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Let $(E,\|.\|_{E})$ and $(F,\|.\|_{F})$ be two normed spaces and let $f:E\longrightarrow F$ a linear map such that $\|f\|=1$. I don't know if this means that $f$ is an isometry $(\|f(x)\|_{F}=\|x\|_{E}, \,\, \forall x\in E)$. Thanks

Kuba Helsztyński
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Serges
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2 Answers2

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It does not. Consider the function $f : \mathbb{R}^2 \rightarrow \mathbb{R}$ defined by $f(x,y)=x$, with both $\mathbb{R}^2$ and $\mathbb{R}$ endowed with the $\| \|_\infty$ norm. Then clearly $\|f\|=1$. However $f$ is not an isometry (it is not even invertible : $f(0,y)=0$ for all $y$).

Albert
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  • If we suppose that E=F, is the result is true or is there any counterexample? – Serges May 29 '13 at 03:33
  • the same counterexample can be adapted : just take $f(x,y)=(x,0)$. Even if the map is invertible, it still doesn't work : take $f(x,y)=(x,y/2)$. – Albert May 29 '13 at 10:08
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There are plenty of endomorphisms of norm $1$ which are not injective. I give an example.

Let $(E,\|\cdot\|_E)=(F,\|\cdot\|_F)=(\ell_\infty, \|\cdot\|_\infty)$ and consider left shift operator $T\colon \ell_\infty\to\ell_\infty$ defined by $T(x_1,x_2,\ldots)=(x_2,x_3,\ldots)$. Clearly, $\|T\|=1$ but there is a whole subspace which goes to $0$ under $T,$ namely $\{(c,0,0,\ldots)\colon c\in \mathbb{K}\}$. Hence not only you have a vector ${\bf v}\in\ell_\infty$ s.t. $$\|T{\bf v} \|_\infty\neq\|{\bf v}\|_\infty$$ but you can even pick it so its norm equals your favourite number, still positive though.

Kuba Helsztyński
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