Let $LC = \{\lambda v \mid \lambda \in L, v \in C\},$ where $C$ is some convex set of the vector space $V.$ If $L$ is an interval, is $LC$ convex?
I thought and assumed so for a while, and some drawings suggested this to be true, however, today I tried to prove it analytically and the proof escaped me.
We start with $\lambda u, \mu v \in L C$ and $t \in (0,1).$ Then $t \mu v + (1-t) \lambda u$...? No clue.
I also tried writing $LC = \bigcup\limits_{\lambda \in L} \lambda C$ which is the union of convex sets; and I know that a tower of convex sets is convex but then I don't know how to prove this family $(\lambda C)_{\lambda \in L}$ is a tower.
Other ideas, suggestions or a counter example appreciated.
In fact, we can continue the same way as you started (assuming, as Izaak van Dongen points out in the comments, that $L$ consists of nonnegative numbers): $$ t\mu v + (1-t)\lambda u = (t\mu + (1-t)\lambda)\left(\frac{t\mu}{t\mu + (1-t)\lambda} v + \frac{(1-t)\lambda}{t\mu + (1-t)\lambda} u\right). $$ Inside the big parentheses, we have a point on the segment $[u,v]$, because $\frac{t\mu}{t\mu + (1-t)\lambda}$ and $\frac{(1-t)\lambda}{t\mu + (1-t)\lambda}$ are two nonnegative weights that add up to $1$. (This is where our assumption about $L$ comes in!) In particular, what we have there is an element of $C$, because $u \in C$ and $v \in C$.
Outside the big parentheses, $(t\mu + (1-t)\lambda)$ is a value somewhere between $\mu$ and $\lambda$. Because both $\mu$ and $\lambda$ are in $L$, we also know that $(t\mu + (1-t)\lambda)$ is in $L$. Therefore the whole expression gives us a point in $LC$.