Let $f : \mathbb{R}^d \to \mathbb{R}$ be defined by $f (x) := \sum\limits_{i \in [d]}{x_i}^2$. How to prove that the following set is convex? $$\left\{ x \in \mathbb{R}^d : f (x) \leq 1 \right\}$$
So far my math is,
If $x, z \in \mathcal{f}$ then $\sum_{i \in [d]} z_i^2 \leq 1$ should be true for $z$. Using the definition of convex set, we can say: \begin{align} \sum_{i \in [d]} (\lambda z_i + (1 - \lambda) x_i)^2 = \sum_{i \in [d]} (\lambda^2 {z_i}^2 + 2\lambda (1 - \lambda) x_i z_i + (1-\lambda)^2{x_i}^2) \\= \lambda^2 \sum_{i \in [d]} {z_i}^2 + 2\lambda (1 - \lambda) \sum_{i \in [d]} x_i z_i + (1-\lambda)^2 \sum_{i \in [d]}{x_i}^2 \end{align}