If $f \in L^{1}[0,2\pi]$ define $\hat{f}(n)$ for $n \in\mathbb{Z}$ by $$\hat{f}(n) = \frac{1}{2\pi} \int_{0}^{2\pi} f(t) \cdot (\cos(nt) -i\sin(nt)) \ dt$$ Suppose $M$ is a closed linear subspace of $L^{1}[0,2\pi]$ such that $\sum |\hat{f}(n)| &lt; \infty$ for each $f \in M$.Then how can one show that there is a constant $C$ such that $$\sum_{n} |\hat{f}(n)| \leq C \int_{0}^{2\pi} |f(t)| \ dt$$ for each $f \in M$. Can anyone help me out with a clear solution.
for $f\in L^1[0,2\pi]$ define $$\hat{f}(n)=\int_{0}^{2\pi} f(t)e^{-int} dt$$ for $n\in\mathbb{Z}$, $X$ is a closed linear subspace of $L^1[0,2\pi]$ such that $\sum_{n} |\hat{f}(n)|&lt;\infty$ for each $f\in X$, we need to show that there is $M&lt;\infty$ such that  $\sum_{n} |\hat{f}(n)|\le M\int_{0}^{2\pi}|f(t)|dt$ for each $f\in X$  please tell me how to solve this one, I have no clue.
I see in my company's IT installations that there are numerous disks with one partition occupying all of the space on disk. There is typically something like a /dev/sdb with a partition /dev/sdb1, and the size of /dev/sdb1 is the size of the whole block device. The partition is then mounted with the required file-system format, mountpoint etc.  I do not understand why we would use a partition rather than the whole disk for such (1-partition) configurations. Is there any reason why a partition would be used in this case rather than a whole disk ? Are there any best practices that recommend this approach ?  For example, we could create a Filesystem on a whole raw block device and mount it without needing any partition:  mkfs.ext4 -E stride=16,stripe-width=64 /dev/xvde mount /dev/xvde /mnt/abc   That creates a mountpoint without a specified partition. As a verification:  # sfdisk -l /dev/xvde  Disk /dev/xvde: 6527 cylinders, 255 heads, 63 sectors/track  # fdisk -l /dev/xvde  Disk /dev/xvde: 53.7 GB, 53687091200 bytes, 104857600 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes   For something with a partition, we get the list of partitions under it:  # fdisk -l /dev/sdb  Disk /dev/sdb: 2000.4 GB, 2000398934016 bytes, 3907029168 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk label type: dos Disk identifier: 0x00082e8c     Device Boot      Start         End      Blocks   Id  System /dev/sdb1            2048  3907026943  1953512448   83  Linux   My fundamental question is this: Is there some advantage in creating a partition and then creating a filesystem on that partition, as compared to creating a filesystem on the whole disk and then mounting the whole disk ? How to remove session entries from LightDM? The energy of a moving object is $E = mv^2\;.$ That is it increases with velocity squared.  I walk at say 3 miles per hour, or lets round that down to 1 meter per second for a slow walk. I weigh less than $100~\mathrm{kg}\;,$ but lets just round that up to $100~\mathrm{ kg}$ for convenience (it is just after Christmas).  So when I walk along the pavement, I have $100~\mathrm{kg\; m^2 s^{-2}}$, 100 joules of kinetic energy.  Now I get on a passenger jet, which is cruising at around 500 knots, call that 250 meters per second.  In my seat I have $100\times 250^2 = 6250000$ joules of kinetic energy. But when I walk down the aisle I have $100\times 251^2 = 6300100$ joules of kinetic energy. The difference between these is: 50100 joules.  It feels the same to me, walking down the pavement as walking down the aisle of the plane. I didn't have to make some huge effort to get up to speed on the plane, yet I needed 500 times the energy to do it.  How is this possible, and where did the energy come from? Is it possible to accept the alternative hypothesis? I'm aware of several related questions here (e.g., , ) but I don't know the definitive answer for my question below.  Suppose a hypothesis test where we want to test whether a coin is fair or not. We have two hypotheses:  $H_0: p(head)=0.5$  $H_1: p(head)\neq0.5$  Suppose we use 5% significance level, there are two possible cases:   When we obtain the data and find that the p-value is less than 0.05, we say "With significance level 5%, we reject $H_0$."  p-value is greater than 0.05, then we say "With 5% significance level, we cannot reject $H_0$."   My question is:      In case 1, is it correct to say "we accept $H_1$"?   Intuitively, and from what I have learned in the past, I feel that "accepting" anything as a result of hypothesis testing is always incorrect. On the other hand, in this case, since the union on $H_0$ of $H_1$ covers the whole "space", "reject $H_0$" and "accepting $H_1$" look exactly the same to me. On another thought, I can also think of the following idea, which says it is incorrect to say "we accept $H_1$":     We have an evidence strong enough to believe that $H_0$ is not true, but we may not have an evidence strong enough to believe that $H_1$ is true. Therefore, "rejecting $H_0$" does not automatically imply "accepting $H_1$"   So, what is the right answer? Evaluate the following limit of finite sum Book set underground everyone lives in pods woman wants to see her son Can anyone help me identify this book please? I read a book in the 80s about a woman living underground in a pod. Everyone had a big tv which gave them news etc (but it was starting to go wrong) and their food was delivered direct to their pod.  They weren't allowed to go to the earth surface as it was supposedly contaminated (post-war?). She gets a message from her son on the other side of the world and she takes the tube train from outside her pod and then some sort of aircraft to see him and eventually gets onto the surface of the earth and breathes the air and meets free people. Finder sidebar flashing text and high CPU load since El Capitan After update to OS X 10.11 I started having these bugs:   Finder sidebar text flashing  Very high CPU load:      I have Macbook Pro Retina 13" Mid 2014  I reset NVRAM.  