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?

  • $\begingroup$ 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 $x,y$. $\endgroup$ – DeepSea May 15 '15 at 1:38

I assume you can prove, by definition, that $x^2$ is strictly convex. Now, you use this result twice:

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


For a twice differentiable function $f$ ; $f''(x)>0$ implies $f$ is strictly convex.. (It is 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 in $\mathbb R\setminus \{0\}$.

Now suppose , $x=0$ & $y\in \mathbb R$. Then , $f(tx+(1-t)y)=f((1-t)y)=(1-t)^4y^4<(1-t)y^4=tf(x)+(1-t)f(y) \text{ , as } 0<t<1 , $

So, $f$ is strictly convex over $\mathbb R$.

  • $\begingroup$ It's actually strictly convex at $0$, too, you just can't use the derivative test to prove it. $\endgroup$ – Michael Grant May 15 '15 at 3:03
  • $\begingroup$ What does it mean to be convex at a point? Does in mean convex in some neighborhood of the point? $\endgroup$ – Nishant May 15 '15 at 3:42
  • $\begingroup$ My point is that it is strictly convex everywhere. The derivative test is insufficient to prove it. $\endgroup$ – Michael Grant May 16 '15 at 4:16

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