convex-analysis
Can someone provide intuition & proof for why a ray is convex, I don't see the sum to $1$ constraints for theta :
A ray of the form $\left\{x_{0}+\theta v \mid \theta \geq 0\right\}$, where $v$ != $0$ is convex, but not affine. It is a convex cone if its base $x_{0}$ is $0.$
From Boyd & Vandenberghe Convex Optimization pg. 27
Hint : In fact you can prove that $C$ lies in boundary of $H$ i.e.,
$$C \subseteq A \subseteq \{x\vert a^Tx= b\}$$
Then clearly $H$ supports $C$ at $x_0$ (even at any point of $C$)
This definition captures the idea that the graph of a convex function is always below the secant joining any two points.
Given two points $u_1 = (x_1, y_1), u_2 = (x_2, y_2)$, the line segment joining the points can be parameterized by the function $$l(t) = (tx_1 + (1-t) x_2, ty_1 + (1-t) y_2)= t u_1 + (1-t) u_2.$$ For example, if $t = .25$ it means we take $25\%$ $u_1$ and $75\%$ $u_2$. If the coefficients don't add to one you may leave the line segment.
Now, let $u_1 = (x_1, f(x_1)$ and $u_2 = (x_2, f(x_2))$. Look at this image taken from Wikipedia.
The requirement is that the curve of $f(x)$ lies below this secant line joining these two points, which is parametrized by $$l(t) = (tx_1 + (1-t) x_2, tf(x_1) + (1-t) f(x_2)),$$ the point indicated on the image for some particular $t$. The point $$ (tx_1 + (1-t) x_2, f(tx_1 + (1-t) x_2)$$ on the curve needs to lie below it.
Another interpretation: $tf(x_1) + (1-t) f(x_2))$ is a weighted average of the outputs of $f$, while $f(tx_1 + (1-t) x_2)$ is output from taking a weighted average of the inputs. So you can say that the requirement for convexity is that
$$f(\text{weighted average of points}) \leq \text{weighted average of }f(\text{points}).$$
This generalizes to Jensen's inequality.
Best Answer
I will prove the case where $x_0 = 0$, since convexity is invariant under translation.
According to the definition, your ray is of the form, for a fixed $\mathbf{x} \in \mathbb{R}^d$,
$$R= \{\lambda\mathbf{x} : \lambda \geqslant 0\}$$
Now let $\mathbf{y},\mathbf{z} \in R$, so $\mathbf{y} = y\mathbf{x}$ and $\mathbf{z} = z\mathbf{x}$ for some $y,z\geqslant 0$ by definition of the ray. As this is a subset of a $2$-dimensional linear (affine, if $x_0 \neq 0$) subspace, all convex combinations will require at most two terms without any redundancy (it is not hard to show for many-termed sums, I will leave this as an exercise to help understand the proof). So for all $t \in [0,1]$, it suffices to show:
$$t\mathbf{y} + (1-t)\mathbf{z} \in R$$
The terms $t\mathbf{y}, (1-t)\mathbf{z}$ are the same as $ty\mathbf{x}$ and $(1-t)z\mathbf{x}$ by the result above, and as $t$ and $(1-t)$ are both at least $0$, $t\mathbf{y}$ and $(1-t)\mathbf{z}$ are in the ray. As such, the above sum is just:
$$(ty + (1-t)z)\mathbf{x}$$
which is in the ray, so the ray is convex.
Generally, proving shapes convex involve showing the conditional:
$$\{\mathbf{x}\}_i \in X \Longrightarrow \sum_i \lambda_i \mathbf{x}_i \in X$$
for $\sum \lambda_i =1$ and each $\lambda_i \geqslant 0$, holds true. You can see how the above proof is essentially just this conditional.
Geometrically and intuitively, a shape is convex when the line between any two points is in the shape. I dare you to find two points in a ray where the line between them is not in the ray.