I cannot understand one small implication in such theorem
Suppose
(a) $M$ is a subspace of a real vector space $X$
(b) $p : X \to \mathbb{R}$ satisfies $$ p(x+y) \leq p(x) + p(y) > \;\text{and}\; p(tx) = tp(x) $$ if $x \in X$ and $y \in X, t \geq 0$
(c) $f : M \to \mathbb{R}$ is linear and $f(x) \leq p(x)$ on $M$.
Then there exists a linear $\Lambda : X \to \mathbb{R}$ such that $$ \Lambda x = f(x) \; , x \in M $$ and $$ -p(-x) \leq \Lambda x \leq p(x) \;,\; x \in X $$
Proof:
If $M \neq X$ we pick $x_1 \in X$,$x \notin M$ and define $$ M_1 = \left\{ x + t x_1 : x \in M, t \in \mathbb{R}\right\} $$ we have $M_1$ is a vector space. Moreover we can derive $$ f(x) + f(y) \leq p(x - x_1) + p(y - x_1) $$ We define also $\alpha$ as the least upper bound of $f(x) - p(x - x_1)$ (by varying $x \in M$. We can write the two inequalities
$$ \begin{array}{ll} f(x) - \alpha \leq p(x - x_1) & (2)\\ f(x) + \alpha \leq p(y - x_1) & (3) \end{array} $$
And on $M_1$ we define $$ f_1(x + tx_1) = f(x) + t\alpha \;\;\; (4) $$
There's this specific bit I don't get
Take $t > 0$ replace $x$ by $t^{-1}x$ in (2) and $y$ by $t^{-1}y$ in (3), and multiply the resulting inequalities by $t$. In combination with (4) this proves that $f_1 \leq p$ on $M_1$.
How exactly doing the mentioned steps we get $f_1 \leq p$ on $M_1$? Doing the steps mentioned I got
$$ f_1(x - t\alpha) + f_1(y + t \alpha) = f_1(x + y) \leq p(x - tx_1) + p(y + t x_1), $$
But I still struggle to arrive at the same conclusion
Thank you
You are unnecessarily adding the two inequalities. If you consider them separately and look at the cases $t>0$ and $t<0$ you will get $f_1 \leq p$ easily.