# Proving $x \mapsto x^4$ is strictly convex

I'm not sure how to prove $f(x) = x^4$ is strictly convex using just the definition of strict convexity: $$f((1-t)x+ty) < (1-t)f(x)+tf(y)$$ for $0<t<1$. Is this just an algebra slog? If so, I can't seem to get it to work. Are there any other ways of proving a function such as this one is strictly convex?

• Try this idea: assume $y , x > 0$ and also $y > x$ . Let $u = \dfrac{y}{x} > 1$, and expand the LHS and divide both sides by $x^4$ to get a function of $u$, and prove it positive. This can be done since you have a homogenous polynomial in $x,y$. May 15, 2015 at 1:38
• Why is no answer accepted? Mar 13 at 14:37

I assume you can prove, by definition, that $$x^2$$ is strictly convex and order-preserving on $$\mathbb R_+$$. Now, you use this result twice:
\begin{aligned} ((1-t)x + ty)^4 &= (((1 - t)x + ty)^2)^2 \\ &\leq ((1-t)x^2 + ty^2)^2 &\quad (x^2 \text{ is strictly convex and order-preserving})\\ &< (1-t)(x^2)^2 + t(y^2)^2 &\quad (x^2 \text { is strictly convex}) \\ &= (1-t)x^4 + ty^4 \end{aligned}
For a twice differentiable function $$f$$, $$f''(x)>0$$ for all $$x$$ implies that $$f$$ is strictly convex. (It is a sufficient condition.)
Here, $$f(x)=x^4$$. So $$f''(x) = 12x^2 > 0$$ for all $$x \in \mathbb R \setminus \{0\}$$. So $$f$$ is strictly convex on $$\mathbb R\setminus \{0\}$$.
Now suppose $$x=0$$ and $$y \in \mathbb R$$. Then for all $$t \in (0, 1)$$ $$f\big(tx+(1-t)y\big) = f\big((1-t)y\big) = (1-t)^4 y^4 < (1-t) y^4 = t f(x)+(1-t) f(y).$$ So $$f$$ is strictly convex on $$\mathbb R$$.
• Isn't the second derivative test only valid on an interval (or more generally, a convex set)? The second derivative test in this case would only tell you that $x^4$ is strictly convex on the intervals $(0,\infty)$ and $(-\infty,0)$, right? Mar 30 at 14:58