# Baby Rudin 5.2: Continuity Required to prove Differentiability?

(From Rudin Principles of Mathematical Analysis, 5.2)

Suppose $$f'(x) > 0$$ in ($$a, b$$). Prove that $$f$$ is strictly increasing in ($$a, b$$), and let $$g$$ be its inverse function.

Prove that $$g$$ is differentiable, and that $$g'(f(x)) = \frac{1}{f′(x)} \quad (a < x < b)$$

Here's an answer I found online:

Let $$g : f(a, b) → (a, b)$$ be the inverse function of $$f$$, i.e., $$g(f(x)) = x$$ for all $$x ∈ (a, b)$$.

We now show that $$g′(y) = \lim\limits_{z→y} \frac {g(z) − g(y)}{z − y}$$ exists for all $$y ∈ f(a, b)$$.

Put $$y = f(x)$$ and $$z = f(t)$$, where $$x, t ∈ (a, b)$$, then since $$f$$ is continuous (by Theorem 5.2), so is $$g$$ (by Theorem 4.17), and $$z → y$$ implies $$t → x$$.

It follows that

\begin{align*} \lim_{z→y} \frac{g(z) − g(y)}{z − y} &= \lim_{t→x}\frac{g(f(t)) − g(f(x))}{f(t) − f(x)} \\ &= \lim_{t→x}\frac{t − x}{f(t) − f(x)} \\ &= \lim_{t→x}\frac{1}{\frac{f(t) − f(x)}{t − x}} \\ &= \frac{1}{f′(x)} \end{align*}

Question:

Why it is necessary to for g to be continuous? The only step that uses continuity is the changing of the limit values ( $$z → y$$ implies $$t → x$$), but that comes from $$f$$ I think?

• $g$ must be continuous at $y = f(x)$ so that $\lim_{y \rightarrow z} g(z) = g(y) = g(f(x)) = x$. We really need to be able to plug in $x$ into the function $f^{-1}(x)$.
– D.B.
Apr 7, 2019 at 15:30
• When we g(y) = g(f(x)), aren't we just plugging values into g? (since we defined y = f(x)). I don't see why the limit is involved to get from g(z)-g(y) to g(f(t)) - g(f(x)) Apr 7, 2019 at 15:53

You do not need to assume that the inverse function $$g\colon f([a,b])\to [a,b]$$ is continuous because that actually does come for free. To see this, it suffices to show that if $$U\subset [a,b]$$ is an relatively open interval, then $$g^{-1}(U)$$ is open in $$f([a,b])$$. Now, $$g^{-1}(U) = f(U)$$, and since $$f$$ is strictly increasing, the image of a relatively open interval in $$[a,b]$$ under $$f$$ is another relatively open interval in $$f([a,b])$$.

In fact, we do need continuity of $$g$$.

If only g is continuous, then $$z→y$$ implies $$f(z)→f(y)$$. But this is unhelpful, since we want $$t→x$$.

Recall the only thing we know is that $$y = f(x)$$ and $$z = f(t)$$.

From this, we know $$g(z) = g( f(t)) = t$$ (since $$g$$ and $$f$$ are inverses).

Similarly: $$g(y) = g( f(x)) = x.$$

Now, suppose $$g$$ is continuous.

Then $$z→y$$ implies $$g(z)→g(y)$$. (def of continuity).

It follows immediately that $$t→x$$.

In some sense you are right that the change of limit values comes from the properties $$f$$. However, proving this is harder than just using a known result that the inverse function of $$f$$ is continuous. But let's do it just for completeness.

What we want to show is that, given $$f(t) \to f(x)$$ as $$t \to x_0$$ for some $$x_0 \in (a,b)$$, it must be the case that $$x_0 = x$$. We argue by contradiction. Suppose $$x_0 \neq x$$. By the continuity of $$f$$, we have that $$f(x_0) = f(x)$$. Assume, without loss of generality, that $$x < x_0$$. Then the mean value theorem tells us there exists $$c \in (x,x_0) \subset (a,b)$$ such that $$f'(c) = \frac{f(x_0) - f(x)}{x_0 - x} = \frac{0}{x_0-x}=0.$$ But this contradicts that $$f'(x) > 0$$ for all $$x \in (a,b)$$. Thus, $$x = x_0$$ must hold. Hence, we have shown that for any $$t,x \in (a,b)$$, $$f(t) \to f(x)$$ implies that $$t \to x$$.