I deleted com.apple.finder.plist, it didn't help.  Also I did this: , no longer flashes. Getting around the FAT32 4GB file size limit Does Windows 7 allow auto-login with a stored password, like TweakUI did? Which JDK version is compatible with Netbeans on Ubuntu? I'm trying to install Netbeans on my Ubuntu. It was successfully installed but was unable to create new projects or new files. I have installed JDK 12 and then I removed it because I found here a solution that recommended removal of latest versions of JDK and set JDK 8 as default. I did what solution said but now I can't install JDK 8. It is showing me this error message:   E: Package 'oracle-java8-installer' has no installation candidate    So I started searching on the Internet about the solution and I found that "Oracle Java (JDK) 8 Installer PPA (DISCONTINUED)".   What should I do now? From : In calculus and other branches of mathematical analysis, an indeterminate form is an algebraic expression obtained in the context of limits. Limits involving algebraic operations are often performed by replacing subexpressions by their limits; if the expression obtained after this substitution does not give enough information to determine the original limit, it is known as an indeterminate form.    The indeterminate forms include $0^{0},\frac{0}{0},(\infty - \infty),1^{\infty}, \ \text{etc}\cdots$   My question is can anyone give me a nice explanation of why $1^{\infty}$ is considered to be an indeterminate form? Because, i don't see any justification of this fact. I am still perplexed. Can this intuition give a proof that an isometry $f:X \to X$ is surjective for compact metric space $X$? A prelim problem asked to prove that if $X$ is a compact metric space, and $f:X \to X$ is an isometry (distance-preserving map) then $f$ is surjective. The official proof given used sequences/convergent subsequences and didn't appeal to my intuition. When I saw the problem, my immediate instinct was that an isometry should be "volume-preserving" as well, so the volume of $f(X)$ should be equal to the volume of $X$, which should mean surjectivity if $X$ is compact. The notion of "volume" I came up with was the minimum number of $\epsilon$-balls needed to cover $X$ for given $\epsilon &gt; 0$. If $f$ were not surjective, then because $f$ is continuous this means there must be a point $y \in X$ and $\delta &gt; 0$ so that the ball $B_\delta(y)$ is disjoint from $f(X)$. I wanted to choose $\epsilon$ in terms of $\delta$ and use the fact that an isometry carries $\epsilon$-balls to $\epsilon$-balls, and show that given a minimum-size cover of $X$ with $\epsilon$ balls, that a cover of $X$ with $\epsilon$-balls could be found with one fewer ball if $f(X) \cap B_\delta(y) = \emptyset$, giving a contradiction. Can someone see a way to make this intuition work? Prove $2^n-1$ is divisible by $(2^m-1)^2$ iff... Prove that for any $m,n\in\mathbb{Z}^+$, the number $2^n-1$ is divisible by $(2^m-1)^2$ iff the number $n$ is divisible by $m(2^m-1)$.  My work so far:  It's an if and only if statement, so the common approach is to split the proof into two "if then" proofs ($A\rightarrow B \text{ and } B\rightarrow A$).  Starting with $A\rightarrow B$, for a direct proof, I would assume $A$ and prove $B$. I can use the definition of divisibility:  Assume $n=m(2^m-1)k$ for some integer $k$  Then I need to prove that $2^n-1=(2^m-1)^2h$ for some integer $h$  So I'm thinking I need to start with the expression $2^n-1$ and transform it to $(2^m-1)^2h$, or vice versa, and I need to use the assumption $n=m(2^m-1)k$ somehow.  I notice that the LS of the equality contains the variable $n$ while the RS contains $m$, and the assumption relates $n$ to $m$, so that could be a way forward. For example, simple substitution gives $2^{m(2^m-1)k}-1$, but I'm not sure what I could do from here. Perhaps there's a divisibility theorem that could help?  Other things I notice: $2^n-1$ and $(2^m-1)^2$ are always odd. Also, the divisors in both cases are multiples of $2^m-1$.  Any help/hints appreciated. How do I tell Maven to use the latest version of a dependency? Do not enumerate a section, but add it into table of contents How do I rid of number of a section in the way that it still would be in the table of contents.  If I use   ... \section{SN1} ...   then I see this section name in table of contents as   1 SN1  2 SN2  3 SN3  ...   If I use   \section*{SN}   I don't see it in the content:   1 SN2  2 SN3   How to get in content just   SN1  1 SN2  2 SN3   ? Individual ColorFunction with RGBColor Referencing bib files with spaces in the filename Split Excel table over multiple PowerPoint slides I have a table in Excel which is too long for one PowerPoint slide. So I would like to split it over 2 or 3 slides, but have the header on each of the slides. What is the easiest way to accomplish that? I am using Excel and PowerPoint 2007. The idea behind the sum of powers of 2 I know that the sum of power of $2$ is $2^{n+1}-1$, and I know the mathematical induction proof. But does anyone know how $2^{n+1}-1$ comes up in the first place.  For example, sum of n numbers is $\frac{n(n+1)}{2}$. The idea is that we replicate the set and put it in a rectangle, hence we can do the trick. What is the logic behind the sum of power of $2$ formula? Counting homomorphisms between $\Bbb Z$ and $\Bbb Z_n$ and also between $\Bbb Z_n$ and $\Bbb Q$