Proving that the sublevel set of a quadratic function is convex

Question

Let

$$C := \{ x \in \mathbb{R} : 2x^2 \le 1 \}$$

Prove that $C$ is convex.

I started with the definition: for $ x_1, x_2 \in C$

$$ \lambda x_1 + (1-\lambda)x_2 = \dots $$

but didn't make any progress. How should I approach this problem?

Answer 1

Hint: I'd rewrite the set as $$C = \{x\in{\Bbb R}\mid -1/\sqrt 2\leq x\leq 1/\sqrt 2\} = \left[-\frac{1}{\sqrt 2};\frac{1}{\sqrt 2}\right].$$

Answer 2

Alternatively, use the fact that the function $f(x)=x^2$ is convex. Then, $$2(\lambda x_1 + (1-\lambda) x_2)^2 \leq 2\lambda x_1^2 + 2(1-\lambda)x_2^2 \leq \lambda + 1 - \lambda = 1,$$ where the second inequality follows from $x_1,x_2 \in C.$

Answer 3

Using the Schur complement, the given quadratic inequality can be rewritten as the following linear matrix inequality (LMI)

$$\begin{bmatrix} \frac 12 & x \\ x & 1\end{bmatrix} \succeq \mathrm O_2$$

whose solution set is a spectrahedron and, thus, convex.

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convex-analysis schur-complement linear-matrix-inequality spectrahedra