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our prof wanted to show that $(C[0,1],\| \cdot \|_2)$ is not complete. So he said $$f_k(x) = x^k$$ is a counterexample. I wonder if this is true. I tried to show that $f_k$ is cauchy sequence. But i ask myself if the limit is $f \equiv 0$ or $f = \mathbb{1}_{\{1\}}(x)$ in the $\| \cdot \|_2$ Norm.

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    $f_k$ converges to $0$ in $(C[0,1],\lVert,\dot,\rVert_2)$. So that isn't actually an example of a Cauchy sequence that doesn't converge. – Daniel Fischer May 19 '16 at 14:01
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    This kind of behavior can be amended to work for, say, $C[-1,2]$, by letting $f_k(x) = \begin{cases} 0 & x \in [-1,0)\ x^k & x \in [0,1) \ 1 & x \in [1,2]\end{cases}$. Substituting $\tilde x = ax+b$ yields the appropriate function for any interval. – Roland May 19 '16 at 14:07
  • I wonder why doesn't it converge to $1_{{1}}$ ? –  May 19 '16 at 14:09
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    Because $1_{{1}} \notin C[0,1]$. For $x_k \rightarrow x$ in $X$ one needs $x \in X$. – Roland May 19 '16 at 14:09
  • @Roland you may add this as an answer and i will accept it, this was what i messed up thinkung about this –  Jul 24 '16 at 07:37

2 Answers2

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Let $a = 0$ and $b = 1$. Define for $n\ge 3$ $$f(x) = \cases{0 & if $x \le 1/2 - 1/n$\cr 1 & if $x\ge 1/2 + 1/n,$}$$ then interpolate linearly in the middle interval. These sequence is Cauchy and does not converge in the $\mathcal{L}^2$ norm.

ncmathsadist
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Having a Cauchy Sequnece is neccesary, but not sufficient to have a converging series. In particular, we say that a seqquence $(x_k)_{k \in \mathbb N}$ is convergent in $(X, \|\cdot\|)$ if there is some $x \in X$ such that $\lim \|x_k - x \| \rightarrow 0$ as $k \rightarrow \infty$.

Here, we have $0 \in X$, but $1_{\{1\}} \notin X$. And while both $\|x_k - 0\|$ and $\|x_k - 1_{\{1\}}\|$ converge to zero, the latter isn't a limit in $X$.

Roland
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