whether dirac delta function a well behaved function? Can u please explain the properties of a   well behaved function..?
19203
Is Dirac delta a function? What is its contribution to analysis?   What I know about it: It is infinite at 0 and 0 everywhere else. Its integration is 1 and I know how does it come.
43637
Say I have a function $f(x)$ on some interval $[a,b]$.  Say it is integrable such that $\displaystyle\int f(x)~dx $ is defined.  Is $\displaystyle\int f(x)~dx $ necessarily continuous?  If I were to know that the integral is integrable itself such that $\displaystyle\int \int f(x)~dx $ is defined. Would that change anything?   If so why?  Thank you
67955
Defining the Characteristic Function $ \quad  \phi(t) := \mathbb{E} \left[ e^{itx} \right] $ for a random variable with distribution function $F(x)$ in order to show it is  uniformly continuous I say  $$ |\phi(t+u) - \phi(t)| = \left |\int e^{itx}(e^{iux} - 1) dF(x) \right| \le \\ \int 1 \cdot|e^{iux} -1|dF(x) \to 0 \quad as  \quad u\to0 $$  Now my question is, does the convergence I state in the last line follow directly, or do I need to be a little carful before I conclude it is true ? (i.e. can I directly use that $|e^{itu} -1| \to 0 ? )$
27861
Hm, this just occurred to me while answering another question:   If I write the Hamiltonian for a harmonic oscillator as $$H = \frac{p^2}{2m} + \frac{1}{2} m \omega^2 x^2$$ then wouldn't one set of possible basis states be the set of $\delta$-functions  $\psi_x = \delta(x)$, and that indicates that the size of my Hilbert space is that of $\mathbb{R}$.  On the other hand, we all know that we can diagonalize $H$ by going to the occupation number states, so the Hilbert space would be $|n\rangle, n \in \mathbb{N}_0$, so now the size of my Hilbert space is that of $\mathbb{N}$ instead.  Clearly they can't both be right, so where is the flaw in my logic?
115954
Are there any general conditions on a function so that its Laplace transform is continuous? e.g. what conditions on $f:\mathbb {R} \rightarrow \mathbb {R}$ guarantee that its Laplace transform  $$F\left(s\right)=\int_{0}^{\infty}f\left(t\right)e^{-st}dt$$ is continuous? It seems to me that if $f$ has compact support and is bounded, then it ought to be. The idea I have is that we get  $$\left|F\left(s'\right)-F\left(s\right)\right|\leq\int_{0}^{y}\left|f\left(t\right)\right|\left|e^{-st}\right|dt$$ by making $s'$ and $s$ close enough, it seems we can make the RHS arbitrarily small. I haven't been able to complete a proof. Additionally, can we change the assumptions? Are there more natural assumptions to get continuity?
96450
Just wanted to confirm that this is a correct solution:    Proof: Suppose $f(x_0) &gt; 0$ for some $x_0 \in [a,b]$. Then, by continuity of $f$, for $\epsilon &lt; f(x_0)$, there exists $\delta &gt; 0$ so that, if $|x-x_0| &lt; \delta$, we have $|f(x) - f(x_0)| &lt; \epsilon$. That is to say, $f(x)$ is non-zero in a neighborhood $B_\delta (x_0)$. Now,   \begin{eqnarray*} U(f,P) &amp;=&amp; \sum_{P([a,b])} M_i \Delta_i \\ &amp;=&amp; \sum_{P(B_\delta(x_0))} M_i \Delta_i \\ &amp;=&amp; \sup_{x \in [x_0-\delta, c_1]} |f(x)| (c_1-(x_0 - \delta)) + \sup_{x \in [c_1,c_2]} |f(x)| (c_2-c_1) + \cdots + \sup_{x \in [c_n, x_0 + \delta]} |f(x)| ((x_0+\delta)- c_n) \\ &amp;&gt;&amp; 0 \end{eqnarray*}  In the same manner,   \begin{eqnarray*} L(f,P) &amp;=&amp; \sum_{P([a,b])} m_i \Delta_i \\ &amp;=&amp; \sum_{P(B_\delta(x_0))} m_i \Delta_i \\ &amp;=&amp; \inf_{x \in [x_0-\delta, c_1]} |f(x)| (c_1-(x_0 - \delta)) + \inf_{x \in [c_1,c_2]} |f(x)| (c_2-c_1) + \cdots + \inf_{x \in [c_n, x_0 + \delta]} |f(x)| ((x_0+\delta)- c_n) \\ &amp;\geq&amp; 0 \end{eqnarray*}    We can conclude from here that $\int_{a}^{b} f(x) dx &gt; 0$.  Is this a valid proof?
