1550251515-Classical_Complex_Analysis__Gonzalez_

340 Chapter^6

and the directional derivative of the given function at z 0 E 'Y* (in the

direction of 'Y) is given by

f' 'Y (z 0 ) = 2to(l 1 + + 2ito 2t~) = zo + zoe-2i8o

where ei^80 = (1 + 2it 0 )/(1 + 4tn^112 •

The existence of the directional derivative off at a point zo of a regular

arc 'Y implies the continuity of f at z 0 along the arc, as shown in the

following theorem.

Theorem 6.11 Let f be defined along a regular arc 'Y· If f has a di-

rectional derivative at zo E 'Y*, then f is continuous at zo along the arc.

However, a continuous function at z 0 E 'Y* (along 'Y) may fail to have a

directional derivative at z 0 •

Proof We have

f(z) = f(zo) + f(z)-f(zo) (z - z 0 )

Z-Zo

for z E -y*. If z = z(t), a :::; t:::; /3, defines the arc"'(, and if z(to) = z 0 ,

letting t --+ t 0 we have z(t) --+ z(t 0 ) = z 0 • Then it follows that

)i..IIJ 0 f(z) = f(zo) + f~(zo) · 0 = f(zo)

ZE'"'f*

which shows that f is continuous at zo along 'Y·

To see that the converse does not hold, consider the following

Example Let f(z) = zsin(l/Rez) for z-=/= 0 and f(O) = 0. This function
is continuous along the line segment z = t +it (0:::; t:::; 1). Yet it does not
have a directional derivative along the segment at z = O, since

f(z) - f(O)

z-0

. 1
= Sln -
t

which has no limit as t --t o+.

Remark It is clear that one-sided derivatives of real functions may be

considered as directional derivatives of such functions at z = x in the

direction of the real line segment z = t (a:::; t:::; b), or else, z = a+ b - t

(a :::; t :::; b ).