# Does a reverse triangle inequality hold for any nonnegative convex function?

Let $$f:\mathbb{R}^2\to\mathbb{R}$$ be nonnegative and strictly convex in the second argument everywhere (that is, $$f(p,\cdot)$$ is strictly convex for all $$p\in\mathbb{R}$$). Moreover, assume $$f(p,p)=0$$ for all $$p$$. A primary example of such function is $$f(p,q)=(q-p)^2$$. If necessary $$f$$ can be twice-continuously differentiable.

Now, take any $$p. Is it always the case that the following "reverse triangle inequality'' holds? $$f(p,q)+f(q,r) If $$f$$ is symmetric (i.e., $$f(p,q)=f(q,p)$$), then I can show this by writing $$q=\lambda p + (1-\lambda)r$$ for some $$\lambda\in(0,1)$$. In particular,

\begin{align*} f(p,q)+f(q,r) & =f(p,q)+f(r,q)\\ & =f(p,\lambda p+(1-\lambda)r)+f(r,\lambda p+(1-\lambda)r)\\ & <\lambda f(p,p)+(1-\lambda)f(r,r)+(1-\lambda)f(p,r)+\lambda f(r,p)\\ & =f(p,r) \end{align*}

But, I would like to show this for nonsymmetric $$f$$, if it is true.

• $f=0$ seems to satisfy your conditions and not satisfy the "strict" triangle inequality. Feb 26 at 0:06
• @Kroki I added the condition that it is strictly convex Feb 26 at 0:57
• @GregMartin Oh shoot, the desired inequality is the other way. I edited the question. Feb 26 at 0:59

Note that if $$f(p,q)$$ is a function with the given properties, then so is $$h(p,q) = f(p,q)g(p)$$ for any positive function $$g$$ whatsoever. But $$h(p,q)+h(q,r) is equivalent to $$f(p,q)g(p)+f(q,r)g(q), or $$g(p) \bigl( f(p,q)-f(p,r) \bigr) < g(q) f(q,r)$$; and it should be easy to choose a function $$g$$ for which this is not always true.
For a specific example, let $$f(p,q)=(q-p)^2$$ and $$g(p) = |p|+1$$ and set $$h(p,q)=f(p,q)g(p)$$. Then, choosing $$0=p, we have $$h(0,q)+h(q,r)-h(0,r) = q^2\cdot1 + (r-q)^2\cdot(q+1) - r^2\cdot1 = q (r-q) (r-q-2),$$ which is undesirably positive if $$q.