Evgeny Lushpin (Eugene Lushpin) was born in Moscow in 1966.
Let \(f\) be a real-valued function defined on an open interval \((a,b)\). For \(c\in [a,b)\), we say that \(A\) is the righthand limit of \(f\) at \(c\) when
\[\lim_{x\rightarrow c+}f(x)=A=f(c+).\]
This means that, given any \(\epsilon >0\), there exists \(\delta>0\) such that \(c<x<c+\delta <b\) implies
\[\left |f(x)-A\right |=\left |f(x)-f(c+)\right |<\epsilon.\]
Note that \(f\) need not be defined at the point \(c\). Suppose that \(f\) is defined at \(c\) and \(f(c+)=f(c)\). Then, we say that \(f\) is continuous from the right at \(c\) (or right-continuous at \(c\)). Lefthand limit and continuity from the left at \(c\) are similarly defined for \(c\in (a,b]\). The notation \(f(c-)\) can be similarly realized. For \(c\in (a,b)\), we see that \(f\) is continuous at \(c\) when
\[f(c)=f(c+)=f(c-).\]
We say that \(c\) is a discontinuity of \(f\) when \(f\) is not continuous at \(c\). More precisely, one of the following conditions is satisfied:
(a) Either \(f(c+)\) or \(f(c-)\) does not exist.
(b) Both \(f(c+)\) and \(f(c-)\) exist with \(f(c+)\neq f(c-)\).
(c) Both \(f(c+)\) and \(f(c-)\) exist with \(f(c+)=f(c-)\neq f(c)\).
In case (c), the point \(c\) is called a removable discontinuity since the discontinuity can be removed by re-defining \(f\) at \(c\) to have the common value \(f(c+)=f(c-)\).
In cases (a) and (b), we call \(c\) an irremovable discontinuity since the discontinuity cannot be removed by re-defining \(f\) at \(c\).
Definition. Let \(f\) be a real-valued function defined on a closed interval \([a,b]\). Suppose that \(f(c+)\) and \(f(c-)\) both exist at some interior point \(c\) of \([a,b]\). Then, we define three different jumps as follows.
- \(f(c)-f(c-)\) is called the lefthand jump of \(f\) at \(c\).
- \(f(c+)-f(c)\) is called the righthand jump of \(f\) at \(c\).
- \(f(c+)-f(c-)\) is called the jump of \(f\) at \(c\).
If any one of these three numbers is not equal to \(0\), \(c\) is called a jump discontinuity of \(f\). \(\sharp\)
Example. We provide many interesting examples of jump discontinuities.
(i) We consider the function
\[f(x)=\left\{\begin{array}{ll}{\displaystyle \frac{x}{|x|}} & \mbox{if \(x\neq 0\)}\\ A & \mbox{if \(x=0\)}.\end{array}\right .\]
Since \(f(0+)=1\) and \(f(0-)=-1\), the function \(f\) has a jump discontinuity at \(0\).
(ii) We consider the function
\[f(x)=\left\{\begin{array}{ll}1 & \mbox{if \(x\neq 0\)}\\ 0 & \mbox{if \(x=0\)}.\end{array}\right .\]
Since \(f(0+)=1=f(0-)\), the function \(f\) has a removable jump discontinuity at \(0\).
(iii) We consider the function
\[f(x)=\left\{\begin{array}{ll} 1/x & \mbox{if \(x\neq 0\)}\\ A & \mbox{if \(x=0\)}.\end{array}\right .\]
Since \(f(0+)\) and \(f(0-)\) do not exist, the function \(f\) has a irremovable jump discontinuity at \(0\).
(iv) We consider the function
\[f(x)=\left\{\begin{array}{ll}\sin (1/x) & \mbox{if \(x\neq 0\)}\\A & \mbox{if \(x=0\)}.\end{array}\right .\]
Since \(f(0+)\) and \(f(0-)\) do not exist, the function \(f\) has a irremovable jump discontinuity at \(0\).
