If $f:X\to Y$ is linear and $y^\ast \circ \ f \in X^\ast$ for all $y^\ast \in Y^\ast$, then $f$ is continuous

Question

Let $X,Y$ be normed over $\mathbb{K}$ (which can be $\mathbb{R}$ or $\mathbb{C}$). If $f:X\to Y$ is linear and $y^\ast \circ \ f \in X^\ast$ for all $y^\ast \in Y^\ast$, then $f$ is continuous.

I came up to a solution to this problem which I know is false, but I can't spot where is the wrong argument. The problem is that my solution only need to use one $y^\ast \in Y^\ast$ to prove $f$ is continuous. I want to know what I'm doing wrong and how is the correct argument to solve this problem. If it makes things better you may consider balls instead neighborhoods in all the arguments below.

Here is my "solution": Given any neighborhood $V$ of $0\in Y$, it's enough to show there is a neighborhood $U$ of $0\in X$ such that $f(U) \subset V$. Start considering any neighborhood $A$ of $0\in\mathbb{K}$ and any $y^\ast \in Y^\ast$, we know $(y^\ast)^{-1}(A)$ is a neighborhood of $0\in Y$, for $y^\ast$ is linear and continuous. Therefore the collection $\{\frac{1}{n}(y^\ast)^{-1}(A)\}_{n\in\mathbb{N}}$ is a local base for $Y$. In particular, exists $n\in\mathbb{N}$ such that $\frac{1}{n}(y^\ast)^{-1}(A) \subset V$.

PS: in this context, local base is a collection of neighborhoods of $0\in Y$ such that, for any neighborhood $W$ of $0\in Y$, there is member in this collection contained in $W$.

We have that $y^\ast\circ f$ is also continuous, so there is a neighborhood $U$ of $0\in X$ such that $y^\ast( f(U)) \subset A$, so we get $f(U)\subset (y^\ast)^{-1}(A)$. Now note that

$$f\Big(\frac{1}{n}U\Big) = \frac{1}{n}f(U) \subset \frac{1}{n}(y^\ast)^{-1}(A) \subset V.$$

Therefore $\frac{1}{n}U$ is the neghborhood of $0\in X$ we are looking for, and from this we can conclude $f$ is continuous.

Answer 1

No, $\{\frac{1}{n}(y^\ast)^{-1}(A)\}_{n\in\mathbb{N}}$ is not a local basis of neighborhoods of $0$ in $Y$. $\ker y^\ast$ is a subspace of $Y$ of codimension $1$, and the entire kernal is contained in every set of that collection.

That is why you need more than one $y^\ast$.

Answer 2

If $X$ and $Y$ are Banach spaces continuity follows from the closed graph theorem. In the normed case you have to show that $f$ maps bounded sets to bounded sets and this follows from the weak continuity since the norm topology and the weak topology on $Y$ have the same bounded sets.