Linear Operator $A$ and $B$ bounded if $(Ax,y) = (x,By)$? Suppose $A,B$ are linear operators on Hilbert Space $\mathcal{H}$, satisfy $$ (Ax,y) = (x,By) $$ for $\forall x,y \in \mathcal{H}$. Show that $A,B$ are bounded.    What I can get so far : $A,B$ are almost 'adjoint', except that we don't know yet if they are bounded.  Thanks for helping.
If a linear operator has an adjoint operator, it is bounded This is a question I'm struggling with for a while:  Let $H$ be a Hilber space. Let  $T,S: H\rightarrow H$ be linear operators (not neccessarily bounded) such that for every $x,y\in H$: $\langle Tx,y\rangle=\langle x,Sy \rangle$.  Prove $T,S$ are bounded.    What I did so far:  First attempt:  $||Tx||^2=\langle Tx,Tx\rangle=\langle x,STx\rangle\le||x||*||STx||\le||x||*||S||*||Tx||$  I'm not entirly sure I'm allowed to do the last inequallity, since $||S||$ might be $\infty$.  It follows $||Tx||\le ||x||*||S||$  and analogly  $||Sx||\le ||x||*||T||$  From the first I'd get $||T||\le ||S||$ and from the second $||S||\le||T||$. Therefore $||S||=||T||$. I don't know where to go from here.  Second Attempt: assume  $T$ isn't bounded, therefore so is $S$, therefore there are series $(x_n),(y_n)\subset H$ such that $||x_n||=||y_n||=1$ and $\lim ||Tx_n||=\infty,\lim ||Sy_n||=\infty$. I'll keep it going soon, but it didn't go so well..    I'd love some guide. Not neccessarily the whole solution, but something that would tell me what i'm missing.  Thanks :)
Prove that, if A and B is similar, then $A^T$ is similar to $B^T$.  Attempt: I try to use the defn of similarity, and then the rule to transpose in different order, but get nowhere. This should be simple...  (Note to self: page 92.) As expected, Nvidia drivers have reduced my customizations. I can't login to Unity 3D session. I can't find any content to help me to uninstall the driver and I don't know what to do as I have never uninstalled them before. Uninstalling/reinstalling QGIS plugin where uninstall is disabled? Plugin Raster->TerrainAnalysis->slope which had initially worked, started showing what another questioner referred to as a "crazy histogram":   Before finding his question and the answer, I endeavored to uninstall and reinstall the plugin.  I found/still find (QGIS 2.0.1, now upgraded to QGIS 2.4, on Ubuntu 14.04), the plugin "uninstall" button grayed out/disabled.  It's working, but I am worried about having a plugin I installed that does not offer an option to be uninstalled.  Failing anything else I would be tempted to use the path information to rename the .so library and see if it disappeared. Is it possible to apply modifiers to multiple objects at once? You can copy modifiers from one object to other objects with CtrlL> Make links > Modifiers, or the Copy attributes addon.  However, is it possible to apply modifiers to many objects at once? A Vista virtual machine I use only has a 10 GB virtual hard disk -- and I'm worried about it running out of space over time. I used  to check for the culprits..  And the primary culprit is that /winsxs or &quot;Windows Side-By-Side&quot; folder.  What is the Windows Side By Side folder? Well, that's .  All of the components in the operating system are found in the WinSxS folder – in fact we call this location the component store.  Each component has a unique name that includes the version, language, and processor architecture that it was built for.  The WinSxS folder is the only location that the component is found on the system, all other instances of the files that you see on the system are “projected” by hard linking from the component store.  Let me repeat that last point – there is only one instance (or full data copy) of each version of each file in the OS, and that instance is located in the WinSxS folder.   So looked at from that perspective, the WinSxS folder is really the entirety of the whole OS, referred to as a &quot;flat&quot; in down-level operating systems.  This also accounts for why you will no longer be prompted for media when running operations such as System File Checker (SFC), or when installing additional features and roles. That explains why the folder starts off big, but not why it gets larger over time – the answer to that question is servicing.   In previous versions of Windows the atomic unit of servicing was the file, in Windows Vista it’s the component.  When we update a particular binary we release a new version of the whole component, and that new version is stored alongside the original one in the component store.  The higher version of the component is projected onto the system, but the older version in the store isn’t touched.  The reason for that is the third part of why the component store gets so large.  Lots of other people :  A coworker recently received his every two year work laptop refresh and was in the middle of loading his new laptop with Vista. For some reason, his hard drive C: was partitioned to only 25GB and the rest of the space reserved for a much larger drive D:. Anyway, after loading many Windows Updates, he was left with hardly any drive space left on C:, and for some reason, the WinSxS directory had ballooned to over 8GB. While I knew what the WinSxS (Windows Side-by-Side) directory was for, I wasn't quite sure why that directory would explode in size so fast with normal operating system updates.  I don't really care about having all these duplicate files for safety, but I do care about the VM exploding due to an out of disk space error. So.. can /winsxs be safely pared down in any way? ..  