(v) We consider the function
\[f(x)=\left\{\begin{array}{ll}x\sin (1/x) & \mbox{if \(x\neq 0\)}\\1 & \mbox{if \(x=0\)}.\end{array}\right .\]
Since \(f(0+)=0=f(0-)\), the function \(f\) has a removable discontinuity at \(0\). \(\sharp\)
Proposition. Suppose that \(f\) is increasing on \([a,b]\). Then \(f(c+)\) and \(f(c-)\) both exist for each \(c\in (a,b)\) satisfying
\[f(c-)\leq f(c)\leq f(c+).\]
At the endpoints, we also have \(f(a)\leq f(a+)\) and \(f(b-)\leq f(b)\).
Proof. Since \(f\) is increasing, the following set
\[A=\left\{f(x):a<x<c\right\}\]
is bounded above by \(f(c)\). Let \(\alpha =\sup A\). Then \(\alpha\leq f(c)\). We shall prove that \(f(c-)=\alpha\). In other words, we want to show that, given any \(\epsilon >0\), there exists \(\delta>0\) such that
\begin{equation}{\label{ma97}}\tag{1}
c-\delta <x<c\mbox{ implies }|f(x)-\alpha |<\epsilon.
\end{equation}
Since \(\alpha =\sup A\), there is an element \(f(x^{*})\in A\) satisfying satisfying \(\alpha -\epsilon <f(x^{*})\leq\alpha\). Since \(f\) is increasing, for any \(x\in (x^{*},c)\), we also have
\begin{align*} \alpha -\epsilon & <f(x^{*})\leq f(x)\\ & \leq f(c)\leq\alpha\\ & <\alpha+\epsilon,\end{align*}
i.e., \(|f(x)-\alpha |<\epsilon\). Therefore, we can take \(\delta =c-x^{*}\) to obtain (\ref{ma97}). We can similarly prove \(f(c)\leq f(c+)\). \(\blacksquare\)
\begin{equation}{\label{map452}}\tag{2}\mbox{}\end{equation}
Proposition \ref{map452}. Suppose that \(f\) is a strictly increasing real-valued function defined on a subset \(S\) of \(\mathbb{R}\). Then \(f^{-1}\) exists and is strictly increasing on \(f(S)\).
Proof. Since \(f\) is strictly increasing, it is one-to-one on \(S\), i.e., \(f^{-1}\) exists. Let \(y_{1},y_{2}\) be two points in \(f(S)\) with \(y_{1}<y_{2}\). Let \(x_{1}=f^{-1}(y_{1})\) and \(x_{2}=f^{-1}(y_{2})\). If \(x_{1}\geq x_{2}\), we obtain \(y_{1}\geq y_{2}\). This contradiction shows that \(x_{1}<x_{2}\), and the proof is complete. \(\blacksquare\)
Proposition. Suppose that the real-valued function \(f\) is strictly increasing and continuous on a compact interval \([a,b]\). Then \(f^{-1}\) is continuous and strictly increasing on the compact interval \([f(a),f(b)]\).
Proof. The result follows immediately from Proposition A in page continuity of functions and Proposition \ref{map452}. \(\blacksquare\)
Definition. Let \(f:[a,b]\rightarrow\mathbb{R}\) be a real-valued function defined on a closed interval \([a,b]\).
- We say that \(f\) is lower semi-continuous at \(y\) when, given any \(\epsilon >0\), there exists \(\delta >0\) such that \[|x-y|<\delta\mbox{ implies }f(y)<f(x)+\epsilon .\]
- We say that \(f\) is upper semi-continuous at \(y\) when, given any \(\epsilon >0\), there exists \(\delta >0\) such that \[|x-y|<\delta\mbox{ implies }f(y)+\epsilon<f(x).\]
The function \(f\) is upper semi-continuous at \(y\) if and only if the function \(-f\) is lower semi-continuous at \(y\).
