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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.

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  • $\begingroup$ Lots of $ missing. $\endgroup$ Apr 14 '16 at 12:47
  • $\begingroup$ Here I corrected typing errors, that you now can someone help me. $\endgroup$
    – J. Jam
    Apr 14 '16 at 19:23
  • $\begingroup$ Isn't this basically Mazur-Ulam theorem? $\endgroup$ Apr 15 '16 at 16:28
  • $\begingroup$ 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. $\endgroup$ Apr 15 '16 at 16:30
  • $\begingroup$ Thanks, I'll try on the basis of proof Mazur-Ulam theorem, solve the problem. If I can, I'll set it to StackEschange. $\endgroup$
    – J. Jam
    Apr 15 '16 at 23:15
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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.

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

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