Show that function's inverse is not continuous at a point I have the following problem:
Show a bijective function $f$ where $f'(0)$ is equal to $1$ and where the $f^{-1}(x)$ is not cont. at $f(1)$.
So far, I know that the derivative of an inverse function $f^{-1}(f(x))$ is $(f^{-1})'(f(x))=\frac{1}{f'(f^{-1}(f(x)))}=\frac{1}{f'(x)}$,
and that $f^{-1}$ not being continuous at $f(0)$ would mean that it is not differentiable there either. Don't know where to go from there however.
Any help would be appreciated!
 A: It's very subtle: Since $f$ is differentiable at $0$, it will be continuous at $0$. But we need a function $f$ that is not continuous on any neighborhood of $0$. Let's try this:
$$f(x) = \begin{cases} \frac x2 + \dfrac1{\left[\frac2x\right]}, & x>0 \\ x, & x\le 0.\end{cases}$$
I leave it to you to check that $f$ is injective. Now, $f$ is differentiable at $0$: From the left, the derivative is $1$. From the right, we look at
$$\lim_{h\to 0^+}\frac{f(h)}h = \lim_{h\to 0^+} \frac{\frac h2+\frac 1{[2/h]}}h = \frac12 + \lim_{h\to 0^+} \frac 1h\frac1{[\frac 2h]}.$$
Now note that
$$\frac 2h-1<\left[\frac 2h\right]\le \frac 2h,$$
so for $0<h\le 2$ we have
$$\frac 1h\frac 1{\frac 2h}\le \frac1h\frac1{\left[\frac 2h\right]} < \frac 1h\frac 1{\frac 2h-1}.$$
Since the left and right quantities approach $1/2$ as $h\to 0^+$, by the squeeze property we have
$$\lim_{h\to 0^+} \frac 1h\frac1{[\frac 2h]}=\frac 12,$$
and, indeed, $f'(0)=1$. Since $f$ has jump discontinuities at $\{1/n: n\in\Bbb N\}$, $f^{-1}$ will not be continuous at $f(0)=0$.
A: A few notes first. Suppose, WLOG, that $f(0) = 0$.
If $f$ is continuous and strictly monotonic then it has a continuous inverse. Hence we need $f$ to be non monotonic or non continuous (or both) in any neighborhood of $0$.
It is easy to verify that if the function is continuous and non monotonic in some neighborhood of $0$, then (by the Intermediate Value Theorem) injectivity is lost. Thus the function must be non continuous in any neighborhood of $0$.
For $f^{-1}$ to be not continuous in $f(0)$, what we need is the existence of $\varepsilon > 0$ such that for any $\delta > 0$ there exists $0<|x|<\delta$ yielding $|f^{-1}(x)|\geq \varepsilon$. If we translate this into a condition on $f$ we obtain the following. There exists $\varepsilon$ such that for any $\delta>0$ there is an $|x|\geq \varepsilon$ such that $0<|f(x)|< \delta$.
In other words, we need the function $f$ to attain values arbitrarily close to $0$ for $x$ belonging to some neighborhood of $\infty$.
First we will construct a function $g(x)$ that satisfies the requirements in the right neighborhood of $0$, and then obtain our function using a reflection.
At this aim, let us start from the identity map $x\mapsto x$ which satisfies the conditions save for the fact that the inverse is indeed continuous in $0$. According to our previous observations, we need to create "holes" in the co-domain (with $0$ as limit point) that will be used as images of "larger" values of $x$. In order to keep differentiability in $0$, we want a function that is bounded for example as follows.
$$h(x) = -x^2+x \leq g(x) \leq x.$$
So if we start, say, from $x= \frac12$, we can decide to replace the identity map with $h\left(\frac12\right) =\frac14.$
Now, of course, to maintain injectivity we have to modifiy the image of $\frac14$. So we let $$g\left(\frac14\right) = h\left(\frac14\right) =  h\circ h\left(\frac12\right) = \frac3{16}.$$
Again this will force us to modify the image of $\frac{3}{16}$. Again we choose to let $$g\left(\frac{3}{16}\right) = h\left(\frac{3}{16}\right) = h\circ h \circ h\left(\frac12\right) = \frac{31}{256}.$$
We can then proceed inductively. This yields
$$g\left(h^{(n)}\left(\frac12\right)\right) = h^{(n+1)}\left(\frac12\right),$$
for $n=0,1,\dots$,
where
$$h^{(n)}(x) =\underbrace{ h\circ h\circ h\cdots \circ h }_{n}(x)$$
for $n>0$ and $h^{(0)}(x) = x$.
Observe that, having started from $\frac12$ we obtained as images rational numbers with some power of $2$ as denominator. See blue dots in the figure below.

