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Given , $X= l_p (p\geq 1)$ , and let $d(x,y) = ( \sum_{k=1}^{\infty} |x_k - y_k |^{p})^{\frac{1}{p}}$ where $x= \{x_k\}_{k\geq 1}$ and $y= \{y_k\}_{k\geq 1}$ are in $l_p$. Let $\{x^{(n)}\}_{n \geq 1}=\{\{x_k ^{(n)}\}_{k\geq 1}\}_{n \geq 1}$ be a sequence in $X$ that converges to $x$. We need to prove that this implies component wise convergence holds true in this case , but vice-versa is not true.

Using the convergence condition , component wise convergence can be established. Coming to the converse part. We let $ x_k ^{(n)} \to x_k$ for each $k$.

Now we let $x_k ^{(n)} = x_k + \delta _{kn}$ where , $\delta _{kn} = 0 $ for $k \neq n$ and $\delta _{kn} = 1 $ for $k = n$.

So we can see , $| x_k ^{(n)} - x_k| = \delta_{kn} = 0 $ for n>k.

And I don't know what happens in the next step. It says :

Consequently , $ x_k ^{(n)} \to x_k$ for each $k$ , however ,

$ d(x^{(n)} , x) = ( \sum_{k=1}^{\infty} |x_k ^{(n)} - x_k |^{p})^{\frac{1}{p}} = 1$ for all $n$. Can anyone explain me this ? How this comes out to be 1 ?

User9523
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    (1).The Kronecker Delta, written $\delta_{k,n}$ is $0$ if $k\ne n$ and $1$ if $k=n$. (2). The point in the proof is that $\delta {k n}=0$ when $n>k$, so for a given $k$ we have $\delta{k,n}=0$ for all $n>k.$ A simple example is to let $x^{(n)}k=\delta{k,n}$ for all $k,n$. Then for a given $k$ we have $x_k^{(n)}=0$ except when $k=n$ so $\lim_{n\to \infty}x_k^{(n)}=0$ (pointwise congence on the co-ordinates). But $|x^{(n)}|=1$ so $x^{(n)}$ does not converge to $0$. In fact $(x^{(n)})_n$ is not a Cauchy sequence – DanielWainfleet Mar 17 '16 at 09:24
  • Got it.. Thanks ! @user254665 – User9523 Mar 17 '16 at 11:32

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This holds by the very definition of $\def\n{{(n)}}x^\n_k$ as $$ x^\n_k = x_k + \delta_{kn} $$ we have \begin{align*} d(x^\n, x) &= \left(\sum_{k=1}^\infty \def\a#1{\left|#1\right|}\a{x^\n_k - x_k}^p\right)^{1/p}\\ &=\left(\sum_{k=1}^\infty \a{x_k + \delta_{nk} - x_k}^p\right)^{1/p}\\ &= \left(\sum_{k=1}^\infty \a{\delta_{nk}}^p\right)^{1/p}\\ &= 1. \end{align*}

martini
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