})} \right )^{\frac{1}{2}}=2^{2n}\times 2^{\frac{-3n}{2}}=2^{\frac{n}{2}}\rightarrow\infty,$$ so everything is settled and the example works.

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$\lVert f_n\rVert_2=2^{2n}\left ( \int_{-\infty}^{\infty} \lvert \chi_{[0,2^{-3n})}\rvert^2 \right )^{\frac{1}{2}}=2^{2n}\left ( \int_0^{2^{-3n}}\lvert 1\rvert^2 \right )^{\frac{1}{2}}=2^{2n}(2^{-3n})^{\frac{1}{2}}=2^{\frac{n}{2}},$

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  • $\begingroup$ it was mistake at write formula in latex $\endgroup$
    – Somaye
    Apr 26, 2013 at 10:14
  • $\begingroup$ Ok, I removed my comments and the downvote. $\endgroup$
    – Martin
    Apr 26, 2013 at 10:15

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