Euclidean Balls \(B(x_c, r)\) are sets of the form \(\{x \mid \lVert x-x_0 \rVert_{2} \leq r \}\). Here \(x_c\) is the center of the ball and \(x, x_c \in R^n\) and \(r \in R\) and \(r \gt 0\). They can also be represented as \(B(x_c, r) = \{ x_c + ru \mid \lVert u \rVert _2 \leq 1 \}\). Here \(x_c \in R^n\) is the center of the ball, \(r \in R\) is the radius of the ball and \(u \in R^n\) is a vector of unit length such that \(ru\) will give us the vector of length \(r\).
Euclidean balls are convex sets.
Proof that euclidean balls are convex:
Let \(x_1\) and \(x_2\) belong to the set \(B(x_c, r)\). Let \(x_3\) be a point such that \(x_3 = \theta x_1 + (1-\theta) x_2\). \(B(x_c, r)\) will be convex if \(x_3\) also belongs to \(B(x_c, r)\), i.e \(\lVert x_3 - x_c \rVert _2 \leq r\).
\[\begin{align*} \lVert x_3 - x_c \rVert _2 &= \lVert \theta x_1 + (1-\theta) x_2 - x_c \rVert _2 \\ &\leq \theta \lVert x_1 - x_c \rVert _2 + (1-\theta) \lVert x_2 - x_c \rVert _2 \\ &\leq \theta r + r - \theta r \\ &\leq r \end{align*}\]Since \(x_3 \leq r\) we can conclude that it also lies in the euclidean ball. Thus we can say that the [Convex Combination] of any points in the euclidean ball will also lie in the euclidean ball. Therefore we can say that the euclidean ball is a [Convex Set] .