Problem. Let $X$, $Y$ be real normed vector spaces and $ f $ isometry space $ X $ in the space $ Y $. Show that there is isomorphism $ A $ spaces $ X $ on the space $ Y $ and vector $ c \in Y $ such that $ f (x) = Ax + c $ for each $ x \in X $.
I know that is true
If $ X $ and $ Y $ normed vector spaces, linear operator $ f: X \to Y $ is called the isometrics if true $ \| F (x) \| = \| x \| $ for each vector $x \in X$. Obviously, then the operator f is bounded and $\|f\|=1$. Each isometrics is injections, because for $ x_1, x_2 \in X $ we have that is worth $$fx_1 = fx_2 \longrightarrow \| x_1-x_2 \| = \| f (x_1-x_1) \| = \| fx_1-fx_2 \| = 0 \longrightarrow x_1-x_2 = 0 \longrightarrow x_1 = x_2 $$
We need to prove that the $ A $ isomorphism space $ X $ on the space $ Y $ ie. We need to show that the $ A $ 1) is a linear operator, 2) bijection. Also, we need to show the existence of a vector c.
But, I have now no idea how to do it.
Does someone can help me? Thanks.