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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.

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    It's not obvious that $\Gamma$ is an isometry. If you're doign this naively (for each $\varphi$ separately, use Hahn-Banach to find an $\overline{\varphi}$), then the map has no reason to be linear, You know that $d(0,\overline{\varphi}) = d(0, \varphi)$, but that doesn't imply $d(\overline{\varphi}_1,\overline{\varphi}_2) = d(\varphi_1, \varphi_2)$ (in the same way that an arbitrary permutation of the unit sphere is not an Euclidean isometry). – 2018-06-12
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    On the other hand, you can try to find $\overline{\varphi}$ coherently, maybe using a Hamel basis, so that $\Gamma$ is an isometry. But then, linearity will be obvious by construction. – 2018-06-12
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    You are right, it's not clear if it is an isometry at all, i'll edit the question,... but the point is that anyway in a strict convex Banach space, for each $\varphi$ there is a unique $\overline\varphi$ that preserves the norm (so there aren't choices to make) and then the question about linearity remains open... – 2018-06-12
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    For uniqueness of the extension, isn't $X'$ the space you want to be strictly convex? – 2018-06-12
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    I'll edit again :) – 2018-06-12
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    The extension is certainly not linear (unless you are in the Hilbert space setting or your $Y$ is in a very special position with respect to the unit ball of $X$). Just think geometrically about how the level hyperplanes of $\bar\varphi$ should be tilted with respect to $Y$ in the case when $X=\mathbb R^3$ with some fancy norm and $Y$ is a $2$-dimensional plane. – 2018-06-14

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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 $p\neq 2$.