Why is the linear functional $f$ bounded with $\|f\| \le 1/r $?

Question

$M$ is a closed subspace of the normed linear space $X$. Let $x_0 \in X\setminus M$ and $Z=M\oplus [x_0]$.
Define a linear functional $f: Z\rightarrow \mathbb{R} $:
$f(m+\lambda x_0)=\lambda$ $ \forall m \in M ,\lambda \in \mathbb{R}$
Since $M$ is closed, there is an $r>0$ for which:
$\|m+\lambda x_0\|=|\lambda |\|(-1/\lambda \cdot m)-x_0\|\ge |\lambda|r$
How can we infer from this, that $f$ with norm $\|f\|=sup[{|f(z)|:z\in Z,\|z\|\le1 }]$ is bounded with $\|f\|\le 1/r$?

Answer 1

If $\|m+\lambda x_0\|=1$, then $1\geqslant|\lambda|r$ and therefore, $|\lambda|\leqslant\frac1r$. On the other hand$$\bigl|f(m+\lambda x_0)\bigr|=|\lambda|\leqslant\frac1r.$$

Answer 2

We have

$$|f(m + \lambda x_0)| = |\lambda| \le \frac{1}{r}\|m+\lambda x_0\|$$

so $f$ is bounded and $\|f\| \le \frac1r$.