Articles of real analysis

Can absolute convergent series be expressed as sum of two series?

Let $C\subset \omega \bigwedge A\bigcap B = \emptyset \bigwedge A\bigcup B = C$. Let $\{x_i\}$ be a sequence of nonnegative reals. Suppose $C$ is infinite and $\sum_{i\in C} x_i$ converges. (Since it converges absolutely, it makes sense to define its sum in this way) Then $\sum_{i\in C} x_i = \sum_{i\in B} x_i + \sum_{i\in A} […]

A strongly pseudomonotone map that is not strongly monotone

Let $K$ be a closed convex subset in $\mathbb{R}^n$ and $F: K\rightarrow \mathbb{R}^n$. We say that $F$ is strongly monotone on $K$ if there exists $\gamma>0$ such that $$ \left<F(y)-F(x), y-x\right>\geq \gamma\|y-x\|^2, \quad \forall x,y\in K. $$ $F$ is strongly pseudomonotone on $K$ if there exists $\gamma>0$ such that $$ \left<F(x), y-x\right>\geq 0 \Longrightarrow \left<F(y), […]

$\lim_{x\to\infty}\int_x^{x+1}f(y)dy=0$ implies $\lim_{x\to\infty}\frac{\int_0^{x}f(y)dy}{x}=0$

Let $f$ be a non negative continuous function on $[0,\infty)$ such that $$\lim_{x\to\infty}\int_x^{x+1}f(y)dy=0.$$ How do we prove that $$\lim_{x\to\infty}\frac{\int_0^{x}f(y)dy}{x}=0.$$ If we see this question using the primitive function, do we have the following result for a continuous function $F$: $$\lim_{x\to\infty}F(x+1)-F(x)=0$$ implies that $$\lim_{x\to\infty}\frac{F(x)}{x}=0.$$

$f: \mathbb{R}^2\to \mathbb{R}^2$ is differentiable, and satisfies an inequality that involves its partials – show that f is a bijection.

Suppose that $f: \mathbb{R}^2\to \mathbb{R}^2$ is differentiable, and the partial derivatives of the components $f_1$, $f_2$ satisfy $$max(|\frac{\ df_1}{dx} -1|, |\frac{df_1}{d_y}|, |\frac{df_2}{d_x}|, |\frac{df_2}{d_y}-1|) <10^{-10}.$$ Prove that f is a bijection. Note: f is not assumed to be continuously differentiable. Any ideas on how to tackle this problem? We don’t have an explicit function given for […]

Determining whether $\int_{0}^{\infty} \frac{x \sin(x)}{1+x^2}dx$ converges and converges absolutely

I would like to check whether $$\int_{0}^{\infty} \frac{x \sin(x)}{1+x²}dx$$ converges and converges absolutely. I have a feeling that neither is true, however none of the methods known to me seem to help. I struggle to find a lower estimate for the function. Any hints and help welcome. I tried using $$\frac{x \sin(x)}{1+x²}\leq \frac{x \sin(x)}{x²}=\frac{ \sin(x)}{x}$$ […]

Behavior at $0$ of a function that is absolutely continuous on $$

The function $f$ on $[0,1]$ is absolutely continuous on $[\epsilon,1]$ for $0<\epsilon<1.$ I further have that $$\int_0^1x|f'(x)|^pdx<\infty.$$ I’m trying to show that $$ \lim_{x\to 0}f(x)\ \text{exists and is finite}\qquad \text{if}\ p>2, $$ $$ \frac{f(x)}{|\log x|^{1/2}}\to 0\ \text{as}\ x\to 0\qquad \text{if}\ p=2,\ \text{and} $$ $$ \frac{f(x)}{x^{1-\frac{2}{p}}}\to 0\ \text{as}\ x\to 0\qquad \text{if}\ p<2. $$ (Of course, for […]

The form $w_p(x,y) = \frac{(x-a)\,dy-(y-b)\,dx}{(x-a)^2+(y-b)^2}$ is exact $\iff$ $\exists$ angle function

Given $p=(a,b)\in\mathbb{R^2}$, show that the $1$-form $w_p:\mathbb{R^2}-\{p\}\to (\mathbb{R^2})^*$, defined by $$w_p(x,y) = \frac{(x-a)\,dy-(y-b)\,dx}{(x-a)^2+(y-b)^2}$$ is closed. Prove that this form is exact in an open $U\subset \mathbb{R^2}-\{p\}$ if and only if there exists a continuous function (necessairly $C^\infty$) $\theta_p:U\to\mathbb{R}$ such that $\cos \theta_p (z) = \frac{x-a}{|z-p|}$ and $\sin \theta_p (z) = \frac{y-b}{|z-p|}$ for all $z=(x,y)\in U$. […]

For which values of $x$ does $\frac {x+x^n}{1+x^n}$ converge and to what?

Let $$s_n(x) = \frac{x+x^n}{1+x^n}$$ for all $n\in\Bbb N$ and $x\in \Bbb R\setminus\{-1\}$. Find each real number $x$ for which the sequence $(s_n)$ is convergent and find the limit of the sequence. I’ve broken it into cases. Here’s what I’ve been able to get so far. $x\in[1,\infty)$ case: $$\left|\frac{x+x^n}{1+x^n}-1\right| = \left|\frac{x+x^n-(1+x^n)}{1+x^n}\right| = \left|\frac{x-1}{1+x^n}\right|\lt \frac{x}{x^n} = \left(\frac […]

Showing that if $f=g$ a.e. on a general measurable set (for $f$, $g$ continuous), it is not necessarily the case that $f=g$.

This is related to a question I just asked, that I now think was based on wrong assumptions. It is true that if $f=a$ a.e. on the interval $[a,b]$, then $f = a$ on $[a,b]$. However, apparently it is not true for a general measurable set $E$ with $m(E) \neq 0$, which confuses me greatly. […]

Theorem 6.12 (e) in Baby Rudin: If $f \in \mathscr{R}\left(\alpha_1\right)$ and $f \in \mathscr{R}\left(\alpha_2\right)$, then $\ldots$

If $f \in \mathscr{R}\left(\alpha_1\right)$ and $f \in \mathscr{R}\left(\alpha_2\right)$, then $f \in \mathscr{R}\left(\alpha_1 + \alpha_2 \right)$ and $$ \int_a^b f d\left(\alpha_1 + \alpha_2 \right) = \int_a^b f d\alpha_1 + \int_a^b f d\alpha_2.$$ This is part of Theorem 6.12 (e) in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition. Here is my proof: Let […]