# Isometry and isomorphism normed spaces

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.

• Lots of $missing. Apr 14 '16 at 12:47 • Here I corrected typing errors, that you now can someone help me. Apr 14 '16 at 19:23 • Isn't this basically Mazur-Ulam theorem? Apr 15 '16 at 16:28 • I have tried to correct some problems you had with MathJax. Coud you please edit your question further, to get what you want to say. Apr 15 '16 at 16:30 • Thanks, I'll try on the basis of proof Mazur-Ulam theorem, solve the problem. If I can, I'll set it to StackEschange. Apr 15 '16 at 23:15 ## 1 Answer I apologize, in setting of problem i have arisen typographical errors. I will repeat text. Problem. Let$X$,$Y$real normed vector spaces and$f$isometry of space$X$in the space$Y$. Show that there is isomorphism A of spaces$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$. Each isometrics is injections, because for$x1,x2 \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 $We need to finde isomorphism$A$space$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.

• As explained on meta it is better to edit your original post than posting the question once again in an answer. Apr 15 '16 at 16:28
• Thank you, from now on I will do so. Apr 15 '16 at 23:41