Proof for the continuity of a min function

Assume a min real function is defined below:

\[f(x)=\min\limits_{x\leq y\leq h(x)}g(y)\]

\(g(x)\) is continuous. \(x\geq 0, h(x)=ax+b, a\geq 1, b\geq 0\). Prove \(f(x)\) is continuous.

_Proof _

For any \(x_0\) in the domain of \(f(x)\), 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 \(\|x-x_0\|<\delta_1\),

\[|g(x)-g(x_0)|<\epsilon_1\tag{1}\]

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

\[|g(h(x))-g(h(x_0))|<\epsilon_2\tag{2}\]

For any two number \(x_1, x_2\) in the domain \([x_0-\delta, x_0+\delta]\), let \(\delta\leq \delta_1\), we can get

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

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

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

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

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

Assume

\[\begin{align*} g(y^\ast_{0})&=\min\limits_{x_0\leq y\leq h(x_0)}g(y)\\ g(y^\ast_{\Delta})&=\min\limits_{x_0-\delta\leq y\leq h(x_0+\delta)}g(y) \end{align*}\]

For any \(x\in [x_0-\delta, x_0+\delta]\),

\[g(y^\ast_{x})=\min\limits_{x\leq y\leq h(x)}g(y)\]

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 \(y^\ast_\Delta\in [x, h(x)]\).

In this case, \(\| g(y^\ast_x)-g(y^\ast_\Delta) \|=0\).

(b). \(y^\ast_\Delta\in [x-\delta, x]\subset [x_0-\delta, x_0+\delta]\).

In this case, let \(\delta\leq \delta_1, \| g(y^\ast_x)-g(y^\ast_\Delta) \|\leq \| g(x)-g(y^\ast_\Delta) \|<2\epsilon_1\). (because of Eq. (3))

(c). \(y^\ast_\Delta\in [h(x), h(x_0+\delta)]\subset [h(x_0-\delta), h(x_0+\delta)]\).

In this case, let \(\delta\leq \delta_2\),

\(\| g(y^\ast_x)-g(y^\ast_\Delta) \|\leq \| g(h(x))-g(y^\ast_\Delta) \|<2\epsilon_2\) (according to Eq. (4)).

Therefore, from the three cases, when \(\delta\leq\min\{\delta_1, \delta_2\}\), we can get

\[| g(y^\ast_x)-g(y^\ast_\Delta) |<\max\{2\epsilon_1, 2\epsilon_2\}\]

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

\[|g(y^\ast_0)-g(y^\ast_\Delta)|<\max\{2\epsilon_1, 2\epsilon_2\}\]

Eq. (6) can be written to be:

\[|f(x)-f(x_0)|<2\max\{2\epsilon_1, 2\epsilon_2\}\]

Let \(\max\{\epsilon_1, \epsilon_2\}=\epsilon\), we can get

\[| f(x) - f(x_0) | < \epsilon\]

The continuity of \(f(x)\) is proved.

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