The possible number of elements in a convex set Let $(X,\|\cdot \|)$ be a normed space over $\mathbb{K}$ ($\mathbb{R}$ or $\mathbb{C}$), and let $M\subseteq X$ be a subspace. 
Let $\phi\in \mathcal L(M,\mathbb{K})$, and let $A$ be the set of all bounded linear functionals $f$ such that $f|_M=\phi$ and $\|f\|=\|\phi\|$. What is the possible number of elements in $A$? Hint: Show that $A$ is a non-empty convex set, and use it to determine the possible number.
What I did: First, I have proven that $A$ is a non-empty convex set. I am not sure how to determine the possible number. Pick $f_1,f_2\in A$ and let $\alpha\in [0,1]$, and write $f_3=\alpha f_1+(1-\alpha) f_2$.  If $f_1=f_2$, then $f_3=f_1$ is the only map in $A$, ie. it is unique, so $A$ must be finite if each element in $A$ are equal to some other. If $f_1\neq f_2$, then $f_3=\alpha f_1+(1-\alpha) f_2$ belongs to $A$ and it is not unique. There are infinitely many such maps, so $A$ is infinite, if at least one of the elements in $A$ is not equal to the rest. 
If my reason is correct, is there a better/rigorious way to formulate?
 A: Fix $f \in \mathcal{L}(M, \mathbb{K})$ and let $A$ be the set of Hahn-Banach extensions of $f$ onto $X$.
$A$ is non-empty: this is a direct consequence of the Hahn-Banach Theorem.
To show that $A$ is convex, consider $f_1, f_2 \in A$ and $\alpha \in [0,1]$. Let's show that $$f_3 = \alpha f_1 + (1-\alpha) f_2 \in A$$
$f_3$ is obviously linear. It extends $f$ on $M$:
$$f_3|_M = (\alpha f_1 + (1-\alpha) f_2)|_M = \alpha f_1|_M + (1-\alpha) f_2|_M = \alpha f + (1-\alpha) f = f$$
$f_3$ is bounded as a linear combination of bounded functionals. We have:
$$\|f_3\| = \|\alpha f_1 + (1-\alpha) f_2\| \le \alpha\|f_1\| + (1-\alpha)\|f_2\| = \alpha\|f\| + (1-\alpha)\|f\| = \|f\|$$
$$\|f_3\| = \sup_{x\in X, \|x\| = 1} |f_3(x)| \ge \sup_{x\in M, \|x\| = 1} |f_3(x)| = \sup_{x\in M, \|x\| = 1} |f(x)| = \|f\|$$
Hence $\|f_3\| = \|f\|$ so $f_3 \in A$. We conclude that $A$ is convex.
Now, $A$ can have one element, this happens when the Hahn-Banach extension of $f$ is unique. For example the zero functional $0 : M \to \mathbb{K}$ has a unique extension, namely $0 : X \to \mathbb{K}$.
On the other hand, $A$ can have at least two distinct elements $f_1$ and $f_2$. We showed that $\alpha f_1 + (1- \alpha) f_2$ is then also a Hahn-Banach extension of $f$, for every scalar $\alpha \in \mathbb{K}$. We conclude that $A$ is in that case infinite (furthermore, it is uncountable).
For an example, consider the subspace $$Y=\{(x,y) \in \mathbb{R^2}:x+y=0\} \le \mathbb{R}^2$$ and the functional $f : Y \to \mathbb{R}$ defined as $f(x,y) = y$ for $(x,y) \in Y$.
You can check that the functionals $f_1, f_2 : \mathbb{R}^2 \to \mathbb{R}$ defined as $f_1(x,y)=y$ and $f_2(x,y)=-x$ for $(x,y) \in \mathbb{R}^2$ are both Hahn-Banach extensions of $f$. Our discussion above shows that any convex combination of $f_1$ and $f_2$ is also a Hahn-Banach extension of $f$.
