Doubt in the proof of Theorem 6.4 in Rudin, Functional Analysis, 1973

Question

Let $X$ a vector space and $(Y, \tau)$ a topological vector space such that $Y \subset X$. Consider $W \subset X$ such that $Y \cap W \in \tau$. If $f \in Y \cap W$, I have to prove that there exists an $\delta > 0$ such that $f \in (1- \delta) W$.

Edit 1: W is a convex and balanced set, that is, $\lambda W \subset W$, for all $|\lambda| \leq 1$.

Edit 2: Y is a vector subspace of $X$.

I don't know if this is true. This doubt appeared while I was studying the proof of Theorem 6.4 in Rudin, Functional Analysis, 1973. There, I think he takes $X = C^{\infty}_0(\Omega)$, $(Y,\tau) = \mathfrak{D}_K$ and $f = \phi - \phi_i$ :

Answer 1

In a topological vector space $Y$ over $\mathbb R,$ for any $v \in Y,$ the function:

$$p_v:\mathbb R\to Y, r\mapsto rv$$

is continuous, where $\mathbb R$ is given the standard topology.

So let $U=Y\cap W \in \tau$ and $f\in U.$ Then, by continuity $V=p_f^{-1}(U)$ is open in $\mathbb R$ and contains $1.$ So there is an $\epsilon>0$ such that $1+\epsilon \in V.$

Setting $\delta=\frac{\epsilon}{1+\epsilon},$ then $1-\delta=\frac{1}{1+\epsilon}.$

Since $p_f(1+\epsilon)=(1+\epsilon)f\in U,$ $$f=(1-\delta)(1+\epsilon)f\in (1-\delta)U.$$

So $f\in(1-\delta)U\subseteq (1-\delta)W.$

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We could break this into a lemma without any reference to $W$:

Lemma: If $Y$ is a topological real vector space and $U\subseteq Y$ is open, and $u\in U,$ then there exists a $\delta>0$ such that $u\in(1-\delta)U.$