Positive property of the mixture of two convex functions Let $f_1, f_2$ be two convex functions on $[0, 1]$ and let $\epsilon>0$. Assume that for every $x\in[0, 1]$, we have
$$
\max\{f_1(x), \, f_2(x)\}>\epsilon.
$$
Then does there exist a $\lambda\in[0, 1]$ such that
$$
\lambda f_1(x)+(1-\lambda)f_2(x)>0,\,\,\,\forall x\in[0, 1]?
$$
I'm not sure if one can prove or disprove this using only calculus, or one can apply some functional analysis argument to obtain it.
 A: Let
\begin{equation}
m = \max(\min_{x\in[0,1]} f_1(x), \min_{x\in[0,1]} f_2(x))
\end{equation}

*

*If $m \ge \epsilon$, we have $\lambda f_1 + (1 - \lambda) f_2\ge \epsilon$ for $\lambda = 0$ or $\lambda = 1$.

*Let us now suppose $m<\epsilon$. Let $[x_0, x_1]$ and $[x_2, x_3]$ be the non-empty subintervals of $[0, 1]$ where $f_1(x)\le m$ and $f_2(x)\le m$ respectively. Their intersection is empty because of the condition $\max\{f_1, f_2\}>\epsilon$ everywhere. Upon perhaps swapping the two functions, we can suppose that $x_1 < x_2$. As $f_1 - f_2$ is negative at $x_1$ and positive at $x_2$, there is a point $a\in (x_1, x_2)$ such that $f_1(a) = f_2(a) = u \ge \epsilon$. Let $\alpha > 0$ and $\beta >0$ be such that
\begin{equation}
\forall x\in [0, 1],\quad\left\{
\begin{array}\cr
f_1(x)\ge u + \alpha (x -a)\cr
f_2(x)\ge u - \beta (x-a)\cr
\end{array}\right.
\end{equation}
that is to say $\alpha$ and $-\beta$ are subderivatives of $f_1$ and $f_2$ at $x=a$. Now let $\lambda = \frac{\beta}{\alpha+\beta}$ and $1-\lambda= \frac{\alpha}{\alpha+\beta}$. One has
\begin{equation}
\lambda f_1(x) + (1-\lambda)f_2(x) \ge u \ge \epsilon
\end{equation}
Hence the result is true, we have shown that the minimum of $\lambda f_1 + (1-\lambda)f_2$ can be chosen larger than $\epsilon$.

Note that we don't even need $\epsilon > 0$ and we have shown that
\begin{equation}
\max_{\lambda\in[0,1]}\min_{x\in[0,1]}(\lambda f_1(x) + (1-\lambda)f_2(x))\ge \min_{x\in[0,1]}\max\{f_1(x), f_2(x)\}
\end{equation}
Edit
Formulated this way, this results appears to be only a special case of Von Neumann's minimax theorem because if we define
\begin{equation}
f(\lambda, x) = \lambda f_1(x) + (1-\lambda) f_2(x)
\end{equation}
then $f(\cdot, x)$ is concave for all $x$ and $f(\lambda, \cdot)$ is convex for all $\lambda\in[0,1]$. The theorem says that in this case,
\begin{equation}
\max_{\lambda\in[0,1]}\min_{x\in[0,1]}f(\lambda, x)= \min_{x\in[0,1]}\max_{\lambda\in[0,1]} f(\lambda, x)
\end{equation}
which is stronger than what we proved above.
