The following is a homework problem (I have solved the majority of it, but need help with the last part)
Suppose $f$ : $V \rightarrow \mathbb{F}$ is a non-zero continuous linear functional on a Banach space $V$ over $\mathbb{F}$.
I. Show that W = $\{x \in V : f(x) = 1\}$ is non-empty, closed and convex.
II. Show that $\inf_{x \in W} \|x\| = \frac{1}{\|f\|}$ III. Show that if the set $ Y = \{x \in V : \|x\| = 1\}$ is compact then there exists $x_0 \in W$ such that $\|x_0\|$ is equal to the infimum in part II.
I'm not sure how to do part III. My teacher suggested the following: consider a sequence ${x_n}$ in $W$ such that $\|x_n\|$ is decreasing to $\displaystyle \inf_{x \in W} \|x\|$. Then $\frac{x_n}{\|x_n\|}$ is a sequence in $Y$. Since $Y$ is compact, this sequence has some subsequence converging in $Y$, i.e. $\frac{x_{n_k}}{\|x_{n_k}\|} \rightarrow y \in Y$ Then since $f$ is continuous: $f\left(\frac{x_{n_k}}{\|x_{n_k}\|}\right) \rightarrow f(y)$ Using the fact that $f$ is linear and that $f(x_{n_k}) = 1$, $\frac{1}{\|x_{n_k}\|} \rightarrow f(y)$
The sequence $\frac{1}{\|x_{n_k}\|}$ is increasing to $\displaystyle \frac{1}{\inf_{x \in W} \|x\|}$, so $f(y) = \displaystyle \frac{1}{\inf_{x \in W} \|x\|}$.
I am not sure how to proceed from here. Ultimately I have to show that $\exists x_o \in W$ such that $\displaystyle \|x_0\| = \inf_{x \in W} \|x\|$, but I can't see where this comes from.