Closed subset of polynomials in function space

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?

• 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.) Jan 2, 2019 at 23:13
• A proper subspace of a normed space is never open. In fact, it even has empty interior. Jan 3, 2019 at 12:16

$$\|g\| < \delta \sum_{n=1}^{\infty} \frac 1 {2^{n}}= \delta$$. 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 $$n$$].
(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 $$t=1/4$$.