About linearity in Hahn-Banach extension under strict-convex hypothesis Let $X$ be a Banach space such that $X'$ is strictly convex. Let $Y$ be a closed proper subspace of $X$.
Then: $$\forall \varphi \in Y', \exists!\overline\varphi \in X', (\overline\varphi_{|Y}=\varphi) \land(\lVert\overline\varphi\rVert=\lVert\varphi\rVert).$$
Define:
$$\Gamma:Y'\rightarrow X', \varphi \mapsto\overline\varphi.$$
Then $\Gamma$ is norm preserving but it's not clear if it is an isometry and, also if this is the case, the lack of surjectivity prevents us from using Mazur-Ulam theorem to show that $\Gamma$ is linear... so the question: is $\Gamma$ linear (and then, being norm preverving, also an isometry)? Thanks in advance.
 A: The function $\Gamma$ is not linear.
We can construct the following counterexample.
Note that most of the details and calculations are not explained and only a rough sketch of the whole thing is given.
We choose $X=\mathbb R^3$ and define
$$
\varphi(x) = x_1+ x_2,
\quad
\psi(x) = x_1+ x_3,
\quad
\lambda(x) = x_1+x_2+x_3.
$$
Also, we define $Y=\ker \lambda$.
As a norm, we use the $\ell^p$-Norm on $\mathbb R^3$ for $p\in (1,\infty)$.
It can be calculated that
$$
\| \varphi\|_{Y'}=\|\psi\|_{Y'}
= 2 ( 2+2^p )^{-\frac1p}.
$$
and
$$
\| \varphi+\psi\|_{Y'}
= \| \varphi+\psi-\lambda\|_{Y'}
= 2(2^p+2)^{-\frac1p}.
$$
Next, it follows from the definitions, that $\Gamma\varphi = \varphi+\alpha\lambda$
and $\Gamma\psi = \psi+\alpha\lambda$.
for some $\alpha\in\mathbb R$.
We have
$$
\|\Gamma\psi\|_{X'}
=
\|\Gamma\varphi\|_{X'}
= \| (1+\alpha,1+\alpha,\alpha) \|_{\ell^q}
= ( 2|1+\alpha|^q+|\alpha|^q)^{\frac1q}
$$
where $q$ defined via $1=1/p+1/q$.
Hence, $\alpha$ has to be chosen in such a way that
$$
( 2|1+\alpha|^q+|\alpha|^q)^{\frac1q}
= 2 ( 2+2^p )^{-\frac1p}.
$$
It can be seen that this is the case if
$\alpha = -2^{p-1}(2^{p-1}+1)^{-1}$.
We also have
$$
\|\Gamma\varphi+\Gamma\psi\|_{X'}
= \| (2+2\alpha,1+2\alpha,1+2\alpha) \|_{\ell^q}
= ( |2+2\alpha|^q + 2|1+2\alpha|^q )^{\frac1q}.
$$
Now we assume that $\Gamma(\varphi+\psi)=$.
This implies
$$
( |2+2\alpha|^q + 2|1+2\alpha|^q )^{\frac1q}
= 2(2^p+2)^{-\frac1p}.
$$
After some rearranging and substituting $\alpha$, this is equivalent to
$$
2(2^{p-1}-1)^q+2^q = 2^p+2.
$$
This is true for $p=2$, but it is false when we choose $p=3$.
In fact, if you plot these functions, you can see that The left-hand side is always larger than $2^p+2$ if $p\neq 2$.
