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Given $X$=$c_0$, null sequence space with sup norm. Consider a subspace $Y$ of $c_0$ consisting of elements of $c_0$ as, $Y=\{x\in c_0 : x_{2i}=i.x_{2i-1}, i \geq 1\}$. I need to find the set of best approximations from an element $x \in c_0$ to $Y$.

Fact: A best approximation, say $y_0 \in Y$ from a point $x \in X$ to a closed subspace $Y \subset X$ satisfies $||x-y_0||=\inf ||x-y||, \forall y \in Y$.

My attempt: An element, say $y$ of $Y$ will be of the form $y=\{(y_1, y_1, y_3, 2y_3, y_5, 3y_5,...)\}$. Now let $x =(x_1, x_2, x_3,...) \in c_0$ be an element. Then $\inf ||x-y||$=$\inf ||(x_1-y_1, x_2-y_1,x_3-y_3,x_4-2y_3,...)||$. I am confused about what to do next. Please help me. I appreciate any help you can provide.

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  • $\begingroup$ You had better say $y = (y_1, y_1, y_3, 2y_3, y_5, 3y_5, \dots)$ to avoid confusion. $\endgroup$
    – Dirk Werner
    Commented Jan 11, 2023 at 16:34
  • $\begingroup$ Yes sir, sorry about the confusion. $\endgroup$
    – PPB
    Commented Jan 12, 2023 at 5:21
  • $\begingroup$ But now $x-y$ is not what you have written. $\endgroup$
    – Dirk Werner
    Commented Jan 12, 2023 at 17:02
  • $\begingroup$ It appears to me that the distance is the sup of the absolute values of the quantities $(x_{2j}-jx_{2j-1})/(j+1)$. Just minimise each successive pair of differences. $\endgroup$
    – Derek
    Commented Jan 17, 2023 at 16:44

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