Consider sequence spaces of the form,  $$\ell^p=\Big\{x=\left(x_j\right)_{j=1}^\infty  \mathrel{}\big|\mathrel{} \sum_{j=1}^\infty \left\lvert x_j \right\rvert ^p\lt\infty\Big\}$$  for $1\le p\lt\infty$, equipped with the norm,  $$\left\|x\right\|_p:=\left(\sum_{j=1}^\infty\left\lvert x_j \right\rvert ^p\right)^{\frac{1}{p}}$$  for $x\in\ell^p$.  Now, in the case that $n=2$, the familiar norm on $\ell^2$ is obtained from the inner product, that is to say that the norm is induced by the inner product. And since $\ell^2$ is a infinite-dimensional Banach space, it follows that it is in fact an infinite dimensional Hilbert space.  Why is it, that when $p\neq2$, the space $\ell^p$ is not an inner product space?
Can anybody please help me to prove this:  Let $p$ be greater than or equal to $1$. Show that for the space $\ell_p=\{(u_n):\sum_{n=1}^\infty |u_n|^p&lt;\infty\}$ of all $p$-summable sequences (with norm $||u||_p=\sqrt[p]{\sum_{n=1}^\infty |u_n|^p}\ )$, there is an inner product $&lt;\_\,|\,\_&gt; $ s.t. $||u||^2=&lt;u\,|\,u&gt;$ if and only if $p=2$.
Is there a way to simulate multiple iphones using xcode/iphone sim? Breaking single polygon into multiple polygons? I created voronoi polygons after which I dissolved them on the basis of one attribute. Now there are cases where in for one single attribute "abc" I have3 polygons at different places with same object id. I would like to break these into 3 polygons as they are at different locations.  Is there any tool in ArcGIS Desktop or QGIS which can help me in doing that? Geometric interpretation of Electromagnetism If there are two twins. such that one of them goes on a space voyage maintaining a constant velocity, and that one never returns to earth, and the twins decide to never meet but send information about their age to each other by some light pulse or any other method.  If then one knows that he is younger, won't he know he is the one travelling thus violating the postulate of relativity that all inertial frames are equal w.r.t. all phenomena and that one can never know if he is moving with a constant velocity. Totally bounded, complete $\implies$ compact Add increase/decrease tab indent buttons to the editor toolbar These should work by adding spaces or removing spaces from the beginning of the current line or every line between the start and end of a text selection (inclusive).  (edited this question to supplement ) Bathroom vanity light with no junction box Ignoring invalid input features with Intersection tool in QGIS? In QGIS 2.18, when I used the Vector > Geoprocessing Tools > Intersection tool, I was able to check an option which said "Ignore NULL geometries [optional]".  Now I've downloaded QGIS 3.2 and that option no longer exists. But when I try to create the intersection between two layers I'm working on, I get a message saying that there is an invalid geometry somewhere and that I should either fix it it or "change the Processing setting to the 'Ignore invalid input features' option."   But that option doesn't exist! Or if it does exist, I just can't find it anywhere.  Does anyone know how/where to change this setting in QGIS 3.2? How to have separated but vinculated multiple Ubuntu One accounts? Why can a function modify some arguments as perceived by the caller, but not others? I'm trying to understand Python's approach to variable scope. In this example, why is f() able to alter the value of x, as perceived within main(), but not the value of n?  def f(n, x):     n = 2     x.append(4)     print('In f():', n, x)  def main():     n = 1     x = [0,1,2,3]     print('Before:', n, x)     f(n, x)     print('After: ', n, x)  main()   Output:  Before: 1 [0, 1, 2, 3] In f(): 2 [0, 1, 2, 3, 4] After:  1 [0, 1, 2, 3, 4] Elementary properties of closure Hi everyone I'd like to know if the following is correct. I really appreciate any suggestion. (Honestly the only one that matters me is the second property the others are easy, I think) Thanks.  Definition: Let $X\subset \mathbb{R}$ and let $x'\in \mathbb{R}$, we say that $x'$ is an adherent point of $X$ iff $\,\forall\varepsilon&gt;0\,\exists x\in X \text{ s.t.} \; d(x',x)\le \varepsilon$. We say that $\overline{X}$ is the closure of $X$ iff contain all the adherent points of $X$.  Lemma (Elementary properties of Closures): Let $X$ and $Y$ be arbitrary subset of $\mathbb{R}$. Then $X\subset \overline{X}, \; \overline{X\cup Y}=\overline{X}\cup\overline{Y},\; \overline{X\cap Y}\subset\overline{X}\cap\overline{Y}$. If $X\subset Y$ then $\overline{X}\subset \overline{Y}$.  Proof:  Clearly if $x\in X$,  $x$ is $\varepsilon$-adherent to $x$ (indeed is $0$-close to itself) for every $\varepsilon$, hence $x\in \overline X$.   Let $z'$ be an adherent point of $X\cup Y$, then there exists some $z\in X\cup Y$ such that $d(z',z)$ for a given $\varepsilon&gt;0$. Suppose  $z\in X$, hence $z'$ is an adherent point of $X$, i.e., $z'\in \overline{X}\subset \overline{X}\cup \overline{Y}$. Similarly when  $z\in Y$, $z'\in \overline{Y}\subset \overline{X}\cup \overline{Y}$. Thus $ \overline{X\cup Y}\subset\overline{X}\cup \overline{Y}$. We  know that $X\cup Y \subset\overline{X\cup Y}\subset\overline{X}\cup \overline{Y}$. Then $\overline{\overline{X\cup Y}}=\overline{X}\cup \overline{Y}$. Since $\overline{\overline{X\cup Y}} = \overline{X\cup Y}$ we're done.  