# Linear combination of convex functions is constant

Let $$\Phi : \mathbb{R}_{++}\to \mathbb{R}$$ be a convex function (not necessarily differentiable). Fix an $$\alpha \in (0,1)$$ and define

$$g(t) = \alpha t .\Phi\left(\frac{1}{t} + 1\right) + 2(1-\alpha)t .\Phi\left(\frac{1}{t}+\frac{1}{2}\right),\quad \text{for }\;t\in[1,2].$$ Does there exist any such functions $$\Phi$$ such that $$g(\cdot)$$ is a constant as a function of $$t$$? I tried to tackle the problem using Jensen's inequality but nothing seems to work. For example, I have obtained $$\alpha t \Phi\left(\frac{1}{t} + 1\right) + 2(1-\alpha)t \Phi\left(\frac{1}{t}+\frac{1}{2}\right)\geq t(2-\alpha)\Phi((1+t)\alpha + (2+t)(1-\alpha)).$$ But this seems like it is not going to get me anywhere. A couple of facts that could be useful is that $$t.\Phi\left(\frac{1}{t}+1\right)$$ and $$2t .\Phi\left(\frac{1}{t}+\frac{1}{2}\right)$$, both are convex functions in $$t$$. Any help will be appreciated. Thanks.

• For a fixed $\alpha$, let $\Phi(t) = K + t$ where $K$ is a constant to be determined later. Then $g(t)$ is affine linear, and the linear coefficient is $(K + 1)\alpha + (1-\alpha)(2K+1)$. For suitable choice of $K$ depending on $\alpha$, this coefficient can be made $0$. (If my math is correct, it's $K = \frac1{\alpha-2}$.) – Todd Trimble Oct 4 '20 at 23:14

define $$\Phi$$ by the expression below $$\Phi(x) = Id(x) = x.$$ Then $$\Phi(\frac{1}{t} + 1) = \frac{1}{t} + 1.$$ And $$\Phi(\frac{1}{t} + \frac{1}{2} ) =\frac{1}{t} + \frac{1}{2} .$$ Hence $$g(t) = \alpha t(\frac{1}{t} + 1) +2(1-\alpha)t(\frac{1}{t} + \frac{1}{2} )= \alpha + 2.$$
• I think that $g(t) = t + (2 - \alpha)$, per @ToddTrimble's comment. – LSpice Oct 29 '20 at 23:23
Just out of curiosity, we may characterize the convex functions $$\Phi:\mathbb{R}_+\to\mathbb{R}$$ satisfying that condition as those convex functions that are affine on the interval $$[1,\frac32]$$ and on the interval $$[\frac32,2]$$ and satisfy $$\alpha\Phi(2)+(2-4\alpha)\Phi \big(\frac32 \big)-(4-4\alpha)\Phi(1)=0.$$
For $$t \in[1,2]$$ put $$s=\frac1t+\frac12$$ so that $$t=\frac2{2s-1}$$ and the condition writes, for some constant $$g$$ $$\alpha \Phi(s+\textstyle\frac12) =(s-\frac12)g- 2(1-\alpha ) \Phi(s)\quad \text{for }\;s\in\big[1,\frac32\big].$$ This implies that $$\Phi$$ is affine on both $$[1,\frac32]$$ and $$[\frac32,2]$$, thus $$\textstyle \Phi(s)=\cases{2\big(\Phi (\frac32)-\Phi(1)\big)(s-\frac32)+\Phi(\frac32)\quad \text {for}\; 1\le s\le \frac32,\\ \\\textstyle 2\big(\Phi(2)-\Phi (\frac32)\big)(s-\frac32)+\Phi(\frac32)\quad \text {for}\; \frac32\le s\le2. }$$
If we plug this in the preceding equation and eliminate the constant $$g$$, we find the stated linear relation between $$\Phi(1), \Phi \big(\frac32 \big), \Phi(2)$$.