How to show $f(x) \leq (1-t)f(x)+tg(x) \leq g(x)$

Question

Given $f(x) < g(x)$ and $f(x), g(x), t \in[0,1]$, how can I show $f(x) \leq (1-t)f(x)+tg(x) \leq g(x)$?

It is easy to show $f(x) \leq (1-t)f(x)+tg(x)$. Indeed, $(1-t)f(x)+tg(x) = f(x)-tf(x)+tg(x) = f(x)+t(g(x)-f(x))$. Given that $f(x) \leq g(x)$ it follows $f(x) \leq (1-t)f(x) +g(x).$

However, I am having difficulties to show $(1-t)f(x)+tg(x) \leq g(x)$. One argument would be to show at $t=0$ we have $(1-t)f(x) \leq g(x)$ and $t=1$ we will get $g(x) \leq g(x)$. But I am not sure if this is good enough since we need to check for $t \in (0,1)$.

I would like an answer that does not depend on the specific choices of $t$. When I showed $f(x) \leq (1-t)f(x)+tg(x)$, the argument did not depend on the choice of $t$ and some sort of squeeze theorem.

Answer 1

$(1-t)f(x)+tg(x)\leq g(x) \iff (1-t)f(x) \leq (1-t)g(x) \iff f(x) \leq g(x).$

The $\iff$ is not quite since $(1-t)$ could possibly be $0$. It just outlines the idea.

Answer 2

You could also just let $k=1-t$. Then $k \in [0,1]$ and

$$\begin{align} (1-t)f(x)+tg(x) &= k f(x) + (1-k) g(x) \\ &= g(x) + k \left[ f(x) - g(x) \right] \\ &\le g(x) \end{align}$$