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Is there a topological vector space such that, for every $x\in X$, there is a proper neighbourhood $V$ of $x$ in $X$ which is convex, but the whole space is not locally convex (i.e. $X$ has a local base consisting of convex sets in each of its point)?

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    $\begingroup$ I think you need to narrow it down a little bit, because the whole space is obviously a convex neighbourhood of each of its points, so any non-locally convex space will satisfy your query. $\endgroup$ – kahen Aug 27 '15 at 22:15
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    $\begingroup$ for a topological vector space $V$ is a neighbourhood of $x$ iff $V-x$ is a neighbourhood of $0$, so the existence of a neighbourhood for all $x$ is equivalent to the existence of a neighbourhood of $0$, $\endgroup$ – Hamza Aug 27 '15 at 23:36
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For $0 < p < 1$, let

$$\ell^p(\mathbb{N}) = \Biggl\{ x \colon \mathbb{N}\to \mathbb{C} : \sum_{n = 0}^\infty \lvert x(n)\rvert^p < +\infty\Biggr\}.$$

We endow it with the $p$-seminorm

$$s_p(x) = \sum_{n = 0}^\infty \lvert x(n)\rvert^p$$

and the metric $d_p(x,y) = s_p(x-y)$ derived from it.

Note the difference from the case $p \geqslant 1$. If $p > 1$, one needs to take the $p^{\text{th}}$ root to have the triangle inequality, and one obtains a normed, hence locally convex, space. For $p < 1$, taking the $p^{\text{th}}$ root would destroy the triangle inequality and thus we obtain a space that isn't locally convex.

$s_p$ is similar to a norm, but it satisfies $s_p(\lambda x) = \lvert\lambda\rvert^p\cdot s_p(x)$ instead of the homogeneity a norm has.

The triangle inequality follows from

$$a^p + b^p \leqslant (a+b)^p\tag{1}$$

for $a,b \geqslant 0$ and $0 < p < 1$. For $a = 0$ or $b = 0$ that is immediate, and for $a,b > 0$ we have $\frac{a}{a+b}, \frac{b}{a+b} \in (0,1)$, so

$$1 = \frac{a}{a+b} + \frac{b}{a+b} < \biggl(\frac{a}{a+b}\biggr)^p + \biggl(\frac{b}{a+b}\biggr)^p,$$

whence $(1)$. The translation-invariance of $d_p$ and the $p$-homogeneity ensure that the induced topology is a vector space topology:

$$d_p(x+y,x_0+y_0) = s_p\bigl((x-x_0)+(y-y_0)\bigr) \leqslant s_p(x-x_0) + s_p(y-y_0) = d_p(x,x_0) + d_p(y,y_0)$$

shows that addition is uniformly continuous, and

\begin{align} d_p(\lambda x,\lambda_0 x_0) &= s_p(\lambda x - \lambda_0 x_0) \\ &= s_p\bigl((\lambda-\lambda_0)x_0 + \lambda_0(x-x_0) + (\lambda - \lambda_0)(x-x_0)\bigr)\\ &\leqslant s_p\bigl((\lambda-\lambda_0)x_0\bigr) + s_p\bigl(\lambda_0(x-x_0)\bigr) + s_p\bigl((\lambda-\lambda_0)(x-x_0)\bigr)\\ &= \lvert\lambda -\lambda_0\rvert^p\cdot s_p(x_0) + \lvert\lambda_0\rvert^p\cdot d_p(x,x_0) + \lvert\lambda-\lambda_0\rvert^p\cdot d_p(x,x_0) \end{align}

shows the continuity of scalar multiplication at $(\lambda_0,x_0)\in \mathbb{C}\times \ell^p(\mathbb{N})$.

These spaces are not locally convex: Define the "standard unit vectors" $e_k$ by

$$e_k(n) = \begin{cases} 1 &, n = k\\ 0 &, n \neq k.\end{cases}$$

Then $s_p(e_k) = 1$ for all $k$, and by the $p$-homogeneity $y_k := (\varepsilon/2)^{1/p}\cdot e_k \in B_\varepsilon(0)$ for $\varepsilon > 0$. But with $x_N = \frac{1}{N}\sum_{k = 0}^{N-1} y_k$, we have

$$s_p(x_N) = N\cdot\frac{\varepsilon}{2N^p} = N^{1-p}\frac{\varepsilon}{2} \xrightarrow{N\to\infty} +\infty,$$

so no $d_p$-ball contains the convex hull of any other $d_p$-ball.

Yet, there are plenty of convex open sets, since the inclusion $\iota_p \colon \ell^p(\mathbb{N}) \hookrightarrow \ell^1(\mathbb{N})$ is continuous.

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