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

Question

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?

Answer 1

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 $\mathbb R$.

Answer 2

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:

$$ \begin{aligned} ((1-t)x + ty)^4 &= (((1 - t)x + ty)^2)^2 \\ &\leq ((1-t)x^2 + ty^2)^2 &\quad (x^2 \text{ is strictly convex and order-preserving})\\ &< (1-t)(x^2)^2 + t(y^2)^2 &\quad (x^2 \text { is strictly convex}) \\ &= (1-t)x^4 + ty^4 \end{aligned} $$