$c_0$ is the Banach space of sequences converging to zero and $c_0 \oplus c_0$ is its algebraical direct sum with itself equipped with the norm $\|(\xi,\eta)\| := \|\xi\|+\|\eta\|$.
How to prove that this spaces is not linearly isometric (though they are isomorhic)?
Thanks for any help.
This is my own solution. Comments are very welcome.
We know that each continuous linear functional on $c_0$ $f$ is of the form $x = (x_n) \mapsto \sum_{n=1}^\infty x_n p_n$ for $p = (p_n) \in \ell^1$ and $||f||=||p||=\sum_{n=1}^\infty|p_n|$. Thus $f$ clearly attains its norm on the unit ball iff $(p_n)$ is finitely supported. Such functionals form a linear subspace in $(c_0)^* = \ell^1$.
We will show that this is not true for $c_0 \oplus c_0$. If there was an isometry between $c_0$ and $c_0\oplus c_0$, it would induce an isometry between the dual spaces preserving the linear structure of set of norm-attaining functionals. So this will end the proof.
First, observe the structure of $(c_0 \oplus c_0)^*$ with the following
Lemma. For normed spaces $X$ and $Y$, $(X \oplus_1 Y)^*$ is isometric to $X^* \oplus_\infty Y^*$, where $\oplus_1$ and $\oplus_\infty$ equip the direct sums with the norms $||(x,y)|| := ||x||+||y||$ and $||(f,g)|| := \max \{||f||,||g||\}$, respectively.
Proof. Construct an isometry $T\colon (X \oplus_1 Y)^* \rightarrow X^* \oplus_\infty Y^*$ in that way: $Tf(x,y) = (f(x,0),f(0,y))$. Clearly $T$ is bijection. Then, $$ ||f||=\sup\limits_{||x||+||y||=1} f(x,y) = \sup\limits_{\substack{||x||=||y||=1 \\ \lambda,\mu \geq 0 \\ \lambda+\mu = 1}} f(\lambda x, \mu y) = \sup\limits_{\substack{||x||=||y||=1 \\ \lambda,\mu \geq 0 \\ \lambda+\mu = 1}} \lambda f(x,0) + \mu f(0,y) = \sup\limits_{\substack{\lambda,\mu \geq 0 \\ \lambda+\mu = 1}} \lambda ||f(\cdot,0)|| + \mu ||f(0,\cdot)|| = \max \{||f(\cdot,0)||,||f(0,\cdot)||\} = ||Tf|| $$ Q.E.D.
So each functional $f \in (c_0\oplus c_0)^*$ is of the form $(x,y) = ((x_n),(y_n)) \mapsto f(x,0)+f(0,y) = \sum_{n=1}^\infty x_n p_n + \sum_{n=1}^\infty y_n q_n$, where $p = (p_n), q = (q_n) \in \ell^1$ and $||f||=\max\{||p||,||q||\}$. The functional $f$ attains its norm on the unit ball of $c_0\oplus c_0$ iff one of $f(\cdot,0)$ and $f(0,\cdot)$ with greatest $(c_0)^*$-norm among them corresponds to finitely supported $\ell^1$-sequence. Finally, that set of norm-attaining functionals is not a subspace.