real analysis - Discontinuous function with continuous inverse?

Here is something that is confusing me.

The function $\begin{equation}
f(x)=\begin{cases}

x, & \text{if $x \in[-1,0]$}\\
x+1, & \text{if $x \in (0,1]$}
\end{cases}
\end{equation}$

Is clearly discontinuous at $x=0$ yet its inverse

$\begin{equation}
f^{-1}(x)=\begin{cases}
x, & \text{if $x \in[-1,0]$}\\

x-1, & \text{if $x \in (1,2]$}
\end{cases}
\end{equation}$

is continuous on the separate counterparts. Now, clearly $f(x)$ doesn't output any values in the range $0 < p \leq 1$ so I don't know if the fact $f^{-1}(x)$ isn't defined in the interval $(0,1]$ is enough to say it's discontinuous or not, because clearly these values aren't of interest for $f(x)$ either. Any clarification would be greatly appreciated!

Thanks

Answer

You must think in terms of intervals.

If f is continuous and injective on an interval, then it has an inverse which is continuous also.

Indeed, if $f$ is continuous and injective on an interval, then it is monotonic and its image is an interval. So the inverse is monotonic on an interval and its image is an interval. Hence it is continuous.

In your case, $f$ is continuous and injective on $[-1,0]$ and on $(0,1]$.

So $f^{-1}$ is continuous on $f([-1,0])=[-1,0]$ and on $f((0,1])=(1,2]$.

The fact that the initial intervals $[-1,0]$ and $(0,1]$ had a common point in their closures does not mean that their images under $f$ have the same property. As your example shows.

The discontinuity of $f$ at $0$ is reflected by a gap in the domain of $f^{-1}$.

real analysis - Convergence from $L^p$ to $L^infty$

If $f$ is a function such that $f \in L^\infty \cap L^ {p_0}$ where $L^\infty$ is the space of essentially bounded functions and $ 0 < p_0 < \infty$. Show that $ || f|| _{L^p} \to ||f || _{L^\infty} $ as $ p \to \infty$. Where $|| f||_{L^\infty} $ is the least $M \in R$ such that $|f(x)| \le M$ for almost every $x \in X$.

The hint says to use the monotone convergence theorem, but i can't even see any pointwise convergence of functions.
Any help is appreciated.

Answer

Hint: Let $M\lt\|f\|_{L^\infty}$ and consider
$$
\int_{E_M}\left|\frac{f(x)}{M}\right|^p\,\mathrm{d}x
$$
where $E_M=\{x:|f(x)|\gt M\}$. I believe the Monotone Convergence Theorem works here.

Further Hint: $M\lt\|f\|_{L^\infty}$ implies $E_M$ has positive measure. On $E_M$, $\left|\frac{f(x)}{M}\right|^p$ tends to $\infty$ pointwise. MCT says that for some $p$, the integral above exceeds $1$.

real analysis - continuous functions on $mathbb R$ such that $g(x+y)=g(x)g(y)$

Let $g$ be a function on $\mathbb R$ to $\mathbb R$ which is not identically zero and which satisfies the equation $g(x+y)=g(x)g(y)$ for $x$,$y$ in $\mathbb R$.

$g(0)=1$. If $a=g(1)$,then $a>0$ and $g(r)>a^r$ for all $r$ in $\mathbb Q$.

Show that the function is strictly increasing if $g(1)$ is greater than $1$, constant if $g(1)$ is equal to $1$ or strictly decreasing if $g(1)$ is between zero and one, when $g$ is continuous.

Answer

For $x,y\in\mathbb{R}$ and $m,n\in\mathbb{Z}$,
$$
\eqalign{
g(x+y)=g(x)\,g(y)
&\implies
g(x-y)={g(x) \over g(y)}
\\&\implies
g(nx)=g(x)^n
\\&\implies

g\left(\frac{m}nx\right)=g(x)^{m/n}
}
$$
so that $g(0)=g(0)^2$ must be one (since if it were zero, then $g$ would be identically zero on $\mathbb{R}$), and with $a=g(1)$, it follows that $g(r)=a^r$ for all $r\in\mathbb{Q}$. All we need to do now is invoke the continuity of $g$ and the denseness of $\mathbb{Q}$ in $\mathbb{R}$ to finish.

For example, given any $x\in\mathbb{R}\setminus\mathbb{Q}$, there exists a sequence $\{x_n\}$ in $\mathbb{Q}$ with $x_n\to x$ (you could e.g. take $x_n=10^{-n}\lfloor 10^nx\rfloor$ to be the approximation of $x$ to $n$ decimal places -- this is where we're using that $\mathbb{Q}$ is dense in $\mathbb{R}$). Since $g$ is continuous, $y_n=g(x_n)\to y=g(x)$. But $y_n=a^{x_n}\to a^x$ since $a\mapsto a^x$ is also continuous.

Moral: a continuous function is completely determined by its values on any dense subset of the domain.

algebra precalculus - Range of Compositions of Functions

Is there an efficient method to find the range of compositions of functions?

