$c_0$ is not isometric to $c_0 \oplus c_0$ $c_0$ 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.
 A: 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.
