Let f: R → R be a function satisfying f(xy) = f(x) f(y) for all x, y ∈ R. If the function f is continuous at x = 1, show that it is continuous for all non-zero x ∈ R.
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Sol. We have,
f(xy) = f(x) f(y) ∀ x, y ∈ R
⇒ f(1) = f(1) . f(1) [Replacing both x and y by 1]
⇒ f(1) = 0 or f(1) = 1
If f(1) = 0, then for any x ∈ R, we have
f(x) = f(x .1) = f(x) . f(1) = 0, which is everywhere continuous.
If f(1) = 1
f is continuous at x = 1,
⇒ 
f(x) = 1 = 
f(x)
⇒ 
f (1+ h) = f(1) = 
f(1 –h)
Let a be any non- zero real number, then
f(x) = 
f(a + h) = 
f{a 
}
= 
f(a). f 
= f(a) 
f 
= f(a) . 1 = f(a)
and 
f(x) = f(a –h) = 
f(a 
) = 
f(a). f.
= f(a) 
f 
= f(a) .1 = f(a)
∴ 
f(x) = f(a) = 
f(x).
⇒ f(x) is continuous at x = a.
Since a is an arbitrary non- zero real number.
∴ f(x) is continuous for all non- zero x ∈ R.
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