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Let $X$ be a normed space, $S$ a subset of $X$ and let $x_0\in X$. Consider the two propositions:

(1) There exists $r>0$ such that $B(x_0,r)\subset S$. (That is, $x_0$ is an interior point of $S$.)

(2) For every $y\in X$, there exists $\varepsilon_y>0$ such that $x_0+ty\in S$ for all $|t|<\varepsilon_y$.

Surely (1) implies (2). (We can take $\varepsilon_r=r/||y||$.) I found out that (1) does not always holds if (2) holds.

I ask the following question: if $S$ is convex, does it follow that (2)$\implies$(1) ?

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When $S$ is convex and closed and $X$ is a Banach Space, then yes. Condition (2) is known as being in the "core" or "algebraic interior" of the set. There's a simple Baire Category Theorem argument that the two conditions are equivalent in these circumstances.

Basically, without loss of generality, you can assume $x_0 = 0$ by translating as required. Then, $nS$ is a cover of $X$ by countably many closed sets. By the BCT, one such set must have an interior point. But, since all sets are just scalings of each other, this means all such sets must have an interior point. In other words, there must be some $x_1 \in S$ and $r > 0$ such that $B[x_1; r] \subseteq S$.

From the core condition (2), there must be some $t > 0$ such that $$x_0 + t(x_0 - x_1) \in S.$$ Let the above point be $x_2$. Then $$x_0 = \frac{t}{t + 1} x_1 + \frac{1}{t + 1} x_2,$$ and a simple convexity argument, using $B[x_1; r] \subseteq S$, shows us that $$B\left[x_0; \frac{tr}{1+t}\right] \subseteq S.$$ To see this, suppose $y \in B\left[x_0; \frac{tr}{1+t}\right]$. Let $$z = \frac{(t + 1)y - x_2}{t}.$$ Then $$\|z - x_1\| = \frac{t + 1}{t}\left\|y - \frac{1}{t+1}x_2 - \frac{t}{t+1}x_1\right\| = \frac{t + 1}{t}\|y - x_0\| \le r,$$ hence $z \in B[x_1; r] \subseteq S$. On the other hand, $$y = \frac{t}{t+1}z + \frac{1}{t + 1}x_2 \in S,$$ thus showing $B\left[x_0; \frac{tr}{1+t}\right] \subseteq S$ as required, hence $x_0 \in \operatorname{int} S$.


As for counterexamples when the above do not hold, consider an infinite-dimensional real normed linear space $X$. Then $X$ admits a discontinuous linear functional $\phi$.

Consider the epigraph $E$ of $\phi^2$, in the space $X \times \mathbb{R}$. It's not hard to verify that $E$ is convex, using the fact that $\phi$ is linear. I claim that $(0, 1)$ is in the core of $E$, but not the interior.

To see that $(0, 1)$ is in the core, consider the restriction of $\phi$ to any one-dimensional subspace of $X$. This restriction becomes a parabola, the epigraph of which is contained in $E$, and in which the point $(0, 1)$ is an interior point. Thus, we can always travel some distance in any given direction and still remain inside $E$.

Now, suppose that $(0, 1)$ is an interior point of $E$. Then, there must exist some $r > 0$ such that $$B[(0, 1); r] \subseteq E.$$ In particular, if $x$ is in the unit ball of $X$, then $$(rx, 1) \in B[(0, 1); r] \subseteq E \implies \phi(x) \le \frac{1}{r}.$$ That is, $\phi$ is bounded (with $\|\phi\| \le 1/r$). This is a contradiction, so $(0, 1)$ is not an interior point of $E$.

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    $\begingroup$ It is possible to make the counterexample a little bit more easy: Just consider the preimage $\phi^{-1}([-1,1])$ in $X$. It is easy to check that $0$ belongs to the core but not to the interior. $\endgroup$ – gerw Jan 11 at 8:11

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