Page 159 - DMTH401_REAL ANALYSIS

Unit 11: Continuity

Theorem 3: Let f and g be any real functions both continuous at a point a  R. Then, Notes
(i) f defined by (f) (x) = f(x), is continuous for any real number ,
(ii) f + g defined by (f + g) (x) = f(x) + g(x) is continuous at a,
(iii) f – g defined by (f – g) (x) = f(x) – g(x) is continuous at a,

(iv) fg defined by (fg) (x) = f(x) g(x) is continuous at a,
f(x)
(v) f/g defined by (f/g) (x) = , is continuous at a provided g(a)  0.
g(x)
Proof: Let x be an arbitrary sequence converging to a. Then the continuity of f and g imply that
n
the sequences f(x ) and g(x ) converge to f(a) and g(a) respectively. In other words, lim f(x ) = f(a),
n n n
lim g(x ) = g(a).
n
Using the algebra of sequences, we can conclude that
lim f(x ) = f(a),
n
lim (f + g) (x ) = lim f(x ) + lim g(x ) = f(a) + g(a),
n n n
lim (f – g) (x ) = lim f(x ) – lim g(x ) = f(a) – g(a),
n n n
lim (f  g) (x ) = lim f(x ) lim g(x ) = f(a) g(a).
n n n
If infinite number of x ’s are such that g(x ) = 0, then g(X ) – g(a) implies that g(a) = 0, a
n n n
contradiction.
This proves the parts (i), (ii), (iii) and (iv). To prove the part (v) we proceed as follows:
Since g(a)  0, we can find a > 0 such that the interval ]g(a) – , g(a) + [ is either entirely to the
right or to the left of zero depending on whether g(a) > 0 or g(a) < 0. Corresponding to a > 0, there
exists a  > 0 such that |x – a| <  implies |g(x) – g(a)| < , i.e., g(a) –  < g(x) < g(a) + at. Thus,
1 1
for x such that |x – a| <  , g(x)  0. If (x ) converges to a, omitting a finite number of terms of the
i n
f(x )
sequence if necessary, then we can assume that g(x )  0, for all n. Hence, n converges to
n g(x )
f(a) f n
and so is continuous at a. This completes the proof of the theorem.
g(a) g
In part (v) if we define f by f(x) = 1, then it follows that if g is continuous at ‘a’ and g(a)  0, then
its reciprocal function 1/g is continuous at ‘a’.
Now, we prove another theorem, which shows that a continuous function of a continuous
function is continuous.
Theorem 4: Let f and g be two real functions such that the range of g is contained in, the domain
of f. If g is continuous at x = a, f is continuous at b = g(a) and h(x) = f(g(x)), for x in the domain of
g, then h is continuous at a.
Proof: Given  > 0, the continuity of f at b = g(a) implies the existence of an  > 0 such that for

|y – b| < , |f(y) – f(b)| <  ...(l)
Corresponding to  > 0, from the continuity of g at x = a, we get a  > 0 such that
|x – a| <  implies |g(x) – g(a)| <  ...(2)
Combining (1) and (2) we get that
|x – a| <  implies that
|h(x) – h(a)| = |f(g(x)) – f(g(a))|

= |f(y) – f(b)| < ,

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