35500
I was wondering on the following and I probably know the answer already: NO. Is there another number with similar properties as $e$? So that the derivative of $\exp(x)$ is the same as the function itself. I can guess that it's probably not, because otherwise $e$ wouldn't be that special, but is there any proof of it?
41182
How to prove   that      $$\mathcal{F}\left\{\int_{-\infty}^{t}f(\tau)\, d\tau\right\}=\frac{1}{i\omega}F(\omega)+\pi F(0)\delta(\omega),$$   where $F(\omega)$ is Fourier transform of $f(t)$.  Could anyone explain to me how to prove this?  thanks.
120696
This is somewhat linked to a  which was looking to see if a proof of mine regarding the Dirichlet function was correct (it wasn't). I now have an answer to the question which can be answered without using directly using the Baire category theorem or the likes; as it is sufficiently different to my original approach, I feel that a new question would be the best way of going about this.  The question is to     Construct a function $f:[0,1] \rightarrow \mathbb{R}$ which is not the pointwise limit of any sequence $(f_n)$ of continuous functions   to which I will have an answer below (while I didn't come up with much of this myself, I feel it's an interesting result to discuss).  Finally, as a warning to those who are seeing this as a result of searching for answers to their example sheet/homework questions, please have a think about the question before reading the answer below.
122942
Let $\Delta=\{z\in \mathbb C: |z| &lt; 1\}$. Let $f: \Delta \rightarrow \mathbb C$ be a one-to-one analytic function fixing the origin. Prove that there is a one-to-one analytic function $g: \Delta \rightarrow \mathbb C$ such that $[g(z)]^2=f(z^2)$. Further, show that such a function is odd.  The only idea which comes to my mind is the following (not sure whether it is the right direction though). Let $h: \Delta \rightarrow \Delta \rightarrow \mathbb C,\ z\mapsto z^2\mapsto f(z^2)$. Note that $h'(0)=0,\ h''(z)=2f'(z^2)+4z^3f''(z^2)$; since $f$ is one-to-one and analytic, its derivative never vanishes, so $h''(0)\ne 0$. So $h$ is an analytic function of $\Delta$ with a zero of order two at the origin. Then there is an analytic function $g$ defined in an open disc about the origin with a simple zero at the origin such that $[g(z)]^2=f(z^2)$.   This is the only thing I came up with, but it seems to be a little bit irrelevant since I don't see any way of extending this $g$ to the whole $\Delta$, and also the injectivity and oddness of such a $g$ are unclear.
6376
Is it true that a positive harmonic function on $\mathbb{R}^n$ must be a constant? How might we show this? The mean value property seems not to be the way...for that we would need boundedness.
103505
I was wondering if there are functions for which $$f'(x) &gt; f(x)$$ for all $x$. Only examples I could think of were $e^x - c$ and simply $- c$ in which $c &gt; 0$. Also, is there any significance in a function that is always less than its derivative?    Edit: Thank you very much for all the replies. It seems almost all functions that apply are exponential by nature...  Are there more examples like - 1/x?  Again are there any applications/physical manifestations of these functions? [for example an object with a velocity that is always greater than its position/acceleration  is always greater than its velocity]
44288
I know some of it, but I would like to learn other ones. I think this question is useful, as it will be helpful for other beginners. Such as: uvw, AM GM, Cauchy–Schwarz inequality, Jensen inequality. Could you please add something? Maybe books containing it?
