$f$ is continuous function on $[a, b]$. If there exists $K&gt;0$ s.t. $|f(x)|\leq K\int_a^x|f(t)|dt$ for all $x$, then $f(x)=0$. I am trying to use Mean Value Theorem for integrals, but it is hard to prove it. Can you give me some hints?
I have been stuck for several days on this old Analysis problem (I am doing some study on my own).  I have tried several things (which I'll indicate below), but I cannot seem to figure it out.  Here is how the problem is presented: Problem: &quot;Let $f$ be a continuous real-valued function on $[a,b]$.  Suppose there exists a constant $M \geq 0 $ such that $$|f(x)| \leq M \int_a^x |f(t)| dt$$ for all $x \in [a,b]$.  Show that $f(x)=0$ for all $x \in [a,b]$.&quot; My Thoughts: I have tried using the mean value theorem iteratively, but that seems to always lead me down a dead end road.  I deduced that $f(a)=0$.  If only $f$ were assumed to be differentiable, then maybe I could play with trying to get the derivative to be $0$, but unfortunately it's only continuous.  My other thought was to (somehow) use the condition to show that $\int_a^b |f(x)| dx = 0$.  I also played around a bit with contradiction, but to no avail.  Even if one of these hair-brained thoughts is correct, I am not really sure what to do next. If you have any ideas, suggestions, or solutions, I would really appreciate it if you are willing to share them.  Thank you for your time.
There is a length limit to Windows file extensions? Diamond is not shown for mod name under wiki posts If a diamond mod is an author of a wiki post then the diamond is not displayed besides the mod name. E.g. :        Even when the name refers to the user profile page (in case of more then one author) like in :        Is it intentional or a bug? $f(z)=\frac{z}{1+|z|} $ Is 1-1? What are the dangers of accepting gifts from unknown accounts on Steam? A user named 'steamfairy1' gifted me a game on Steam. I don't know this user, their profile shows no games, no friends and zero hours played. The profile of 'steamfairy1' was created the day I received the gift.  It very much looks like a scam, but I can't figure out how the scam is supposed to work. What's the worst thing that could happen if I accept the gift? Any idea what the trick here is? It could be just a random act of kindness and I'm just paranoid…  I contacted the Steam support, but they were unhelpful as usual: responding in the wrong language and with a canned one-liner that didn't make a whole lot of sense or had much to do with my question.  Keywords (not really tags, but I want this to be searchable): steamfairy, steam fairy Grant group traversal rights on a folder Batch/Multi rename only files with a specific number range I have a situation where I have the following:  file01.ext . . . file99.ext file100.ext . . .   And I want to (using rename ) insert the '0' for the files 01-99 (so that they become 000-099) but not modify files larger than 100. Which expression should I use and what is a good source on learning regular expressions such as these?   On Windows I would have used Total Commander's multi rename function which has a very simple renaming syntax. Are there some good alternatives for that tool (multi-rename) for Ubuntu? Does a word rhyme with itself? Given $\int_0^{\infty}e^{-x^2}dx = \frac{\sqrt{\pi}}{2}$, evaluate $\int_0^{\infty}e^{-a^2x^2-\frac{b^2}{x^2}}dx $ Given $$\int_0^{\infty}e^{-x^2}dx = \frac{\sqrt{\pi}}{2}$$ evaluate:   $$\int_0^{\infty}e^{-a^2x^2-\frac{b^2}{x^2}}dx. $$   I can find that  $$\left(ax+\frac{b}{x}\right)^2 = a^2x^2+2ab+\frac{b^2}{x^2}$$  therefore: $$\int_0^{\infty}e^{-a^2x^2-\frac{b^2}{x^2}}dx = e^{2ab}\int_0^{\infty}e^{-\left(ax+\frac{b}{x}\right)^2}dx$$  but I can't find any clue then. Hibernate @Enumerated mapping How to generate world files for raster X? Upgrade to version newer than repositories I have  about how to upgrade ghostscript to a version that is available on 15.04 but not on 14.04LTS (which I have). On this site there are numerous almost identical question regarding different packages, a few examples include:         So I thought it would be very useful in general to know: is there  a general, automated way to upgrade a package, including all its dependencies, to a version that is newer than Ubuntu's repos? Is this more easily done if the newer version exists in more recent Ubuntu releases? Which Jedi have served in the Rebellion? Besides Luke Skywalker, Kanan Jarrus, Ezra Bridger and Ahsoka Tano, which other Jedi, regardless of whether they were survivors of Order 66 or became one at a later point, served within the Rebellion's ranks during the Galactic Civil War?  