Dtyjlui

Assume a min real function is defined below:

$$
f(x)=minlimits_{xleq yleq h(x)}g(y)
$$

$g(x)$ is continuous. $xgeq 0$, $h(x)=ax+b$, $ageq 1, bgeq 0$. Prove $f(x)$ is continuous.

This is a problem I recently encountered when reading a paper. There may be several scenarios regarding the location of $y^ast$ ($g(y^ast)=minlimits_{xleq yleq 2x}g(y)$) in the domain $[x-delta, 2x+2delta]$. And we may prove the continuity by discussing the different scenarios. My proof is below:

$textit{Proof}.$

We must prove that for any $epsilon>0$, there exists $delta>0$, such that when $|x-x_0|<delta$,

$$|f(x)-f(x_0)|<epsilon$$

Because $g(x)$ is continous, for any $epsilon_1>0$, there exits $delta_1$, when $|x-x_0|<delta_1$,

$$|g(x)-g(x_0)|<epsilon_1tag{1}$$

$g(x)$ is also continous on $x=h(x_0)$. There exists $delta_2$, when $|h(x)-h(x_0)|<delta_2$,

$$|g(h(x))-g(h(x_0))|<epsilon_2tag{2}$$

For any two number $x_1$, $x_2$ in the domain $[x_0-delta, x_0+delta]$, let $deltaleq delta_1$, we can get

$$
begin{aligned}
|g(x_1)-g(x_2)|&=|g(x_1)-g(x_0)+g(x_0)-g(x_2)|\
&leq |g(x_1)-g(x_0)|+|g(x_0)-g(x_2)|<2epsilon_1
end{aligned}tag{3}
$$

For any two number $x_1$, $x_2$ in the domain $[x_0-delta, x_0+delta]$, because $h(x)=ax+b$,

$$
begin{aligned}
a(x_0-delta) +bleq h(x_1)leq a(x_0+delta)+b\
a(x_0-delta) +bleq h(x_2)leq a(x_0+delta)+b
end{aligned}
$$

When $deltaleq delta_2$, then $h(x_1), h(x_2)in[h(x_0)-delta_2, h(x_0)+delta_2]$, by Eq. (2), we can get

$$
begin{aligned}
|g(h(x_1))-g(h(x_2))|&=|g(h(x_1))-g(h(x_0))+g(h(x_0))-g(h(x_2))|\
&leq |g(h(x_1))-g(h(x_0))|+|g(h(x_0))-g(h(x_2))|<2epsilon_2
end{aligned}tag{4}
$$

Assume
$$g(y^ast_{0})=minlimits_{x_0leq yleq h(x_0)}g(y)$$

$$g(y^ast_{Delta})=minlimits_{x_0-deltaleq yleq h(x_0+delta)}g(y)$$

For any $xin [x_0-delta, x_0+delta]$,

$$g(y^ast_{x})=minlimits_{xleq yleq h(x)}g(y)$$

Apparently, $g(y^ast_{Delta})leq g(y^ast_{x})$, $g(y^ast_{Delta})leq g(y^ast_0)$.

And $f(x)=g(y^ast_x)$, $f(x_0)=g(y^ast_0)$.

So,

$$
begin{aligned}
|f(x)-f(x_0)|&=|g(y^ast_x)-g(y^ast_0)| \
&= |g(y^ast_x)-g(y^ast_Delta)+g(y^ast_Delta)-g(y^ast_0)|\
&leq |g(y^ast_x)-g(y^ast_Delta)|+|g(y^ast_0)-g(y^ast_Delta)|
end{aligned}tag{5}
$$

According the location of $y^ast_Delta$, there are three cases possible for $| g(y^ast_x)-g(y^ast_Delta) |$ :

(a). $y^ast_Delta= y^ast_x$, if $y^ast_Deltain [x, h(x)]$.

In this case, $| g(y^ast_x)-g(y^ast_Delta) |=0$.

(b). $y^ast_Deltain [x-delta, x]subset [x_0-delta, x_0+delta]$.

In this case, let $deltaleq delta_1$, $| g(y^ast_x)-g(y^ast_Delta) |leq | g(x)-g(y^ast_Delta) |<2epsilon_1$. (because of Eq. (3))

(c). $y^ast_Deltain [h(x), h(x_0+delta)]subset [h(x_0-delta), h(x_0+delta)]$.

In this case, let $deltaleq delta_2$,

$| g(y^ast_x)-g(y^ast_Delta) |leq | g(h(x))-g(y^ast_Delta) |<2epsilon_2$ (according to Eq. (4)).

Therefore, from the three cases, when $deltaleqmin{delta_1, delta_2}$, we can get

$$| g(y^ast_x)-g(y^ast_Delta) |<max{2epsilon_1, 2epsilon_2}$$

Because $g(y^ast_0)$ is a special situation of $g(y^ast_x)$ when $x=x_0$, there is also

$$|g(y^ast_0)-g(y^ast_Delta)|<max{2epsilon_1, 2epsilon_2}$$

Eq. (6) can be written to be:

$$|f(x)-f(x_0)|<2max{2epsilon_1, 2epsilon_2}$$

Let $max{epsilon_1, epsilon_2}=epsilon$, we can get $|f(x)-f(x_0)|<epsilon$. The continuity of $f(x)$ is proved.
$$hspace{300pt}Box$$