Let $z'\in \overline{X\cap Y}$, i.e., $z'$ is an adherent point of $X\cap Y$. So, there is some $z\in X\cap Y \subset X$ such that $d(z',z)\le \varepsilon$ for any given $\varepsilon&gt;0$. It follows that $z'$ is an adherent point of $X$, i.e., $z'\in \overline{X}$. Similarly since $z\in X\cap Y \subset Y$ we have $z'\in \overline{Y}$. Hence $z'\in \overline{X}\cap\overline{Y}$.   Let $x'\in \overline{X}$ we'd like to show that $x'\in \overline{Y}$. Let $\varepsilon&gt;0$ be arbitrary then $\exists x\in X\subset Y$ such that $d(x',x)\le \varepsilon$, so $x'$ is an adherent point of $Y$, i.e., $x'\in \overline{Y}$ as desired. $\Box$     Claim 1: If $X\subset Y \subset \overline{X}$. Then $\overline{Y}=\overline{X}$.    Proof claim: Let $y'$ be an adherent point of $Y$, i.e., $y'\in \overline{Y}$ and let $\varepsilon&gt;0$ be given. Thus, $\exists y \in Y\subset \overline{X}$ s.t. $d(y,y')\le \varepsilon$. So, $y$ is an adherent point of $X$. Thus $\exists x\in X$ s.t. $d(x,y)\le \varepsilon$ and therefore  we must have $d(x,y')\le 2\varepsilon$ which shows that $y'$ is an adherent point of $X$, i.e., $\overline{Y}\subset \overline{X}$. Conversely if $x'$ is an adherent point of $X$. Then $\exists x\in X\subset Y$ s.t. $d(x,x')\le \varepsilon$, and so is adherent to $Y$. $\Box$     Claim 2: $\overline{\overline{X}} = \overline{X}$    Proof: It will suffice to show that if $x''\in \overline{\overline{X}}$, then $x''$ is an adherent point of $X$. Let $\varepsilon &gt;0$ be arbitrary and let $x''$ be an adherent point of $\overline{X}$. Then there is some $x' \in \overline{X}$ such that $d(x'',x')\le \varepsilon$. Since $x' \in \overline{X}$ by definition is an adherent point of $X$, so there is some $x\in X$ for which $d(x,x')\le \varepsilon$. Thus $d(x,x'')\le d(x,x')+d(x',x'')\le 2\varepsilon$ it follows that $x''$ is an adherent point of $X$ as desired. $\Box$ Flag apron shows superfluous closing anchor tag I awoke this morning to a spam flag on a site I moderate, and noticed this when I went to read the post.    Notice answered&lt;/a&gt;, which I imagine isn't meant to have the &lt;/a&gt; there.  I also confirmed that this displays asked&lt;/a&gt; for other types of flags (namely, closed without comment).   I looked around for a duplicate of this but can't seem to find it, so maybe it's a recent change? I'm looking for a way to insert continuous colour legend for a raster layer.  I managed to create discrete symbols with values along the bar, but I would like to have a continuous bar with some values on the bar, similar to the old plugin "One-Band Raster", that it doesn't work with the QGIS version higher than 1.7  I tried to set in properties->style, where I selected the  Render Type "Singleband pseudocolor" and see the "generate new color ramp" box.  I tried both "Continuous" or "Equal interval". However, in the Composer the legend always is displayed with discrete symbols.  Is there any way that can make QGIS Composer to display a nice continuous band for a raster layer? How can I make hibernate my desktop in UEFI with Secure boot on.  I used command "sudo pm-hibernate" to make hibernate my desktop but my machine did nothing (my desktop is booted up on UEFI with Secure boot on). But, if I boot my desktop on UEFI with Secure boot off, Hibernation is normal.  Sorry for my English, thanks, C# compiler error: "cannot have instance field initializers in structs" Why does the series $\sum_{n=1}^\infty\frac1n$ not converge? Mouse gestures for Unity desktop This started to happen immediately after I had rebooted the first time after doing a system upgrade. It first starts with a dialogue that says "System program problem detected". Then when I try to hit 'report problem' not much happens. I am led through a dialogue that always ends up the problem cannot be solved.  I am aware this is not a lot of information, however I'm not sure which information I need to publish and how should I obtain it to debug this problem.  Here's a screenshot! Maximum temperature laptop internals can handle? Cooling options? So I have a Compaq CQ50-215CA and my sister was using her laptop on her bed and I was just using it quickly to check my email. The surface of the laptop was really hot and I opened Speccy to see that the CPU and GPU temperature was 96 C. A couple months ago, my laptop was on the desk without a cooling pad and I was playing a game. Using GPU-Z back then, I had a reading of 104 C. Even with a cooling pad on a desk, running several processes/games keeps me around 88-90 C.  Is this normal for a laptop and what's the highest temperature the internals can handle? Besides a cooling pad, are there any other more effective ways of cooling a laptop? How do I show that the equation $x^3+10x^2-100x+1729$$=0$ has at least one complex root $a$ such that $|a|$$&gt;$$12$.