A bounded real linear functional on a subspace of a real normed space extends with the same norm

Define $p:X\to \mathbb{R}$ by $p(x):=\mathrm{\parallel }{f}_0\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }\mathrm{.}$ The triangle inequality and real homogeneity of the norm make $p$ sublinear. For $m\in M$, [L3] and [L4] give $f_0(m)\le \mathrm{\mid }{f}_0(m)\mathrm{\mid }\le \mathrm{\parallel }{f}_0\mathrm{\parallel }\mathrm{\parallel }m\mathrm{\parallel }=p(m),$ so $f_0\le p$ on $M$.

By [L1], there exists a linear functional $F:X\to \mathbb{R}$ extending $f_0$ and satisfying $F(x)\le p(x)$ for every $x\in X$.

Apply step 2.1 to $x$ and to $-x$. Since $p(-x)=p(x)$, one gets $-\mathrm{\parallel }{f}_0\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }\le F(x)\le \mathrm{\parallel }{f}_0\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel },$ hence $\mathrm{\mid }F(x)\mathrm{\mid }\le \mathrm{\parallel }{f}_0\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }(x\in X)\mathrm{.}$ So $F$ is bounded in the sense of [L2], and [L3] gives $\mathrm{\parallel }F\mathrm{\parallel }\le \mathrm{\parallel }{f}_0\mathrm{\parallel }$.

Since $F{\mathrm{\mid }}_M=f_0$, for every $m\in M$ with $\mathrm{\parallel }m\mathrm{\parallel }\le 1$ one has $\mathrm{\mid }{f}_0(m)\mathrm{\mid }=\mathrm{\mid }F(m)\mathrm{\mid }\le \mathrm{\parallel }F\mathrm{\parallel }$. Taking the supremum over the unit ball of the normed subspace $M$ and using [L3] and [L4] yields $\mathrm{\parallel }{f}_0\mathrm{\parallel }\le \mathrm{\parallel }F\mathrm{\parallel }$.

Steps 3.1 and 3.2 give $\mathrm{\parallel }F\mathrm{\parallel }=\mathrm{\parallel }{f}_0\mathrm{\parallel }$, so $F$ is the required norm-preserving extension.