Is there a normed space which is isometric to a proper subspace of itself?

Question

Already that the normed vector space here cannot be real number R as every isometry from R will sent to R itself. Cannot be N either as N itself is not a normed vector space at all. Then how can I approach the problem?

Answer 1

Note that a normed space is necessarily a vector space. The Mazur-Ulam theorem tells us that for any isometry $f:X \to Y$, the function $g:X \to Y$ defined by $g(x) = f(x) - f(0)$ will be a (bijective) linear isometry. That is, any two isometric normed spaces will necessarily be isomorphic as vector spaces. Using this fact, we can conclude that if $X$ is a finite-dimensional normed space, then no proper subspace $Y$ of $X$ will be isometric to $X$. In particular, we would necessarily have $\dim(Y) < \dim(X)$, which means that $X$ and $Y$ are not isomorphic as vector spaces and are therefore not isometric as normed spaces.

So, to get a suitable example, we need an infinite dimensional normed space. Probably the most intuition-friendly of these spaces is the sequence space $\ell^2$. In particular, the elements of $\ell^2$ are sequences $x = (x_i)_{i \in \Bbb N} = (x_1,x_2,x_3,\dots)$. More specifically, we have $$ \ell^2 = \left\{(x_i)_{i \in \Bbb N} : \sum_{i=1}^\infty x_i^2 < \infty\right\}. $$ We define a norm on this set by $\|x\| = \sqrt{\sum_{i=1}^\infty x_i^2}.$ You might find it a useful exercise to show that $\ell^2$ satisfies the definition of a normed space.

With that established, consider the space $X = \ell^2$ and its subspace $Y$ defined by $$ Y = \{(x_i)_{i \in \Bbb N} \in X : x_1 = 0\}. $$ Verify that $Y$ is a subspace of $X$ and that there is indeed an isometry between these two spaces.