Is infimum on nonnegative function $>0$ in an example Let $f,g,h$ be bounded funtions on the domain $[0,T]$ with

*

*$0<f<1$

*$g, h >0$

*$\inf_{t\in[0,T]}((1+f(t))g(t)-h(t))>0 \enspace (1)$

*$\inf_{t\in[0,T]}(h(t)-(1-f(t))g(t))>0 \enspace (2)$
Question Is $\inf_t f(t)>0$?
My progress $(1) \Rightarrow 0< \inf_t ((1+f(t))g(t)) - \inf_t h(t) \leq \inf_t (1+f(t))\cdot \inf_t g(t) \Rightarrow \inf_t f(t)>-1$, but this does not help...
 A: Since $g$ is bounded, let $g(x)<M$ for all $x \in [0,T]$. Note that $M>0$. We also have $$
\inf_{[0,T]} ((1+f(t))g(t) - h(t)) >0 \\
\inf_{[0,T]} (h(t) - (1-f(t))g(t)) >0 \\
$$
In particular, there exist constants $C_1,C_2>0$ such that $$
((1+f(t))g(t) - h(t)) > C_1 \\
(h(t) - (1-f(t))g(t)) > C_2
$$
for all $t \in [0,T]$. Adding these expressions for each $t$ gives $$
2f(t)g(t)>C_1+C_2
$$
for all $t \in [0,T]$. Since $g$ is a positive function, we get $$
2f(t) > \frac{C_1+C_2}{g(t)} > \frac{C_1+C_2}{M}
$$
for all $t \in [0,T]$. This amounts to $\inf_{t \in [0,T]} f(t) > \frac{C_1+C_2}{2M} > 0$, as desired. Note that considering $\sup_{t \in [0,T]} f(t) \leq 1$ will also give you that $\inf_{t \in [0,T]} g(t)>0$ because $$
2g(t) \geq 2g(t)f(t) > C_1+C_2 \implies g(t) > \frac{C_1+C_2}{2}
$$ for all $t \in [0,T]$ i.e. $\inf_{t \in [0,T]} g(t) \geq \frac{C_1+C_2}{2} >0$. Finally, note that $(1-f(t))g(t) \geq 0$ so $$
\inf_{t \in [0,T]} h(t) \geq \inf_{t \in [0,T]} [h(t) - (1-f(t))g(t)]>0
$$
so $\inf_{t \in [0,T]} h(t)>0$ as well. That is, from the hypothesis, each of $$
\inf_{t \in [0,T]} f(t), \inf_{t \in [0,T]} g(t) , \inf_{t \in [0,T]} h(t)
$$
is a positive number.
