A function $f$ such that $f \in L_1$ but $f \notin L_p$ for $p>1$ I want find a function $f: [0,1] \mapsto \mathbb{R}$ such that $f \in L_1[0,1]$ but $f \notin L_p[0,1]$ for all $p&gt;1$.  My attempts:  First I thought in the family of functions $\frac{1}{x^\alpha}$ but this function belongs to $L_q$ iff $\alpha \cdot q \leqslant 1$ so I need find $\alpha$ such that: $\alpha &lt;1 $ and $\alpha \cdot q \geqslant 1$ for all $q&gt;1$ but this its impossible!!   After other attempts using variations and combinations of $1/x$, $ln x$ and $e^x$ I researched in the mathstack and found this questions:   The kingkongdonutguy's question is exactly what I was looking for, but I do not understand very well the Tomas' (and of Davide) hint... My interpretation:  Choice two sequences $\{a_n\}_{n \in \mathbb{N}}$ and $\{t_n\}_{n \in \mathbb{N}}$ com $a_n,t_n \to 0$ now make a sequence os disjoint intervals $\{I_n\}_{n \in \mathbb{N}}$ such that, for each $n$,$0 &lt; m(I_n) &lt; t_n$ and $\bigcup I_n = [0,1]$. Define a function: $$f(x)= \sum\limits_{n=1}^{\infty} a_n \cdot \chi_{I_n}(x)$$ Make a simple calculation: $$\int\limits_{0}^{1} f(x)dx = \sum\limits_{n=1}^{\infty} \int_{I_n} a_n dx = \sum\limits_{n=1}^{\infty} a_n\cdot m(I_n) \leqslant \sum a_n \cdot t_n$$ So I need choice $\{a_n\}$ and $\{t_n\}$ such that $\sum a_n \cdot t_n$ converges but $\sum a_n ^{p} \cdot t_n$ not converges for $p&gt;1$. The problem: using  we have $$\lim_{n \to \infty} \frac{a_n^p \cdot t_n}{a_n\cdot t_n} = \lim_{n \to \infty} a_n^{p-1}=0 $$(because $p&gt;1$) so don't is possible this choice ...  Also found this question  but I could not adapt for a finite domain..  Someone can give me a (other) hint to construct this function??
$f \in L^1$, but $f \not\in L^p$ for all $p > 1$ "Find an $f \in [0,1]$ such that $f \in L^1$ but $f \not\in L^p$ for any $p &gt; 1$."  I've thought about doing something like $$f(x) = \frac{1}{x}$$ where $|f|^p = \frac{1}{x^p}$ doesn't converge when $p &gt; 1$.  But this function isn't itself in $L^1$.  Could someone please give me a hint for how to solve this problem?  I wish there were a situation where you had convergence on the closed half disc $[0,1]$ and divergence on $(1, \infty)$, rather than my current predicament where I have convergence on the open half-disc $[0, 1)$ and divergence on $[1, \infty)$.
An inequality involving 4 real or complex numbers Solve this integral for free WiFi I saw this today, I checked in Mathematica and the integral comes out to $\pi$, but I have no idea how to solve it.     FREE Wi-Fi: The Wi-Fi password is the first $10$ digits of the answer.   $$\int_{-2}^2\left(x^3\cos\frac x2+\frac12\right)\sqrt{4-x^2}\ dx$$ What is a loss function in decision theory? My notes define a loss function as the 'cost' incurred when the true value of $\theta$ is estimated by $\hat\theta$. What kind of cost is it talking about? monetary cost? or is it something related to errors? What are the various units (ex, em, in, pt, bp, dd, pc) expressed in mm? How can one determine the length of the different units (measured 1ex, 1em, 1in, 1pt, 1bp, 1dd, 1pc) in mm? Is there a way (eg. a binary hack into Finder) to stop it from creating .DS-Stores on LOCAL Volumes?   I am using Leopard, therefore these don't work:    (requires Lion)  (requires Snow Leopard)   Also,  slows down my computer. On November 11, my American girlfriend flew over to the UK to visit me. She was detained and questioned, and refused entry under the belief she would not leave within those six months. This was due to her having no return flight booked, as we were going to wait until we were together in the UK to book a return flight for either herself, or both of us back to the USA once we decided what exact dates I would be able to travel. Lesson learned. A hard lesson, but we learned!  Anyway, they said although she'll flag up in the future, it won't be a straight ban from entering and she could even try to come back the next day if she so pleased, although it would be unwise if her circumstances had not changed.  Fast forward to December 18. She got herself a job in the States to prove she intends to leave at the end of her stay, and has since applied for a UK Standard Visitor Visa. She posted her documents off to New York earlier today.  Will they send her a letter back refusing to issue her a visa since, as an American citizen, she does not even need one? She wanted to get one to be safe as I told her not to fly to the UK even with return flights booked if she doesn't have a visa, as she risks being denied once again after a long flight. It's hard to get this information on the Internet, and some stories I read said that they were sent a letter back from the UK Embassy denying their visa simply because they don't need one in the first place - even after that person was denied entry.  We're in a bit of a pickle granted, so any help would be greatly appreciated.   Cheers. Edited photos have reverted to originals on iphone What is the use of the square brackets [] in sql statements? I currently have Windows 7 on my computer and I want to install Ubuntu alongside Windows safely.  Could someone please provide clear instructions? I'd like to get a notification when a question that I've voted to close (but isn't closed yet) is edited One of the first things ever taught in a differential calculus class:   The derivative of $\sin x$ is $\cos x$. The derivative of $\cos x$ is $-\sin x$.   