Now that we have created a "hole" in the co-domain, we can fill it, for example by letting
$$g(2) = \frac12.$$
This in turn creates a "hole" in $2$ that can be filled by letting $$g(4) = 2.$$
Proceding further yields, for $n=2,3,\dots$,
$$g\left(2^n\right) = 2^{n-1}.$$
Refer again to the blue dots in the figure below.

We can now go back to the neighborhoods of $0$ and create another "hole" in $\frac13$, using the same procedure. Note that these further modifications do not interfere with the previous ones since now the images have, as denominator, some power of $3$ (red dots in the first figure above). Again the "hole" will be filled by letting $g(3) = \frac13$, and the process will continue in correspondence with the powers of $3$.
Hence if we let $p_k$, for $k=1,2,\dots$ be the sequence of primes
and, for $x\geq 0$,
$$g(x) =\begin{cases}
h(x) & \left(\mbox{if}\ \ x=h^{(n)}\left(\frac1{p_k}\right),\ \ \mbox{for some}\ k>0\ \mbox{and some}\ n\geq 0\right)\\
\frac1{p_k} & \left(\mbox{if}\ \ x= p_k\ \mbox{for some}\ k>0\right)\\
p_k^{n-1} & \left(\mbox{if}\ \ x=p_k^n,\ \mbox{for some}\ k>0\ \mbox{and some}\ n> 0\right)\\
x & (\mbox{otherwise})
\end{cases} 
$$
we get a function with the desired properties for $x\geq 0$.
Finally let
$$f(x) = \begin{cases}
g(x) & (x\geq 0)\\
-g(-x) & (x<0).
\end{cases}
$$
Since $-x^2+x \leq g(x) \leq x$, and $x\leq -g(-x)\leq x^2+x$, we have
$$-x^2+x\leq f(x)\leq x^2+x,$$
wich guarantees, by the Squeeze Theorem, that
$$f'(0) = \lim_{x\to 0} \frac{f(x)}x = 1,$$
as required.
Consider now the null sequences
$$a_k = 2^{-k}$$
and
$$b_k = \frac1{p_k}.$$
We have that $f^{-1}(a_k) \to 0$ whereas $f^{-1}(b_k) = p_k \to \infty$, thus  $$\lim_{x\to 0} f^{-1}(x)$$
is not defined and the inverse function is not continuous in $0$.

EDIT
One can even think of a function having the required properties and also a connected and compact domain and range, as suggested in the plot below.

The dashed lines represent the parabolas with equation
$$y=-x^2+x$$
and
$$x=y^2-y+1,$$
and the straight line
$$r: \ \ y=1-x.$$
We can exploit the simmetry with respect to $r$ by first defining for $0\leq x\leq \frac12$ the injective function
$$
g_1(x)=
\begin{cases}
h(x) & \left(\mbox{if}\ \ x=h^{(n)}\left(\frac1{p_k}\right),\ \ \mbox{for some}\ k>1\ \mbox{and some}\ n\geq 0\right)\\
x & (\mbox{otherwise}).
\end{cases}
$$
Then, for $0\leq x \leq 1$,
$$
g(x)=\
\begin{cases}
g_1(x) & \left(0\leq x\leq \frac12\right)\\
\frac1{p_k} & \left(\mbox{if}\ \ x=1-\frac1{p_k},\ \ \mbox{for some} \ k>1\right)\\
1-g_1^{-1}(1-x) & (\mbox{otherwise}).
\end{cases}
$$
