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As in the title, I'm trying to show that in a normed linear space $X$, if $K$ is convex then $\mathcal{D}_K(x)=\{y\in X\colon||y-x||=d(x,K)\}$ is convex.

My attempt : for $\alpha,\beta\in\mathcal{D}_K(x)$ and $t\in[0,1]$,

$$||t\alpha+(1-t)\beta-x||=||t(\alpha-x)+(1-t)(\beta-x)||\le t||\alpha-x||+(1-t)||\beta-x||=d(x,K)$$

I need to show the other direction : $$d(x,K)\le||t\alpha+(1-t)\beta-x||\forall t\in[0,1]$$

but I got stuck here. I'm not sure how to use the convexity of $K$ here. Any hint would be helpful.

Jimmy
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    I don't agree with your assertion: for $X=\Bbb R$, $K={0}$ and $x=1$ we have ${y\in\Bbb R,:, \lvert y-1\rvert=d(1,{0})}={0,2}$. –  Sep 18 '21 at 14:46
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    The set $D_K(x)$ as defined here is just the sphere centered at $x$ with radius $d(x,K)$, which is not convex (at least not for the usual definition in linear algebra) – Didier Sep 18 '21 at 14:48
  • oh yes that's true, but I got this problem in the problem book of real and functional analysis by Torchinsky, chapter $8$, problem $24$, Im not sure the problem may be wrong or some hypothesis may be missing there! – Jimmy Sep 18 '21 at 14:52

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As Saucy's comment points out, the claim in your question is false. However, if you change your definition of the set $\mathcal D_K(x)$ to $$\mathcal{D}_K(x)=\{y\in K\colon||y-x||=d(x,K)\},$$ then the claim becomes true. Indeed, for $\alpha,\beta\in\mathcal D_K(x)$ and $t\in[0,1]$, we have $t\alpha+(1-t)\beta\in K$, so \begin{align*} d(x,K)&\leq\|t\alpha+(1-t)\beta-x\|\\ &\leq t\|\alpha- x\|+(1-t)\|\beta-x\|\\ &=td(x,K)+(1-t)d(x,K)\\ &=d(x,K). \end{align*} It is thus my hypothesis that there is a typo in the book, and that the definition of $\mathcal D_K(x)$ should be this one, not the stated definition.

Aweygan
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