A question about the proof of Hahn-banach theorem

Question

I am reading the proof of Hahn-banach on Bass(https://www.math.wustl.edu/~victor/classes/ma5051/rags100514.pdf) Theorem 18.5. When proving that the functional can be extended for 1 more dimension, why "we need to verify that the norm of f_1 is less than or equal to 1" instead of equal to 1, given that the statement of Theorem 18.5 says the extended functional has the same operator norm?

Answer 1

The Hahn-Banach Theorem as presented in your attached document:

If $M$ is a subspace of a normed linear space $X$ and $f$ is a bounded real linear functional on $M$, then $f$ can be extended to a bounded linear functional $F$ on $X$ such that $\|F\| = \|f\|$.

At the relevant point in the proof, you are assuming that $\|f\|=1$. $M_1$ is defined to be the linear span of $M$ and $\{x_0\}$, and the extension $f_1$ to $M_1$ is defined by $f_1(x+\lambda x_0):=f(x)+\lambda\alpha$.

So for each $\epsilon>0$ there exists $x\in X$ with $\|x\|\leq 1$ and $|f(x)|>1-\epsilon$. Then $x\in M_1$, $\|x\|\leq 1$ and $|f_1(x)|=|f(x)|>1-\epsilon$ gives that, so $\|f_1\|>1-\epsilon$. Since this holds for arbitrary $\epsilon>0$, we have that $\|f_1\|\geq 1$. Now it suffices to prove that $\|f_1\|\leq 1$.