Let $(X,\|\cdot\|_X)$ and $(Y,\|\cdot\|_Y)$ be Banach spaces. Denote $T:X \rightarrow Y$ as an onto linear isometry.

For a convex subset $C$ of $X,$ we say that $x \in C$ is an extreme point of $C$ if for any $x^{\prime},x^{\prime \prime} \in C$ and any $0 < t < 1$ with $$x = tx^{\prime} + (1-t)x^{\prime \prime},$$ we must have $x = x^{\prime} = x^{\prime \prime}.$

Question: $x$ is an extreme point of a unit ball in $X$ if and only if $T(x)$ is an extreme point of unit ball in $Y$.

My attempt:

Clearly unit ball is convex, so the question makes sense.

Suppose that $x$ is an extreme point of unit ball $B_X$ in $X.$ Observe that $T$ is surjective from unit ball $B_X$ in $X$ to unit ball $B_Y$ in $Y.$ Suppose that for any $T(y), T(z) \in B_Y$ where $y,z \in B_X$ and $0 < t < 1$ with $$T(x) = tT(y) + (1-t) T(z).$$ Since $T$ is linear, we have $$T(x) = T(ty + (1-t)z).$$ Since an isometry is injective, in particular, it is injective on unit ball, so we have $$x = ty+(1-t)z.$$ Since $x$ is an extreme point of $B_X,$ we have $$x = y = z.$$ Therefore, $T(x) = T(y) = T(z).$

Since $T$ is onto and isometry, it is a bijection, thus $T^{-1}$ exists. One can show easily that $T^{-1}$ is an isometry.

I think the converse can be shown similarly as above.

Is my proof correct?

  • $\begingroup$ Everything looks alright. $\endgroup$ – Aweygan Jul 7 '17 at 18:42
  • $\begingroup$ As you mentioned, the inverse function of $T$ is also a linear isometric isomorphism, so proving one direction proves both, and there is no additional work for a "converse." $\endgroup$ – Jonas Meyer Jul 8 '17 at 0:25

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