I know this: The domain of the composition of functions is the INTERSECTION of the domain of the INSIDE function and the domain of the RESULTING function.

However, I'm struggling to find the range of compositions of functions. Is the range the INTERSECTION of the range of the OUTSIDE function and the range of the resulting function? I don't think this assumption is correct though. Can someone please help?

For example, let's define two functions: $f(x)= \sqrt{x}$ and $g(x)= x+1$. The resulting function, $f(g(x))$ is $\sqrt{x+1}$. This resulting function has a domain of greater than or equal to $-1$, and it has a range of greater than or equal to $0$.

In the example, $f(x)$ has a domain of greater than or equal to $0$, $g(x)$ has a domain of all real numbers, $f(x)$ has a range of greater than or equal to $0$, $g(x)$ has a range of all real numbers.

The domain of the composition is the INTERSECTION of the domain of the INSIDE function (in this case, $g(x)$ has a domain of all real numbers), and the domain of the RESULTING function (in this case, $f(g(x))$, or $\sqrt{x+1}$, has a domain of greater than or equal to $-1$). Hence, the final domain of the composition is GREATER THAN OR EQUAL TO $-1$.

What is the range??

Thank you!

calculus - How to prove that $log(x)1$?

It's very basic but I'm having trouble to find a way to prove this inequality

$\log(x)

when $x>1$

($\log(x)$ is the natural logarithm)

I can think about the two graphs but I can't find another way to prove it, and, besides that, I don't understand why should it not hold if $x<1$

Can anyone help me?

Thanks in advance.

real analysis - How to prove that $limlimits_{xtoinfty}e^xtext{arccot}(x)=infty$?

How to prove that
$\lim\limits_{x\to\infty}e^x\text{arccot}(x)=\infty$?

I already figured that $\frac{\text{d}}{\text{dx}}[\text{arrcot}(x)]=\frac{\text{d}}{\text{dx}}\left[\arctan\left(\frac{1}{x}\right)\right]=-\frac{1}{x^2+1}$. Now I wanted to use L'Hospitals rule after doing some algebra:$$\lim\limits_{x\to\infty}e^x\text{arccot}(x)=\lim\limits_{x\to\infty}\frac{e^x}{\frac{1}{\text{arccot}(x)}}=\lim\limits_{x\to\infty}\frac{e^x}{\dfrac{1}{\left(x^2+1\right)\operatorname{arccot}^2\left(x\right)}}$$ using it twice didn't work out aswell, what am I supposed to do?

Answer

$$\lim_{x\to\infty}\dfrac{\text{arccot} x}{e^{-x}}=\lim_{x\to\infty}\dfrac{-\dfrac1{1+x^2}}{-e^{-x}}=\lim_{x\to\infty}\dfrac{e^x}{1+x^2}$$

$$=\lim_{x\to\infty}\dfrac{1+x+\dfrac{x^2}2+\dfrac{x^3}{3!}+O(x^4)}{1+x^2}$$

Divide numerator & denominator by $x^2$

analysis - The limit of these sequences

In my analysis homework, I have the following two True/False questions about limits:

(c) if

$$\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = 1$$

Then $(a_n)$ converges

(d) if

$$\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = 1$$

Then $(a_n)$ diverges

The thing is: I'm not sure how to interpret these questions. In the answer key, both are given as False; for (c) they give the example where $a_n = n$, and for (d) they give $a_n=\frac{1}{n}$.

Here's where I get confused:

1. For both (c) and (d), how do I interpret the question exactly? It seems that they give counterexamples where it couldn't be the case that $(a_n)$ converges / diverges.





2. The way I interpret it now is that the limit of the absolute value of $\frac{a_{n+1}}{a_n}$ as n tends to infinity is 1. So if we have $a_n =1$, how is this not the case for the sequence $(a_n)$?

Answer

There's a default convention behind such questions. Informally speaking: any such statement with some object in it is understood to be true if it is always true, i.e. if it is true for any such object.

For example, the statement

(c) If $\displaystyle \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|=1$, then $(a_n)$ converges,

actually is meant to say

(c) For any sequence $(a_n)$, if $\displaystyle \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|=1$, then $(a_n)$ converges,

and that's what you need to classify as true or false.

• The statement is true if it is true for all such sequences $(a_n)$;


• The statement is false if at least one counterexample exists (regardless of the fact that there may be other examples that satisfy it).

The answer key for this question gives you such a counterexample: the sequence $a_n=n$ satisfies the "if" part since $\displaystyle \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to\infty} \frac{n+1}{n}=1$, and yet $\displaystyle \lim_{n\to\infty}a_n$ doesn't exists since these numbers tend to infinity.

Question d) should be interpreted similarly.

And then if you put them together, then as @Peter said, we see that the fact $\displaystyle \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|=1$ gives us no information about the behavior of the sequence $(a_n)$: some such sequences, like the counterexample for c), diverge; and some such sequences, like the counterexample for d), converge.