Is the Limit Function of Continuous Functions Measurable
2.3: Limits and Continuous Functions
- Page ID
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If \(f(z)\) is defined on a punctured disk around \(z_0\) then we say
\[\lim_{z \to z_0} f(z) = w_0\]
if \(f(z)\) goes to \(w_0\) no matter what direction \(z\) approaches \(z_0\).
The figure below shows several sequences of points that approach \(z_0\). If \(\lim_{z \to z_0} f(z) = w_0\) then \(f(z)\) must go to \(w_0\) along each of these sequences.
Many functions have obvious limits. For example:
\[\lim_{z \to 2} z^2 = 4 \nonumber\]
and
\[\lim_{z \to 2} \dfrac{z^2 + 2}{z^3 + 1} = 6/9. \nonumber\]
Here is an example where the limit doesn't exist because different sequences give different limits.
Show that
\[\lim_{z \to 0} \dfrac{z}{\overline{z}} = \lim_{z \to 0} \dfrac{x + iy}{x - iy} \nonumber\]
does not exist.
Solution
On the real axis we have
\[\dfrac{z}{\overline{z}} = \dfrac{x}{x} = 1, \nonumber\]
so the limit as \(z \to 0\) along the real axis is 1. By contrast, on the imaginary axis we have
\[\dfrac{z}{\overline{z}} = \dfrac{iy}{-iy} = -1, \nonumber\]
so the limit as \(z \to 0\) along the imaginary axis is -1. Since the two limits do not agree the limit as \(z \to 0\) does not exist!
Properties of limits
We have the usual properties of limits. Suppose
\[\lim_{z \to z_0} f(z) = w_1 \text{ and } \lim_{z \to z_0} g(z) = w_2\]
then
- \(\lim_{z \to z_0} f(z) + g(z) = w_1 + w_2\)
- \(\lim_{z \to z_0} f(z) g(z) = w_1 \cdot w_2\).
- If \(w_2 \ne 0\) then \(\lim_{z \to z_0} f(z)/g(z) = w_1 /w_2\)
- If \(h(z)\) is continuous and defined on a neighborhood of \(w_1\) then \(\lim_{z \to z_0} h(f(z)) = h(w_1)\) (Note: we will give the official definition of continuity in the next section.)
We won't give a proof of these properties. As a challenge, you can try to supply it using the formal definition of limits given in the appendix.
We can restate the definition of limit in terms of functions of \((x, y)\). To this end, let's write
\[f(z) = f(x + iy) = u(x, y) + iv (x, y)\]
and abbreviate
\[P = (x, y), P_0 = (x_0, y_0), w_0 = u_0 + iv_0.\]
Then
\[\lim_{z \to z_0} f(z) = w_0 \text{ iff } \begin{cases} \lim_{P \to P_0} u(x, y) = u_0 \\ \lim_{P \to P_0} v(x, y) = v_0 \end{cases}\]
Note. The term 'iff' stands for 'if and only if' which is another way of saying 'is equivalent to'.
Continuous Functions
A function is continuous if it doesn't have any sudden jumps. This is the gist of the following definition.
If the function \(f(z)\) is defined on an open disk around \(z_0\) and \(\lim_{z \to z_0} f(z) = f(z_0)\) then we say \(f\) is continuous at \(z_0\). If \(f\) is defined on an open region \(A\) then the phrase '\(f\) is continuous on \(A\)' means that \(f\) is continuous at every point in \(A\).
As usual, we can rephrase this in terms of functions of \((x, y)\):
Fact. \(f(z) = u(x, y) + iv(x, y)\) is continuous iff \(u(x, y)\) and \(v(x, y)\) are continuous as functions oftwo variables.
(i) A polynomial
\[P(z) = a_0 + a_1 z + a_2 z^2 + ... + a_n z^n \nonumber\]
is continuous on the entire plane. Reason: it is clear that each power \((x + iy)^k\) is continuous as a function of \((x, y)\).
(ii) The exponential function is continuous on the entire plane. Reason:
\[e^z = e^{x + iy} = e^x \cos (y) + ie^x \sin (y). \nonumber\]
So the both the real and imaginary parts are clearly continuous as a function of (\(x, y\)).
(iii) The principal branch \(\text{Arg} (z)\) is continuous on the plane minus the non-positive real axis. Reason: this is clear and is the reason we defined branch cuts for arg. We have to remove the negative real axis because \(\text{Arg} (z)\) jumps by \(2 \pi\) when you cross it. We also have to remove \(z = 0\) because \(\text{Arg} (z)\) is not even defined at 0.
(iv) The principal branch of the function \(\text{log} (z)\) is continuous on the plane minus the non-positive real axis. Reason: the principal branch of log has
\[\text{log} (z) = \text{log} (r) + i \text{Arg} (z). \nonumber\]
So the continuity of \(\text{log} (z)\) follows from the continuity of \(\text{Arg} (z)\).
Properties of Continuous Functions
Since continuity is defined in terms of limits, we have the following properties of continuous functions.
Suppose \(f(z)\) and \(g(z)\) are continuous on a region \(A\). Then
- \(f(z) + g(z)\) is continuous on \(A\).
- \(f(z) g(z)\) is continuous on \(A\).
- \(f(z) / g(z)\) is continuous on \(A\) except (possibly) at points where \(g(z) = 0\).
- If \(h\) is continuous on \(f(A)\) then \(h(f(z))\) is continuous on \(A\).
Using these properties we can claim continuity for each of the following functions:
- \(e^{z^2}\)
- \(\cos (z) = (e^{iz} + e^{-iz})/2\)
- If \(P(z)\) and \(Q(z)\) are polynomials then \(P(z)/Q(z)\) is continuous except at roots of \(Q(z)\).
Source: https://math.libretexts.org/Bookshelves/Analysis/Complex_Variables_with_Applications_(Orloff)/02%3A_Analytic_Functions/2.03%3A_Limits_and_Continuous_Functions