Quod Erat Demonstrandum


Continuity of composite functions

Filed under: HKALE,Pure Mathematics — johnmayhk @ 7:50 上午

It is well-known that the following is NOT true in general,

\lim_{x \rightarrow a}f(g(x)) = f(\lim_{x \rightarrow a}g(x))

Just give an example, let

f(x) = \frac{x^2 - 1}{x - 1}
g(x) = \frac{\sin x}{x}

Easy to have

\lim_{x \rightarrow 0}f(g(x)) = 2, but
f(\lim_{x \rightarrow 0}g(x)) is undefined, hence

\lim_{x \rightarrow a}f(g(x)) \ne f(\lim_{x \rightarrow a}g(x)) in this case.

So, when does the equality sign hold? Here is a simple theorem in secondary school pure mathematics.

g(x) is continuous at x = a and
f(x) is continuous at x = g(a),
f(g(x)) is continuous at x = a.

It may be a bit confusing when using x as the (independent) variable of both functions f and g, though it is perfectly OK. For educational purpose, may be it is better to re-write the theorem as follows.

g(x) is continuous at x = a and
f(t) is continuous at t = g(a),
f(g(x)) is continuous at x = a.

There is a ‘bridge’ t, which is also a fuction of x, that is t = g(x).

Let’s prove the theorem step by step.

The given that “g(x) is continuous at x = a" can be interpreted in this way.

When the value of x tends to a, the value of g(x) will get closer and closer to that of g(a).

In symbol,

x \rightarrow a \Rightarrow g(x) \rightarrow g(a) – – – – – – (*)

Besides, the given that “f(t) is continuous at t = g(a)“, by the definition of continuity, we have

\lim_{t \rightarrow g(a)}f(t) = f(g(a)) – – – – – – (**)

OK, to show that “f(g(x)) is continuous at x = a“, we need to show

\lim_{x \rightarrow a}f(g(x)) = f(g(a))

Hence, we start with \lim_{x \rightarrow a}f(g(x)).

\lim_{x \rightarrow a}f(g(x))
= \lim_{x \rightarrow a}f(t) [where t = g(x)]
= \lim_{g(x) \rightarrow g(a)}f(t) [by (*)]
= \lim_{t \rightarrow g(a)}f(t)
= f(g(a)) [by (**)]

That is f(g(x)) is continuous at x = a.

Of course, it is extremely easy to give a proof using \epsilon - \delta language. But, again, this is for secondary school educational purpose, right?

OK, the conclusion above is

\lim_{x \rightarrow a}f(g(x)) = f(g(a)) for f, g satisfying certain conditions.

Knowing that g(a) = \lim_{x \rightarrow a}g(x), the result can be written as

\lim_{x \rightarrow a}f(g(x)) = f(\lim_{x \rightarrow a}g(x))

and hence the original question is answered.

[SBA time]
1. Try to find out loopholes of the so-called proof above. Give comments on the weakness of the discussion above.

2. Try to visualize and explore the situation by clicking the following created by a silly teacher named John Ng.


1. Suppose f, g are continuous at x = a, it is easy to know that f+g, f-g, fg and \frac{f}{g} (g(a) \ne 0) are continuous at x = a.

How about f \bigvee g and f \bigwedge g? Are they continuous at x = a as well?

f \bigvee g = \max\{f , g\}
f \bigwedge g = \min\{f , g\}

(e.g. f(x) = x + 3, g(x) = 2x, then (f \bigvee g)(4) = 8, (f \bigwedge g)(4) = 7)

The answer is affirmative. And the proof is extremely simple when knowing that

f \bigvee g and f \bigwedge g can be converted into

f \bigvee g = \frac{1}{2}(f + g + |f - g|)
f \bigwedge g = \frac{1}{2}(f + g - |f - g|)

Try to verify the statements above.

2. Do you know that “continuous" can be refined into “upper semi-continuous" and “lower semi-continuous"? Try to explore the concepts by reading books and searching from the internet.


2 則迴響 »

  1. I wonder if there should be no absolute signs on f + g for maximum and minimum functions…

    迴響 由 皮旦 — 2009/03/19 @ 8:00 上午 | 回覆

  2. Yes, thank you 皮旦!

    迴響 由 johnmayhk — 2009/03/19 @ 8:07 上午 | 回覆

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