Articles of analysis

Limit of square root without L'Hopital's rule.

How might one go about taking the following limit without using L’Hopital’s rule? I am stumped: $$\lim_{x \to \infty} \sqrt{x^2 + x} – x$$

If $f \in L^2$, then $f \in L^1$ and$\|f\|_{L^1} \leq \sqrt{2\pi} \|f\|_{L^2}$

I’m learning about Fourier analysis and need help with the following problem (which is part of a subchapter on $L^p$ spaces): Using the Cauchy-Schwarz inequality show that if $f \in L^2[-\pi, \pi]$, then $(1)$ $f \in L^1[-\pi, \pi]$ and $(2)$ $\|f\|_{L^1} \leq \sqrt{2\pi} \|f\|_{L^2}$. My work and thoughts: $(1)$ We note that if $f \in […]

These unknown uniformly differentiable functions

Let $f$ be defined on $[a,b]$ and there uniformly differentiable ($\,$the $\delta$ in the definition of derivative is independent of the point). Given $\epsilon>0$, choose a partition $P \, : \, a=a_0<a_1<\ldots<a_n=b \,$ of $ \,[a,b] \,$ with $||P|| \lt \delta$, and apply the definition to the points $\,a_0,\ldots,a_{n-1} \,$ getting $$\left|\frac {f(a_{i+1})-f(a_i)}{a_{i+1}-a_i}-f'(a_i) \right|<\epsilon\qquad(i=0,\ldots,n-1)$$ Typically, […]

Continuity of Parametric Integral

Consider a continuous function $f: X \times Y \rightarrow \mathbb{R}_{} \geq 0$, where $X \subset \mathbb{R}^n$ is compact, and $Y \subseteq \mathbb{R}^m$ is closed. Define $\hat{f}:X \rightarrow \mathbb{R}_{\geq 0}$ as the parametric integral $$ F(x) \ := \ \int_Y f(x,y) dy $$ Assume that $X$ is such that $\sup_{x \in X} F(x) < \infty $. […]

Summability of a function

If $u\in W^{1,p}(\Omega)$ where $\Omega$ is an open subset of $\mathbb{R}^n$ and $\xi$ is a smooth compactly supported function in $\Omega$, is it true that $\xi u^{\beta-p+1} \in W^{1,p}_0$ if $\beta >p-1$? (In the end my problem is to say if $u^{\beta-p+1} \in W^{1,p}$ (from this I know it follows the result).) I think if […]

Sufficient condition for isometry

They could give me some suggestions on how to use the data on the derivative, in: If $f:\mathbb{R}^{m} \rightarrow \mathbb{R}^{m}$ class is $C^{1}$ such that $\Vert f^{\prime}(x)v\Vert =\Vert v\Vert$, for all $v \in \mathbb{R}^{m}$. then $f$ is isometry. thank you very much

Measurable set of real numbers with arbitrarily small periods

I am trying to prove the following exercise (exercise 3, chapter 7 of Rudins Book “Real and Complex Analysis”): Suppose that $ E $ is a measurable set of real numbers with arbitrarily small periods. Explicitly, this means that there are positive numbers $ p_i $, converging to $ 0 $ as $ i\rightarrow \infty […]

How to prove this function is integrable??

Let $f(x)=0$ when $x<0$, and $f(x)=1$ if $x\geq 0$. Choose a countable dense sequence $\{r_n\}$ in [0,1]. Then, show that the function $F(x)=\sum_{n=1}^\infty 1/n^2 f(x-r_n)$ is integrable and has discontinuities at all points of the sequence $\{r_n\}$. I will prove this problem by using thm “if {x; f is discontinuous} has measure zero and f […]

Prove that $ \sqrt{\sum_{j=1}^{n}a_j^2} \le \sum_{j=1}^{n}|a_j|$, for all $a_i \in \mathbb{R}, i=\{1,2,…,n\}$

As the title says, prove that $ \sqrt{\sum_{j=1}^{n}a_j^2} \le \sum_{j=1}^{n}|a_j|$, for all $a_i \in \mathbb{R}, i=\{1,2,…,n\}$. I can solve for the case of n=2, but kind of stuck while proving the n-terms version. The attempt is as follow: $\left(\sqrt{\sum_{j=1}^{n}a_j^2}\right)^2 = \sum_{j=1}^n {a_j}^2 \le |\sum_{j=1}^n {a_j}^2| \le \sum_{j=1}^n |a_j|^2 \mbox{by triangle inequality.}$ Alright, I am stuck […]

The definition of $p$ capacity of a set $A\subset\mathbb{R}^n$

I am having a bit of difficult understanding the definition of the $p$-capacity of a set $A\subset\mathbb{R}^n$ and I was wondering if anyone would be able to clarify whether I have the right idea or not. This is what I understand so far. Formerly the definition of the $p$-capacity of a set is presented as […]