About a technique used in the proof of Hahn-Banach Theorem Recall Hahn-Banach (cf. Kreyszig's book) : If $X$ is a real
vector space with a sublinear functional $p$ and if
  $f$ is linear on a subspace $Z$ with $p(z)\geq f(z),\ z\in Z$,
  then there exists an extension $F$ of $f$ on $X$ s.t. $$
 \ p(x)\geq F(x),\ x\in X,\quad\rm{and}\tag{1} $$
$$F(z)=f(z),\ z\in Z.\tag{2}$$
Try and Question : In the proof there exists following
technique :
For $y_1\in X-Z$, define $$ F(y+ay_1):=f(y)+ac $$
This is linear. So we must show (1). To do this, in the book we have
$$ f(y)-f(z)\leq p(y+y_1)+ p(-y_1-z) $$
for all $y,\ z\in Z$. This inequality and remaining part of the proof
 is computational. But I do not know how we consider such inequality
 ? I have a question : How do we make this inequality ? Subsequent argument implies that we must take linearity from a norm (Note that norm is sublinear). Is there more approachable proof ? 
So have $$ f(y)-p(y+y_1)\leq p(-y_1-z)+f(z) )$$
If sup of lefthand side is $m_0$ and inf of righthand side is $m$,
then we have $$ m_0 \leq c\leq m $$
That is, when we extend $f$, we must find suitable bounds $m_0,\
m_1$ :
(1) $X=l^2,\ Z=(e_1)\oplus \cdots \oplus (e_{n-1})$ and $
f(x)=\frac{1}{\sqrt{n}}
                      \sum_{i=1}^{n-1} x_i$ Then we have extensions : $$
                      F_1(x) = \frac{1}{\sqrt{n}}
                      \sum_{i=1}^{n} x_i,\ F_2(x)=\frac{1}{\sqrt{n}}
                      \sum_{i=1}^{n-1} x_i$$
(2) $X=l^\infty,\ Z=(e_1)$ and $ f(x)=x_1$ Then we can have only one
extension $$ F(x)=x_1
$$
In these example, wrt $X$, we have bounds $m_0,\ m_1$. The book,
without detailed calculations or concrete example, gives a concise
proof. Can you explain how we consider the above inequality ? Thank
you.
 A: The way the inequality arises is that you must be able to assign $f(x)$ a value, say $\gamma$, such that the following hold for all $z_1, z_2 \in Z$:
$$
                         f(z_1+x) = f(z_1)+\gamma \le p(z_1+x),\\
                         f(z_2-x) = f(z_2)-\gamma \le p(z_2-x).
$$
That there exists such a $\gamma$ is equivalent to there existing an extension of $f$ to the linear space spanned by $Z$ and $x$ for which $f$ remains bounded by $p$. Does there exist such a $\gamma$? Equivalently, is there a $\gamma$ such that the following holds for all $z_1, z_2 \in Z$?:
$$
                f(z_2)-p(z_2-x) \le \gamma \le p(z_1+x)-f(z_1).
$$
The existence of $\gamma$ is implied if the following holds for all $z_1,z_2\in Z$:
$$
                       f(z_2)-p(z_2-x) \le p(z_1+x)-f(z_1),\\
                        f(z_1)+f(z_2) \le p(z_1+x)+p(z_2-x).
$$
This last inequality is equivalent to yours if $z_2$ is replaced by $-z_2$.
And, of course, the last inequality holds because
$$
               f(z_2+z_1) \le p(z_2+z_1)=p(z_1+x+z_2-x) \le p(z_1+x)+p(z_2-x).
$$
