How to show convexity of an inequality?

Question

How would $x^2 + y^4 + z^2 < 5$ be shown to be convex? What I was thinking was that this is sort of like an ellipsoid and these are convex sets. The traditional definition of convexity for some $\lambda \in [0,1]$ such that a convex combination is also in the set does not seem clear here because how would we consider that against the "<5" part?

Answer 1

The functions $\pi_k: \mathbb{R}^n \to \mathbb{R}$ given by $\phi_k(x) = x_k^m$ are convex for $m$ even and positive. As Robert points out in the comments, the sum of convex functions is convex. Hence $(x,y,z) = x^2+y^4+z^2$ is convex.

If $f$ is convex, then the set $\{x | f(x) < L \}$ is easily shown to be convex.

In your example, $f(x,y,z) = x^2+y^4+z^2$, $L=5$.

Answer 2

Given the specific wording of your question, you seem to be confusing the convexity of sets and the convexity of functions, or perhaps what it means for an inequality to be convex.

When presented with an equation or inequality along with a claim that it is "convex", what is meant is that the points that satisfy that equation/inequality form a convex set. Similarly, a system of equations and inequalities (collectively called constraints) is called "convex" if the points that satisfy all of the constraints simultaneously form a convex set.

A set $X$ is convex if the set contains every line segment that spans any two points in the set: $$\alpha x_1+(1-\alpha) x_2\in X \quad \forall x_1,x_2\in X,~\alpha\in[0,1]$$ Mapped to your example, if $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ both satisfy the inequality, then so must every point in the segment between them: $$(\alpha x_1+(1-\alpha)x_2)^2+(\alpha y_1+(1-\alpha)y_2)^2+(\alpha z_1+(1-\alpha)z_2)^2<5 \quad \forall \alpha\in[0,1]$$

A function $f$ is convex if its domain is a convex set, and if every secant line lies above the function itself: $$\alpha f(x_1+(1-\alpha) x_2\in X) \leq \alpha f(x_1) + (1-\alpha) f(x_2) \quad \forall x_1,x_2\in\mathop{\textrm{dom}} f,~\alpha\in[0,1]$$

Not surprisingly, the notions of convexity for sets and functions are tightly related: a function is convex if and only if its epigraph is a convex set: $$ f~\text{is convex} \quad\Longleftrightarrow\quad \mathop{\textrm{epi}} f \triangleq \left\{ (x,y) ~\middle|~ x\in\mathop{\textrm{dom}} f,~f(x)\leq y \right\}~\text{is convex}$$

With regards to your specific example, copper.hat's post is on point. For an convex function $f$, the constraint $f(x) \leq y$ describes a convex set, whether $y$ is a constant or another variable.