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?
$\begingroup$
$\endgroup$
2
-
$\begingroup$ 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 $x,y$. $\endgroup$– DeepSeaMay 15, 2015 at 1:38
-
$\begingroup$ Why is no answer accepted? $\endgroup$– ViktorSteinMar 13 at 14:37
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
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:
$$ \begin{aligned} ((1-t)x + ty)^4 &= (((1 - t)x + ty)^2)^2 \\ &\leq ((1-t)x^2 + ty^2)^2 &\quad (x^2 \text{ is strictly convex and order-preserving})\\ &< (1-t)(x^2)^2 + t(y^2)^2 &\quad (x^2 \text { is strictly convex}) \\ &= (1-t)x^4 + ty^4 \end{aligned} $$
$\begingroup$
$\endgroup$
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 $\mathbb R$.
-
1$\begingroup$ 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? $\endgroup$ Mar 30 at 14:58