Articles of real analysis

recurrence relations for proportional division

I am looking for a solution to the following recurrence relation: $$ D(x,1) = x $$ $$ D(x,n) = \min_{k=1,\ldots,n-1}{D\left({{x(k-a)} \over n},k\right)} \ \ \ \ [n>1] $$ Where $a$ is a constant, $0 \leq a \leq 1$. Also, assume $x \geq 0$. This formula can be interpreted as describing a process of dividing a […]

Showing that a limit exists and showing $f$ is not integrable.

Show that $\lim\limits_{n \to \infty} \int_{1}^{n} f$ exists while $f$ is not integrable over $[1,\infty)$. Define $f(x)=\frac{\sin(x)}{x}$ for $1 \leq x <\infty$.

Real Analysis, Folland Problem 5.3.29 The Baire Category Theorem

The Baire Category Theorem – Let $X$ be a complete metric space a.) If $\{U_n\}_1^\infty$ is a sequence of open dense subsets of $X$, then $\bigcap_1^\infty U_n$ is dense in $X$. b.) $X$ is not a countable union of nowhere dense sets. The name for this theorem comes from Baire’s terminology for sets: If $X$ […]

Comparison of the change of variable theorem

I would like to compare the change of variable theorem for 1 variable and more. What are the differences, in which case we need stronger assumptions? How do they differ? What is the best way to write the theorems for comparison? Multivariable: Let $\varphi : \Omega \rightarrow \mathbb{R}^n$ (where $\Omega \subset \mathbb{R}^k$ and $k\leq n)$ […]

Prove that, there exists no continuous function $f:\mathbb R\rightarrow\mathbb R$ with $f=\chi_{}$ almost everywhere.

Prove that, there exists no continuous function $f:\mathbb R\rightarrow\mathbb R$ with $f=\chi_{[0,1]}$ almost everywhere.$\textbf(Make\ sure\ that\ your\ proof\ is\ completely\ rigorous)$. I don’t know, which property to use. (It is not allowed to show it with $\epsilon-\delta-criterion$, our last topics were: Lp-Spaces, Radon-Nikodym Theorem, Riesz Representation Theorem,Lipschitz-Functions, Product measures, Fubini Theorem) but i can’t find […]

Convergence of a recursively defined sequence

Let a sequence $(a_n)_{n=0}^\infty$ be defined recursively $a_{n+1} = (1-a_n)^{\frac1p}$, where $p>1$, $0<a_0<(1-a_0)^{\frac1p}$. Let $a$ be the unique real root of $a=(1-a)^{\frac1p}$, $0<a<1$. It is clear $0<a_0<(1-a_0)^{\frac1p}\Leftrightarrow 0<a_0<a$. Prove 1) $a_{2k-2}<a_{2k}<a<a_{2k+1}<a_{2k-1}$ and $a_{2k+1}-a<a-a_{2k}$. 2) $\lim\limits_{n\to\infty}a_n=a$. Define $f(x):=(1-x)^{\frac1p}$. Consider $f^2$. When $p=2$, $a_{n+2}=f^2(a_n)=\big(1-(1-a_n)^{\frac12}\big)^{\frac12}$. $a_{n+2}>a_n\Leftrightarrow (1-a_n)(1+a_n)^2>1\Leftrightarrow a_n<(1-a_n)^{\frac12}$, and the conclusion is proved. But I am having difficulty […]

Possible areas for convex regions partitioning a plane and containing each a vertex of a square lattice.

If the plane is partitioned into convex regions each of area $A$ and each containing a single vertex of a unit square lattice, is $A\in (0,\frac{1}{2})$ possible? If each each vertex is in the interior of its region is $A \neq 1$ possible? More generally if $\rm{ I\!R}^n$ ($n\ge 1$) is partitioned into convex regions, […]

A continuously differentiable function with vanishing determinant is non-injective?

(This question relates to my incomplete answer at https://math.stackexchange.com/a/892212/168832.) Is the following true (for all n)? “If $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ is continuously differentiable and satisfies $\det(f'(x)) = 0$ for all $x$, then $f$ is not injective.” If so, what’s the most elementary proof you can think of? It is clearly true for $n=1$. In […]

If $f$ is midpoint convex, continuous, and two times differentiable, then for any $a, b \in \mathbb{R}$, there exists $c$ such that $f''(c) \geq 0$

One of my analysis texts states this as an exercise If $f$ is midpoint convex, continuous, and two times differentiable, then for any $a, b \in \mathbb{R}$, there exists $c \in [a, b]$ such that $f”(c) \geq 0$. and says as a hint that I shouldn’t have to prove that midpoint convexity and continuity together […]

Showing $ {L^{2}}() $ Is Contained in $ {L^{1}}() $

I’m trying to show that $ {L^{2}}([0,1]) $ is contained in $ {L^{1}}([0,1]) $. This is what I have so far: Since $ f \in {L^{2}}([0,1]) $, then we have that $(\int_0^1 |f|^2)^{\frac{1}{2}}<\infty$. Thus we have $\int_0^1 |f|^2=M<\infty$, making $|f|^2$ integrable over $[0,1]$. This means that we can write $\int_{A_1} |f|^2+\int_{A_2} |f|^2$, where $A_1=\lbrace x\in[0,1]||f|^2>|f|\rbrace$ […]