Administrators should not, for any reason, take it upon themselves to clean out the /winsxs directory - doing so may prevent Windows Update and MSI from functioning properly afterwards.  Preventing accidental deletion from the directory is accomplished by putting a strong security descriptor on the directory that inherits to its children.  If this /winsxs folder gets any larger, I'm going to have to re-image this virtual machine.. which I really don't want to do! Are \( and \) preferable to dollar signs for math mode? Is there a way to merge to two bars in GNOME 3? Best case scenario would be to have one single bottom or top bar that has all the symbols on the left side and the rest like the time indicator, connections, on/off button, etc. on the right side.  Is that possible in any way? Database general error: 2006 MySQL server has gone away This is problem 6.2 from the 3rd edition of Principles of Mathematical Analysis.      Problem 6.2: Suppose $f\geq 0$, f is continuous on $[a, b]$, and $\int_a^b f(x) \, dx = 0$. Prove that $f(x)=0$ for all $x \in [a, b]$.   I'm looking for a critique of my proof. It's a pretty easy problem, but I am always wary of making too bold of assumptions, especially on these low level/fundamental proofs. I'll be using Rudin's notation and refer to theorems from the text (If I should include the text of each theorem, feel free to leave a comment... I'm lazy but could probably use the TeX practice :p)  Proof: Assume, for contradiction, that $f&gt;0$. Then, for any partition $P$ we have the Lower Riemann Sum: $L(P, f)=\sum_{i=1}^n m_i \, \Delta x_i$. At least one $\Delta x_i$ must be positive, since $a &lt; b$, and each $m_i$ must be positive since we have $f&gt;0$ by assumption, so certainly $\sup f &gt; 0$. That means $L(P, f)&gt;0$. Thus, we have: $$0 = 0(b-a) &lt; L(P,f)\leq \sup L(P, f) = \inf U(P,f) =L$$ where the last string of equalities holds because our function is continuous on a compact interval, so is integrable by theorems 6.8 and 6.6. So our integral has value $L&gt;0$. This is in contradiction to our given assumption that $\int_a^b f(x) \, dx = 0$, so we must have that $f=0$ on $[a, b]$. $\Box$  So, I am wondering if my proof is correct (and is presented well). Also if someone could enlighten me as to what Rudin means when he say "Compare this with exercise 1," I'd be appreciative. Is it a hint or is there something else he expects you to notice? there are a lot of things I could compare :)...  Exercise 6.1: Suppose $\alpha$ increases on $[a, b]$, $a \leq x_0 \leq b$, $\alpha$ is continuous at $x_0$, $f(x_0)=1$, and $f(x)=0$ if $x\neq x_0$. Prove that $f$ is Riemann-Stieltjes Integrable and that $\int f \, d\alpha = 0$ I'm trying to use grep to show only lines containing either of the two words, if only one of them appears in the line, but not if they are in the same line.  So far I've tried  grep pattern1 | grep pattern2 | ... but didn't get the result I expected. How to move Applications from phone to SD Card? If a thin metal plate is placed between two charges $+q$ and $+q$, will this cause a change in the electrostatic force acting on one charge due to another? What is the concept behind this? What will happen if the metal plate is thick? What is the reasoning behind the design decision to allow users to create a display name which matches another existing users exactly?    This is opposed to the common practice of enforcing unique user names, and I'd really like to learn something. Prove that $f$ has a fixed point . Show that $2^n > n^2$ whenever n>4. Let $S(n)$ be the proposition "$2^n &gt; n^2$"  Basis step: $S(5)$ is true, since $2^5 = 32$, $ 32 &gt; 25 = 5^ 2$  Inductive step:  Induction hypothesis:   Assume $S(K)$ is true, $2^K &gt; K^2$.  Then  $2^{K+1} = 2.2^K &gt; K^2 + K^2$   I don't know how to proceed after this. How to complete this proof? Is there a "best" way to automatically restart SQL Server on a regular basis? I read that I could create a batch file to net start, net stop the agent and service in a batch file and schedule that, but I was curious if there was a better way of doing this.  Please nevermind how this is indicative of a larger problem of operation, I've been arguing this for a month and we've ended up here.  Thanks :). Why didn't Thanos come to Earth to retrieve Tesseract by himself? In the Avengers movie, why didn't Thanos personally come to Earth instead of Loki? Or, if Loki's skills were required, why didn't he come with Chitauri? He knew that humans were primitive; there wouldn't be any threat and he could personally secure Tesseract.  AFAIK, the movie didn't display much. Is there any other source answering this? Find the center of the symmetry group $S_n$.   Attempt:  By definition, the center is $Z(S_n) = \{ a \in S_n : ag = ga \forall\  g \in S_n\}$. Then we know the identity $e$ is in $S_n$ since there is always the trivial permutation. Suppose $a$ is in $S_n$, but not equal to identity. Now we can imagine the permutation as bijective function that maps from $\{1,2,\dotsc,n\}$ to $\{1,2,\dotsc,n\}$.  So suppose $p$ is a permutation map. Then $p$ maps from a location $i$ to a location $j$. Take $p(i) = j$ where $i\neq j$. Let $k$ be in $\{1,2,\dotsc,n\}$, where $k$, $i$ and $j$ are all different elements. The cycle $r = (jk)$, then we will see if this commutes.  $rp(i) = rj$    Can someone please help me, I am stuck?  Thank you. How do I audit my reputation? If I am concerned that my reputation score is incorrect, how can I audit it, or get a report of a detailed breakdown of my reputation? Define the sequence $aₙ=20+n²$ and $dₙ=\gcd(aₙ , aₙ₊₁)$ Find with proof the values dₙ takes.