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Assume that $(X,\|\cdot\|)$ is a Banach space with $\|\cdot\|$ strictly convex. Define $S=\{x\in X:\|x\|=1\}$. Suppose that $\varepsilon>0$ and $x_0\in S$ and define $$ B_\varepsilon=\{x\in X:\|x-x_0\|\le\varepsilon\}\cap S. $$ Consider the following problem:

Question 1: Can we find $y\in X$ such that $\|y\|<1$ and any line which contains $y$ must intersect $B_\varepsilon$?

Now consider the following solution in the case where $X$ is finite dimensional:

Take a supporting hyperplane $H$ for the convex set $B=\{x\in X:\|x\|\le 1\}$ at the point $x_0$.

Claim: There exists $z\in X\setminus\{0\}$ such the set $$ H_z=\{h+z:h\in H\}, $$ has the property that $H_z\cap B_\varepsilon\neq \emptyset$ and $H_z\cap B_\varepsilon=H_z\cap S$.

Indeed, we claim that there exists $\delta>0$ such that $z=-\delta x_0$. Since $H$ is a supporting hyperplane it follows that if $\delta$ is small, then $H_{-\delta x_0}\cap B_\varepsilon\neq \emptyset$. Now suppose on the contrary that there exists a sequence $\delta_n>0$ such that $\delta_n\downarrow 0$ and a corresponding sequence $x_n\in H_{-\delta_n x_0}\cap S$ satisfying $x_n \notin B_\varepsilon$.

Once $x_n$ is bounded, we can assume without loss of generality that $x_n\to x\in H\cap S$. If $x=x_0$ we get a contradiction since $x_n\in B_\varepsilon$ for bigger $n$. If $x\neq x_0$, we also get a contradiction with the strictly convexity, by observing that the segment $\{tx_0+(1-t)x:\ t\in [0,1]\}$ should belong to $B_\varepsilon$. Therefore the Claim is true.

Now let $Q$ be the region enclosed by $H_z$ and $B_\varepsilon$.

It follows that if $y\in Q$, then any line which contains $x$ must intersect $B_\varepsilon$.

Question 2: What about the infinite dimensional case?

Remark 1: If $X$ is an infinite dimensional Banach space, then every $y$ which solves Question 1 belongs to the weak closure of $B_\varepsilon$.

Remark 2: Cross-posted: math.stackexchange.

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  • $\begingroup$ Could you explain how to get such a $z$ from just strict convexity? $\endgroup$ – Dirk Werner Dec 10 '18 at 22:22
  • $\begingroup$ In fact, in my proof here I were able to find $\delta>0$ in such a way that $z=-\delta x_0$ worked. But now I see that there is a gap here. Let me try to fix it. $\endgroup$ – Tomás Dec 11 '18 at 10:10
  • $\begingroup$ @DirkWerner, I can prove it only when $X$ is finite dimensional. A compactness argument do the work. No idea on how to tackle the infinite dimensional case. $\endgroup$ – Tomás Dec 16 '18 at 21:10
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If you look at the definition of a locally uniformly convex Banach space in, for example, my book Ostrovskii, Metric Embeddings, page 291, you will see that any point $x_0$ at which the condition of this definition is satisfied, has the property described in your Question 1. The converse does not seem to be true, as it seems that you need more than the negation of local uniform convexity, you need a kind of symmetric lack of the local uniform convexity.

As an example of a strictly convex Banach space without the property described in Question 1 I suggest the space $c_0$ with the norm $$||\{a_i\}_{i=1}^\infty||=\max_i|a_i|+\left(\sum_{i=1}^\infty |a_i|^2/2^i\right)^{1/2}.$$

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Actually, a somewhat weaker condition is sufficient, namely that $x_0$ is a strong extreme point, meaning that $\|z_n\|\to0$ whenever $\|x_0\pm z_n\|\to1$. This is the same as saying that for each $\varepsilon>0$ there is $\delta>0$ so that if $t>1-\delta$ and $\|tx_0\pm z\|\le1$, then $\|z\|\le\varepsilon/2$, which implies $\|z-x_0\|\le \varepsilon$.

If $x_0$ is not a strong extreme point, then this argument tells you that $y=tx_0$ will not satisfy your requirement, but some other $y$ might (although I agree with Misha that one should expect some uniformity in the convexity for your property to hold).

If every point in the sphere is a strong extreme point, the space is called midpoint locally uniformly rotund. I think R. Haydon has given an example of a strictly convex space without an MLUR renorming.

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