Find a combination of convex function so that it is positive

Question

A student in my class asked me the following question, I did know what tools will be needed to attack it. But I found it is an interesting question.

Let $f_1,f_2$ be two convex functions on $[0,1]$ and let $\varepsilon>0$. For every $x\in[0,1]$，we have $\max\{f_1(x),f_2(x)\}>\varepsilon$. Does there exist a $\lambda\in [0,1]$ so that $$ \lambda f_1(x)+(1-\lambda)f_2(x)>0, \quad \forall x\in [0,1]? $$

Any hints are welcome! Thanks!

Answer 1

Here is the easiest explanation, two intersecting straight lines satisfy this property, and two intersecting straight lines are the worst-case among all convex functions when fixing the intersection point of the two functions.

And the following proof is an effort to make the above intuition rigorous and mathematical readable.

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The following argument only available for convex function with reasonable regularity

First, as point out by Loic Teyssier, we need to consider the case independently: when $f_1,f_2$ do not intersection. And when $f_{1}, f_{2}$ is not intersect and then we can take $\lambda=0$ or 1 to make $\lambda f_{1}(x)+(1-\lambda) f_{2}(x)=\max \left\{f_{1}, f_{2}\right\}$.

Second, as point out by Dieter Kadelka, we need to consider the endpoint $\{0,1\}$ separate because the prove of N-L formula relay the point $a,b$ is in some interval, in the following argument. This can be down use the convex property to gain a lower bound on $f_1(0),f_1(1),f_2(0),f_2(1)$.

Then, this is true and in fact, we can determine $\lambda$. We need the following 2 lemmas.

In the case $\exists \quad x_{0} \in[0,1] . \quad f_{1}\left(x_{0}\right)=f_{2}\left(x_{0}\right)>0$

$f$ is a continue couvex function on [0.1], then $f$ is lipschitz function.

$f$ is a lipschitz function. then $f^{\prime}(x) \quad$ exist a.e. in $[0,1]$ and $\forall \quad 0\leq a< b\leq 1$, $$f(b)-f(a)=\int_{a}^{b} f^{\prime}(x) d x$$ We can understand the previous N-L formula in distribution to avoid unnecessary trouble（The derivative does not exist, the left and right derivatives are not equal, and the derivative is zero but the function is a strictly monotonic function）

then $\forall x \in[0.1]$, $\begin{aligned} \lambda f_{1}(x)+(1-\lambda) f_{2}(x) &=\lambda\left(f_{1}\left(x_{0}\right)+\int_{x_{0}}^{x} f_{1}^{\prime}(t) d t\right)+(1-\lambda)\left(f_{2}\left(x_{0}\right)+\int_{x_{0}}^{x} f_{2}(t) d t\right) \\ &=\lambda f_{1}\left(x_{0}\right)+(1-\lambda) f_{2}\left(x_{0}\right)+\lambda \int_{x_{0}}^{x} f_{1}^{\prime}(t) d_{t}+(1-\lambda) \int_{x_{0}}^{x} f_{2}^{\prime}(t) d t \end{aligned}$

Now $\quad$ WLOG $\quad$ we assame
$f_{1}(x)>0$ When $0 \leqslant x \leqslant x_{0}$, $f_{2}(x)>0 $ when $ x_{1} \leqslant x \leqslant 1$, and assume $f_{1}^{\prime}\left(x_{0}\right)>0. f_{2}^{\prime}\left(x_{0}\right)<0$.
When $ 1 \geqslant x>x_{0}$. Then,

$$ \begin{array}{l} \int_{x_{0}}^{x} f_{1}^{\prime}(t) d t \geqslant\left|x-x_{0}\right| \cdot f_{1}^{\prime}\left(x_{0}\right) \\ \int_{x_{0}}^{x} f_{2}^{\prime}(t) d t \geqslant\left|x-x_{0}\right| \cdot f_{2}^{\prime}\left(x_{0}\right) \end{array} $$

When $0 \leqslant x<x_{0}$, then, \begin{array}{l} \int_{x_{0}}^{x} f_{1}^{\prime}\left(t) d t \geqslant\left|x-x_{0}\right| \cdot f_{1}^{\prime}\left(x_{0}\right)\right. \\ \int_{x_{0}}^{x} f_{2}^{\prime}(t) d t \geqslant \left|x-x_{0}\right| \cdot f_{2}^{\prime}\left(x_{0}\right) \end{array}

So take $\lambda$ solve $\quad \lambda f_{1}^{\prime}\left(x_{0}\right)+(1-\lambda) f_{i}^{\prime}\left(x_{0}\right)=0$, then $\quad \lambda f(x)+(1-\lambda) f_{2}(x) \geqslant 0 \quad \forall \quad x\in [0,1]$.

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To prove the property is true for the general convex function we may follow the argument above and modified the detail(especially the H-L formula part) by use supporting lines of $f_1,f_2$ and observe convex functions lie above their supporting lines.