Intuition why $l^p(\mathbb{N})$ is not locally convex. Consider the space
$$l^p(\mathbb{N}) = \left\{(a_n)_{n=0}^\infty: \sum_{n=0}^\infty |a_n|^p < \infty\right\}$$
for $0 < p < 1$. Consider the metric
$$d(x,y) := \sum_{n=0}^\infty |x_n-y_n|^p$$
This metric induces a topology $\mathcal{T}$. It is well-known that the pair $(l^p(\mathbb{N}), \mathcal{T})$ is a topological vector space which is not a locally convex vector space.
While I know the proof of this fact, I'd like to know how someone came up with this example. Is there a way to intuitively see that this is not locally convex, by geometric intuition. For example, by looking how the balls around $0$ look like, or studying the behaviour of convex sets for $l^p(S_n)$ where $S_n$ is a sequence of sets that converges to $\mathbb{N}$ in some suitable sense.
Thanks!
 A: The issue here is that your metric is not induced a vector norm, so the distance and topology defined is independant to the linear structure of the vector space. Sequences getting closer to the origin in the vector space are not getting closer to the origin by the metric.
I'm not sure which proof  you're thinking of, but here's one that's pretty simple that can make one understand the problem :

We notice that :
$$d(ku_n,0)=k^pd(u_n,0)$$

The distance is "decreasing" when a sequence is mutliplied by an "increasing" scalar.

Because we're in a metric space, if there's a convex neighborhood $N$
  of $0$ in $(l_p,\mathcal{T})$, then it contains, for a certain
  $\epsilon$, all the sequences at distance less than $\epsilon$ from
  origin. If we call $u_n$ an arbitrary sequence of distance $\delta$ to
  origin, then $(\frac{0.5\epsilon}{\delta})^{-p}u_n$ is in $N$ (because
  of the formula we established earlier). By convexity, all sequences
  $(1-t)(0.5\frac{\epsilon}{\delta})^{-p}u_n$ for $0\leqslant t \leqslant 1$ are in $N$. If $\frac{0.5\epsilon}{\delta}^{-p}$ is
  smaller than one, then it means that $\delta < 0.5 \epsilon$ and
  $u_n\in N$. Else, there exists $t$ such as
  $(1-t)\frac{\epsilon}{\delta}^{-p}=1$, and $u_n$ is in $N$.
Therefore, $0$ doesn't have non-trivial convex neighborhoods.