Looking for both canon and Legends versions.  Edit: ok, just to clarify what would qualify as a Jedi for the answer. Loosely speaking, anyone who follow the ideals of the Jedi philosophy - a practitioner in the ways of the Force, a protector of the galaxy, and an opposer to the dark side - would count.   Only trained Force users, not aligned with the dark side, would count. They need to be able to consciously control their use of the Force to aid the Rebellion. This excludes Leia Members of the Jedi Order of the late Republic would obviously count. However, they need to have survived Order 66 (duh), did not renounce the Jedi ways nor succumbed to the dark side, and become a member of the Rebellion. This excludes the Grand Inquisitor nor Darth Vader, but includes Kanan Jarrus and, in Legends, Rahm Kota. Those who are killed in action after joining the Rebellion also count. However, you don't need to be an official member of the order to be counted. You could be opposed to what was wrong with the late order, but as long as you follow the ideals defined above, and identify with them, you are a Jedi at heart by the count of this question. This counts Ahsoka Tano, Luke Skywalker and Ezra Bridger among its number. They aren't in the old Jedi Order's roster, but they believe in what the Jedi are meant to be, and follow its ways in that regard. Galen Marek in Legends, when he genuinely turned to the ways of the light, even though he was still a user of dark side powers, would be one. It's the thought that counts.   So loosely speaking, that means members of the Rebellion who are trained Force sensitives and genuine good guys. How does a wizard or bard learn new spells? I am traveling from the US to India via London Heathrow on November 6. I will be at LHR on 7th at 6:20 am and departing at 9:25 pm. I'm an Indian national and have a valid US visa.  Do I need a transit visa to stay airside?  I have been scared that British Airways have stopped a few people from boarding flights (no details just rumors).  Do I need a transit visa to stay landside? I want to do some sightseeing. What are the chances of getting past officials using transit without visa concession?  Real stories will help a lot to calm the butterflies in my stomach. Einstein-Hilbert action as an effective field theory I have read the following statement:      "The Einstein-Hilbert action can be seen as an effective field theory: setting $g_{\mu\nu}= \eta_{\mu\nu}+h_{\mu\nu}$, one gets a free Lagrangian for the field $h_{\mu\nu}$ plus interactions."   Now, how is it possible to define all the interaction terms?  EDIT: I ask apologize if my statement was appeared unclear: I have created a  which is more precise; please don't continue to consider seriously this one. Test for convergence  $$\sum_{n=1}^{\infty} \frac{1}{n^n}$$  I'm at a loss on what to do, is this a geometric series $\frac{1^n}{n^n}$? Is it possible to call the base method from a prototype method in JavaScript if it's been overridden?  MyClass = function(name){     this.name = name;     this.do = function() {         //do somthing      } };  MyClass.prototype.do = function() {       if (this.name === 'something') {         //do something new     } else {         //CALL BASE METHOD     } }; How many 5-cards poker hands are there containing at least 3 of the 4 suits?  I don't how know how to properly handle this question.  There 13 ranks of each card, and 4 suits  My Attempt:  Begin with the first 3: C(13,3) X 4 ways to choose suits  Then pick last 3 cards of anything:  C(49,2)  My solution: 1,345,344  Thanks, Given   $$ f_{XY}(x,y) = \frac{1}{2\pi \sqrt{1-\rho^2}} \exp \left( -\frac{x^2 +y^2 - 2\rho xy}{2(1-\rho^2)} \right) $$  $Y = Z\sqrt{1-\rho^2} + \rho X$  And   $$ f_{XZ}(x,z) = \frac{1}{2\pi } \exp \left( -\frac{x^2 +z^2}{2} \right) $$  Show that $P(X&gt;0,Y&gt;0)= \frac{1}{4}+\frac{1}{2\pi}(\arcsin \rho) $  I'm supposed to use the fact that X and Z are independent standard normal random variables, but I don't quite understand how. Any help would be greatly appreciated. Is this group solvable? Let $G=H\times K$ (a direct product of $H , K$) where $H$ is an abelian 2-group and $K$ is a non-abelian simple group. Is $G$ solvable? why?  These three answers are true, and I thought so. But in a book I found a remark that made me confused: T.M. Gagen, Topics in Finite Groups, London Math. Soc. Lecture Note Ser., vol. 16, Cambridge Univ. Press, Cambridge, 1976, Remark, Theorem A on p. $40 $ and Definition $11.3$ on p. $39$. Please see it and say me your ideas.