This leads to a rather neat (and convenient?) chain of derivatives:   sin(x) cos(x) -sin(x) -cos(x) sin(x) ...   An analysis of the shape of their graphs confirms some points; for example, when $\sin x$ is at a maximum, $\cos x$ is zero and moving downwards; when $\cos x$  is at a maximum, $\sin x$ is zero and moving upwards.  But these "matching points" only work for multiples of $\pi/4$.  Let us move back towards the original definition(s) of sine and cosine:  At the most basic level, $\sin x$ is defined as -- for a right triangle with internal angle $x$ -- the length of the side opposite of the angle divided by the hypotenuse of the triangle.  To generalize this to the domain of all real numbers, $\sin x$ was then defined as the Y-coordinate of a point on the unit circle that is an angle $x$ from the positive X-axis.  The definition of $\cos x$ was then made the same way, but with adj/hyp and the X-coordinate, as we all know.  Is there anything about this basic definition that allows someone to look at these definitions, alone, and think, "Hey, the derivative of the sine function with respect to angle is the cosine function!"  That is, from the unit circle definition alone.  Or, even more amazingly, the right triangle definition alone.  Ignoring graphical analysis of their plot.  In essence, I am asking, essentially, "Intuitively why is the derivative of the sine the cosine?" Proving the summation formula using induction: $\sum_{k=1}^n \frac{1}{k(k+1)} = 1-\frac{1}{n+1}$ I am trying to prove the summation formula using induction:  $$\sum_{k=1}^n \frac{1}{k(k+1)} = 1-\frac{1}{n+1}$$  So far I have...  Base case:   Let n=1 and test   $\frac{1}{k(k+1)} = 1-\frac{1}{n+1}$  $\frac{1}{1(1+1)} = 1-\frac{1}{1+1}$    $\frac{1}{2} = \frac{1}{2}$   True for n=1   Induction Hypothesis:   Assume the statement is true for the n-th case   $\sum_{k=1}^n  \frac{1}{k(k+1)} = 1-\frac{1}{n+1}$  Inductive Step:   Prove, using the Inductive Hypothesis as a premise, that   $$\sum_{k=1}^{n+1}  \frac{1}{k(k+1)} = \sum_{k=1}^{n}  \frac{1}{k(k+1)} + \frac{1}{(n+1)(n+2)} = 1-\frac1{n+1} + \frac{1}{(n+1)(n+2)} = \frac{(n+1)(n+2)}{(n+1)(n+2)}+\frac{-2-n}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)} = \frac{(n+1)(n+2)-2-n+1)}{(n+1)(n+2)} = \frac{(n+1)(n+2)-n-1}{(n+1)(n+2)} = \frac{n^2+2n+1}{(n+1)(n+2)} = \frac{(n+1)(n+1)}{(n+1)(n+2)} = \frac{n+1}{n+2}$$  To prove $$ 1-\frac{1}{n+2} = \frac{n+1}{n+2} $$ Multiply both sides by $n+2$ to get an equivalent expression. $$ (1-\frac{1}{n+2}) * (n+2) = (\frac{n+1}{n+2}) * (n+2)  $$ $$ n+1=n+2−1 $$  Does this all make sense? How can this be improved upon? How do I list the default installed packages? Proving that $\sum_{k=0}^{n}\frac{(-1)^k}{{n\choose k}}=[1+(-1)^n] \frac{n+1}{n+2}.$ I'm downloading some images from a service that doesn't always include a content-type and doesn't provide an extension for the file I'm downloading (ugh, don't ask).  What's the best way to determine the image format in .NET?  The application that is reading these downloaded images needs to have a proper file extension or all hell breaks loose. How to fix java.lang.UnsupportedClassVersionError: Unsupported major.minor version I was able to install Ubuntu 12.04 using the alternative version of the image, (64 bit) but when I start the system it tells me the GPU  locked up, and instead presents a text only screen.  I understand that this site is not for bug reports, but at the same time, I know that there are some versions of the installer than have to be compatible with my graphics card (Nvidia GTS 360M).  I assumed that the mere fact that the alternative installer ran was an indication that I could get some sort of graphical interface running after the install.  Is there some way of getting some type of graphical interface running on my machine? Even better would be is there a way to get the rest of the system running as normal (i.e. GPU up and running)?  If an internet connection is required to download driver support or something similar, directions for how to configure a wireless connection with WPA2-Personal protection from the text-only interface (bash).  Any help would be appreciated.  Note: I was able to find  which tells me that the card is supported.  Note 2: In the install software stage of the install, I was unable to install anything (no internet connection)  Update: The exact error message is [8.591394] [drm] 0000:01:00.0: GPU Lockup - switching to software fbcon  Update 2: If I try to install with the regular version of Ubuntu, the screen goes staticy. This also happens if I install with Wubi (distasteful I know) and then try to use Ubuntu.  Update 3: I've tried xforcevesa, nomodeset and xforcevesa nomodeset as my wubi boot options.  Update 4: Thanks to jokerdino, I can now connect to the internet, so internet tools would now be a possibility. Understanding the -exec option of `find` Correctly indexing multiple domains with same content in Google and others Can I apply for two Schengen visas if I have two subsequent trips within a short period? I am traveling to Germany on May 1st for an exhibition for which I have already obtained a visa and its validity is up to May 13th. I am returning on May 12th and then traveling on May 17th to UK, and from UK to Switzerland for vacation with family. As time is very short for applying for a Schengen visa again after returning on the 12th with May 13-14 being a Saturday and Sunday, can I apply for Schengen visa again and can I get 2 valid Schengen visa?