Let $X$ be a linear space and let $\|.\|_1$, $\|.\|_2$ be two norms on $X$ such that $X_i:=(X,\|.\|_i)$ is separable for $i=1,2$. Then also $Y:=(X,\|.\|_1+\|.\|_2)$ is separable.
I tried the following: we know that then $X_1\times X_2$ equipped with the same norm as the norm of $Y$ is separable and $Y$ can be seen as a subspace of this cartesian product, hence separable.
Is this approach usable ? If yes what are the details of the proof ?
Thanks for your help.
Yes, I think your idea works.
Let $X_1 = (X, ||\cdot||_1)$ and $X_2 = (X, ||\cdot||_2)$ be separable normed vector spaces. They are metric spaces as well by equipping them with induced metric. Therefore, we can form the product metric space $X_1 \times X_2$ with the $\ell_1$-norm such that $$d_{X_1 \times X_2}((a_1, a_2), (b_1, b_2)) = ||b_1 - a_1||_1 + ||b_2 - a_2||_2$$ Let $Y = (X, ||\cdot||_1 + ||\cdot||_2)$ be another normed vector space. Its induced metric is $$d_Y(a, b) = ||b - a||_1 + ||b - a||_2$$ and thus $Y$ can be embedded as a metric subspace via the diagonal map $y \mapsto (y, y)$.
Finally, because product of two separable metric spaces is separable and every subspace of a separable metric space is separable, $Y$ is separable.