62009
I know that for every function $f\in L^1(R)$ its Fourier transform has the following properties: it is continuous, it's bounded, and we have the limit $$ \lim_{\omega\to\pm\infty} \hat f (\omega) = 0 $$ which is the Riemann-Lebesgue lemma. Now suppose the function $f$ also has these properties (and it is in $L_1$). Will its Fourier transform be also in L1? That is, if $$ f\in L^1(R),  \mbox{ f is continuous and  } \lim_{x\to\pm\infty}  f (x) = 0$$ does it mean that $$ \hat f(\omega)\in L^1(R)$$ or not? My first intuition is no, but I don't have a counter example.  Any ideas?
80751
Physicist here. I am puzzled by a question: looking at a continuous function $g :\mathbb{R} \rightarrow \mathbb{R}$ that goes to zero at infinity, I am interested in the behavior of its derivative $\beta = g'$. Precisely, does it go to zero too? By writing on paper it looks like: $$ \beta(+ \infty)=\text{lim}_{x\rightarrow \infty}\text{lim}_{h\rightarrow 0} \frac{g(x+h)-g(x)}{h}$$ And if I can invert the two limits, I get what I expect: $\beta(+\infty)=0$, so I was curious about the hypothesis behind this permutation. Is the requirement of continuity enough? Thanks.
32010
This should be a very easy question but it might just be that I'm confusing myself. So we have the definition of a function $f$ on $S$ being continuous at $x_0$:     For any $\epsilon$>0, there exists $\delta&gt;0$ such that: whenever $|x-x_0|&lt;\delta$, we have $|f(x)-f(x_0)|&lt;\epsilon$   And I assume the negation is     There exists $\epsilon$>0 such that for all $\delta&gt;0$, $|x-x_0|&lt;\delta$ yet $|f(x)-f(x_0)|\ge \epsilon$.   Now I want to show that the function $f(x)=\sin(\frac{1}{x})$ together with $f(0)=0$ cannot be made into a continuous function at $x=0$. So I need to show that there exists $\epsilon&gt;0$ such that for all $\delta&gt;0$, $|x|&lt;\delta$ yet $|f(x)|\ge\epsilon$.  Let $\epsilon = \frac{1}{2}$. Then no matter what $\delta$ we choose, let $|x|&lt;\frac{1}{2}$. It is certainly possible that $|f(x)|\ge \frac{1}{2}$, because, well, $\frac{1}{x}$ can really take on arbitrarily large value as $x$ is small.  Now, what confuses me is that, as $x$ gets small, $f(x)$ can certainly be greater than $\frac{1}{2}$ for infinitely many times, but it will be less than that infinitely many times, too. But I suppose it doesn't really matter. So I think there's something wrong with my negation but I couldn't figure out where.  Update: The correct version can be found . Watch for Lemma 4.6
93945
I'm currently stuck on a problem where I have to integrate on a particular set defined through a dirac delta function. If I understood correctly it all boils down to using the curved analogous of  $$  \int_{\mathbb{R}^n} f(\vec{x}) \delta(\phi(\vec{x}))|\nabla\phi|d^nx \tag{1} $$  my guess is that in GR  this expression becomes either  $$ \int_\Omega f(x) \delta(\phi(x))\sqrt{g^{\mu\nu}\frac{\partial \phi(x)}{\partial x^\mu}\frac{\partial \phi(x)}{\partial x^\nu}} \sqrt{ -g}\,d^4 x \tag{2}$$  or  $$ \int_\Omega f(x) \delta(\phi(x))\sqrt{g^{\mu\nu}\frac{\partial \phi(x)}{\partial x^\mu}\frac{\partial \phi(x)}{\partial x^\nu}}\,d^4 x \tag{3}$$  given that in GR $$\delta^4 (\xi) =\frac{\delta^4(x)}{\sqrt{-g}}\tag{4}$$ where the $\xi$ are flat coordinates, but here I have a $\delta^{1}$, not a $  \delta^{4}! $ So I'm thinking the first formula is the correct one. Am I right?  Which one is correct?  PS. It would be awesome if someone had any sort of useful reference on the topic!