Suppose $X,Y$ are two real normed spaces,$T:X\to Y$ is the bijective map such that $||Tx+Ty+Tz|| = ||x+y+z||$ for any $x,y,z\in X$.Is $T$ a linear map?
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Not necessarily. – Rushabh Mehta Nov 08 '19 at 17:31
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Does there exist a counter-example? – math112358 Nov 08 '19 at 17:37
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I don't have an example off the top, but why do you believe this to be true? – Rushabh Mehta Nov 08 '19 at 17:38
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By assumption $$ \|T(x) + T(y) + T(-x-y)\| = 0, $$ so $$ T(x) + T(y) = -T(-x-y). $$ In particular, with $x=y=0$ it follows $T(0)=0$, and with $y=0$, $$ T(x) = -T(-x). $$ Hence $$ T(x) + T(y) = -T(-x-y) =T(x+y), $$ and $T$ is additive. Let me show that $T$ is continous. Let $x_n\to x$. Then $$ \|T(x) - T(x_n) \| = \|T(x) + T(-x_n)\|=\|x-x_n\|\to0. $$ So $T$ is continuous and additive, hence it is linear, see Continuous and additive implies linear
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I see -- that argument is for functions $\mathbb{R} \to \mathbb{R}$ but I see that you can adapt it without much difficulty to work in this case – user125932 Nov 08 '19 at 21:22