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Not for homework, I am trying to self study functional analysis and encountered the following problem.

We let $C[0,1/2]$ the continous functions defined on that subset of the real line. We look at a subspace of $C[0,1/2]$, consisting of all polynomials on $[0,1/2]$ and call it $W$. Given $\delta>0$, set.

$g(x)=\delta\sum_{n=1}^{\infty}\frac{1}{n+1}x^n$

Where $x\in[0,1/2]$. First, we are tasked with showing that $g$ is in the open ball $B(0,\delta)$. I suppose this is done by showing that the norm of $g$ must be less than delta (we are using the supremum norm inherited from $C[0,1/2]$). But from that result, we are tasked with using it to conclude that $W$ cannot be an open subset of $C[0,1/2]$. So I must show that for some $w\in W$, there is no $\epsilon$, such that an open ball $B(v,\epsilon)$ is in $W$? Not sure how that follows from the previous?

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    $\begingroup$ To answer your first question, yes, you need to show the norm of $g$ is less than $\delta$. As for your second question, note that if you define $g_k(x) = \delta \sum_{n=1}^k x^n/(n + 1)$ then $g_k \in W$. If $W$ is closed, do you know anything about how the limit of the sequence $(g_k)_{k \geq 1}$ must behave? (This is just a characterisation of closedness different to the one you have provided.) $\endgroup$
    – Riley
    Jan 2, 2019 at 23:13
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    $\begingroup$ A proper subspace of a normed space is never open. In fact, it even has empty interior. $\endgroup$ Jan 3, 2019 at 12:16

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$\|g\| < \delta \sum_{n=1}^{\infty} \frac 1 {2^{n}}= \delta$. This proves the first part. Now,suppose $W$ is open . Since the zero polynomial is in $W$ there must exist $\delta >0$ such that $B(0,\delta) \subset W$. Consider the $g$ corresponding to this $\delta$. Then $g \in B(0,\delta)$ so we must have $g \in W$. Can you see that this is a contradiction? [It is a known fact that if $\sum a_n x^{n}$ converges for $|x| \leq r$ and if the sum is zero for all such $x$ the $a_n=0$ for all $n$].

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(1). Suppose that $Y$ is a vector subspace of a normed linear space $X$ and that $Y$ has non-empty interior. Then $ Y=X:$ For some $r>0$ we have $B(0,r)\subset Y,$ but then ( because $Y$ is a vector space), $Y\supset \cup_{n\in \Bbb N}\{nv: v\in B(0,r)\}=\cup_{n\in \Bbb N}B(0,nr)=X.$

The reason $B(0,r)\subset Y$ for some $r>0$ is that for some $y\in Y$ and some $r>0$ we have $B(y,r)\subset Y,$ and since $Y$ is a vector space we have $Y\supset \{y'-y:y'\in B(y,r)\}=B(0,r). $

(2). $W$ is a vector subspace of $C[0,1/2],$ so to prove that $W$ has empty interior in $C[0,1/2]$, it suffices to show $W\ne C[0,1/2].$ Let $f(t)=|t-1/4|$ for $t\in [0,1/2]$. Then $f\in C[0,1/2],$ but $f$ is not a polynomial because $f(t)$ is not differentiable at $t=1/4$.

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