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$X=\{(a,b)\mid a \in C[0,1],b \in C[0,1]\}$, and its norm is $\|(a,b)\|=\|a\|_\infty+\|b\|_\infty.$

$Y=\{(a,a')\mid a \in C^1[0,1], \ a'(t)=\frac{da}{dt} \},\ Z=\{(0,b)\mid b \in C[0,1]\}.$

Under these conditions, Please show that $Y+Z=\{y+z \mid y \in Y, z \in Z\}$ is not a closed subset of $X$

(I have already known that $X$ is Banach space, and $Y,Z$ are closed subsets of $X$.)

I have to say that there is one sequence $\{y_n+z_n \mid y_n \in Y, z_n \in Z\}$ which converges to $(y,z) \notin Y+Z$. But what kind of sequence will suffice above?

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  • $\begingroup$ Is your $T$ supposed to be $Z$? $\endgroup$ – Ian Jul 29 '15 at 0:33
  • $\begingroup$ Notice $Y+Z = C^1[0,1]\times C[0,1]$ and that $C^1[0,1]$ is not closed in $C[0,1]$. $\endgroup$ – user251257 Jul 29 '15 at 0:54
  • $\begingroup$ I fixed $T \rightarrow Z$. Thanks for indicating my error. $\endgroup$ – yluerr Jul 29 '15 at 1:49
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For any $(a,a')\in Y$, we have that $(0,-a')\in Z$. So $(a,0)\in Y+ Z$. In other words, $Y+Z$ contains the subspace $W=\{(a,0):\ a\in C^1[0,1]\}$.

So now we need a Cauchy sequence in $W$ that is not convergent in $W$. For instance $\{(a_n,0)\}$, where $a_n(t)=(t+1/n)^{1/2}$.

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  • $\begingroup$ it must converge in $X$. it does not converge in $W$ $\endgroup$ – user251257 Jul 29 '15 at 1:01
  • $\begingroup$ It does converge in $X$. It converges to $a(t)=t^{1/2}$. $\endgroup$ – Martin Argerami Jul 29 '15 at 1:13
  • $\begingroup$ So $(a_n(t),a_n'(t))$ that is not convergent in $Y$ is in $Y$?(which means Y is not a closed subset of X???) $\endgroup$ – yluerr Jul 29 '15 at 4:21
  • $\begingroup$ The sequence $(a_n,a_n') $ is in $Y $, but it is not Cauchy. It tells you nothing about the closedness of $Y $. $\endgroup$ – Martin Argerami Jul 29 '15 at 5:06

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