I'm trying to solve
Let consider the space $\Bbb{R}^2$ with the norm $\|(x,y)\|_p=(|x|^p+|y|^p)^{1/p}$ and let $Z$ be the subspace $$Z=\{(t,mt);t\in\Bbb{R}\},$$ for some $m\in\Bbb{R}$. Let $f\colon Z\to\Bbb{R}$ be defined by $f(t,mt)=t$.
(a) For $1<p<\infty$ find the unique linear extension $\tilde{f}$ of $f$ defined over $\Bbb{R}^2$ such that $\|\tilde{f}\|=\|f\|$.
(b) For $p=1$, find two distincts extensions of $f$. Do the same for $p=\infty$.
I'm stucked in (a), and neither try (b) yet. I was thinking that the extension which works in (a) is $\tilde{f}(x,y)=y/m$. But I'm with big difficult to see what to do with the norms.
For any $(t,mt)\in Z$, we have that $$\|(t,mt)\|_p=|t|(1+|m|^p)^{1/p},$$ and $$|f(t,mt)|=|t|\leqslant|t|(1+|m|^p)^{1/p}=\|(t,mt)\|_p,$$ hence $\|f\|\leqslant1$. So the candidate to be the norm of $f$ is $1$. But I can't find the vector in $Z$ that does the work, neither usying some sequence in $Z$ that makes the norm be $1$.
So, I'm thankful for any help.
First of all, as it has correctly been observed, $\|f\|_{Z^*}=\frac{1}{(1+|m|^p)^{1/p}}$.
Note that $Z$ is a line that passes through the origin. Take the vertical line to $Z$ which is the subspace $W=\{(-mt,t):t\in\mathbb{R}\}$. In order to define an extension of $f$ on $\mathbb{R}^2$ it suffices to define a value for the extension at the point $(-m,1)$, because $(-m,1)$ and $(1,m)$ are linearly idependent and span $\mathbb{R}^2$. The extensions are in one to one correspondence with the possible values of $(-m,1)$, so let's take an extension that assigns the value $c\in\mathbb{R}$ to the point $(-m,1)$. Let's denote this (unique) extension by $F_c$, so $F_c(-m,1)=c$ and $F_c(1,m)=f(1,m)=1$.
Write a point $(x,y)$ of $\mathbb{R}^2$ as a linear combination of $(1,m)$ and $(-m,1)$, say $(x,y)=a\cdot(1,m)+b\cdot(-m,1)$. We have to solve the system (solve for $a,b$) $$a-mb=x$$ $$ma+b=y$$ and by Cramer's rule we get that $a=\frac{x+my}{1+m^2}$ and $b=\frac{y-mx}{1+m^2}$. Therefore we have that $$F_c(x,y)=\frac{x+my}{1+m^2}F_c(1,m)+\frac{y-mx}{1+m^2}F_c(-m,1)=\frac{x+my+c(y-mx)}{1+m^2}$$ Now $$\|F_c\|=\sup_{(x,y)\neq0}\frac{|F(x,y)|}{\|(x,y)\|}=\frac{1}{1+m^2}\cdot\sup_{x,y\neq0}\frac{|(1-mc)x+(c+m)y|}{(|x|^p+|y|^p)^{1/p}}\;\;\;\;(\star).$$
Question 1
Let $q$ be the conjugate exponent of $p$, i.e. $1/p+1/q=1$. Then Using the H"older inequality we have that $|(1-mc)x+(c+m)y|\leq\|(1-mc,c+m)\|_q\cdot\|(x,y)\|_p$, so $$\|F_c\|\leq\frac{\|(1-mc,c+m)\|_q}{1+m^2}$$. On the other hand, take the vector $(x,y)=((1-mc)\cdot|1-mc|^{q-2},(c+m)\cdot|c+m|^{q-2})$. Note that $\|(x,y)\|_p=\|(1-mc,c+m)\|_q^{q/p}$ so $$\|F_c\|\geq\frac{|F_c(x,y)|}{\|(x,y)\|_p}=\frac{1}{1+m^2}\cdot\frac{\|(1-mc,c+m)\|_q^q}{\|(1-mc,c+m)\|_q^{q/p}}=\frac{\|(1-mc,c+m)\|_q}{1+m^2} $$ So we conclude that $$\|F_c\|=\frac{\|(1-mc,c+m)\|_q}{1+m^2}$$ Now in order to have an isometric extension we must choose $c$ so that $\|F_c\|=\|f\|_{Z^*}$, i.e. $$\frac{\|(1-mc,c+m)\|_q}{1+m^2}=\frac{1}{(1+|m|^p)^{1/p}},$$ or equivalently $$|1-mc|^q+|c+m|^q=\bigg(\frac{1+m^2}{(1+|m|^p)^{1/p}}\bigg)^{q} $$ So one has to verify that this equation has a unique solution for $c\in\mathbb{R}$.
Question 2
Suppose that $p=1$. Then the obvious inequality $|(1-mc)x+(c+m)y|\leq\max\{|1-mc|,|c+m|\}\cdot(|x|+|y|)$ together with $(\star)$ for $p=1$ yield $$\|F_c\|\leq\frac{\max\{|1-mc|,|c+m|\}}{1+m^2}.$$ On the other hand, applying $F_c$ at $(1,0)$ and at $(0,1)$ yield $\|F_c\|\geq\frac{|1-mc|}{1+m^2}$ and $\|F_c\|\geq\frac{|c+m|}{1+m^2}$, so we have that $$\|F_c\|=\frac{\max\{|1-mc|,|c+m|\}}{1+m^2}.$$
we want the extension to be isometric, so we want $\|F_c\|=\|f\|_{Z^*}=\frac{1}{1+|m|}$, so we want $$\frac{\max\{|1-mc|,|c+m|\}}{1+m^2}=\frac{1}{1+|m|}$$
so one has to verify that this equation has at least two solutions for $c\in\mathbb{R}$.
Suppose that $p=\infty$. In that case $\|f\|_{Z^*}=\frac{1}{\max\{1,|m|\}}$. Also, $$|F_c(x,y)|=\frac{|(1-mc)x+(c+m)y|}{1+m^2}\leq\frac{(|1-mc|+|c+m|)\max\{|x|,|y|\}}{1+m^2}$$ so $\|F_c\|\leq\frac{|1-mc|+|c+m|}{1+m^2}$. On the other hand, $$\|F_c\|\geq\frac{|x(1-mc)+y(c+m)|}{1+m^2}$$ for all $x,y\in\{-1,1\}$. Depending on the order of $1,m,c$ we can choose proper values of $x,y\in\{-1,1\}$ to establish the inequality $$\|F_c\|\geq\frac{|1-mc|+|c+m|}{1+m^2}$$ so we conclude that $$\|F_c\|=\frac{|1-mc|+|c+m|}{1+m^2}.$$ We want an isometric extension, so we are looking for the values of $c$ so that $$\frac{|1-mc|+|c+m|}{1+m^2}\equiv\|F_c\|=\|f\|_{Z^*}\equiv\frac{1}{\max\{1,|m|\}}$$
and one has to verify that this equation has at least two solutions for $c\in\mathbb{R}$.
A final comment: This seems too complicated to me. If I were you I would only care for $m=0$, since any other line can be obtained from the canonical axis system with a mere rotation. Anyway, I am unsure how one would solve the equations that arise in each case, all those absolute values make things somehow complicated. I believe the idea and the norm computations are correct though. If I have any mistakes, please point them out.