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7

Because $\max$ only exists if the set contains its $\sup$. The set $\{0.9, 0.99, 0.999, 0.9999, 0.99999,...\}$ has sup 1 but no max.


4

No, you can't. For example, consider the situation $$ F(s) = \delta(s) + \delta(s-1) + \delta(s+1)\\ G(s) = e^{-s^2}\cos\left(\frac{\pi}{2} s\right) $$ Counterexample that I'm not sure of: $$ F(s) = \delta(s) + \delta(s-1) + \delta(s+1)\\ G(s) = \delta(s) + \delta(s-2) + \delta(s+2) $$


3

You have on differentiating both sides, $2f(x)f'(x)=f(x)$ for all $x$. Choose some $x$ for which $f(x)\neq0$ then $2f'(x)=1\implies f(x)=\dfrac{x}{2}+K$ where K is another constant. Note that $f(0)=K$ in our solution. In your given equation, setting $x=0$, we have $f(0)^2+C=0\implies K^2+C=0\implies K=\sqrt{-C}$ Note that $C\leq0$ so that it makes sense. ...


3

hint:$I = 4\displaystyle \int_{0}^1 x^2\sqrt{1-x^2}dx$. Let $x = \sin \theta, 0 \leq \theta \leq \dfrac{\pi}{2}$. Can you continue ? the same trick used for evaluating $J$.


3

$$ f(x) \geq \sqrt{\int_{0}^{x}f(t)\,dt} \geq \sqrt{\int_{0}^{x}t\cdot \inf_{t\in(0,1)}\frac{f(t)}{t}\,dt}=\frac{x}{\sqrt{2}}\cdot\sqrt{\inf_{t\in(0,1)}\frac{f(t)}{t}}$$ leads to: $$ \frac{f(x)}{x}\geq \sqrt{\frac{1}{2}\inf_{t\in(0,1)}\frac{f(t)}{t}}$$ from which: $$ \inf_{t\in(0,1)}\frac{f(t)}{t}\geq\frac{1}{2} $$ follows. As you noticed, equality is ...


2

Another easy test you can use: the ratio or d'Alembert test. $$\frac{a_{n+1}}{a_n}=\frac{(n+1)e^{-(n+1)a}}{ne^{-an}}=\frac{n+1}n\cdot\frac1{e^a}\xrightarrow[n\to\infty]{}\frac1{e^a}<1$$


2

I think that the best way is to use the root test:http://en.wikipedia.org/wiki/Convergence_tests. Indeed you have $(ne^{-an})^{\frac{1}{n}}\rightarrow e^{-a}<1$.


2

$$f(x):=\frac{x^6-x^2\sin\frac1{x^2}}{x^4}=x^2-\frac1{x^2}\sin\frac1{x^2}$$ Now, for example, choose the sequences $$x_n=\frac1{\sqrt{2n\pi}}\;,\;\;y_n=\frac{\sqrt2}{\sqrt{(4n+1)\pi}}$$ Observe that both sequences tend to zero when $\;n\to \infty\;$ , so if the function's limit existed it'd be the same on both sequences, yet: ...


2

For the first question: Correct -- a one-dimensional space cannot have any intrinsic curvature; the Riemann tensor always vanishes. For the second: In 3-dimensional hyperbolic space, the Gaussian curvature of a geodesic plane will be $-1$, because it it itself a 2-dimensional hyperbolic space. This is most easily seen by considering the hyperplane that is ...


2

Well there isn't always a maximum. Consider $S=\{r\in\Bbb Q:r<2\}$. This is a bounded set with no maximum element. But we can say that $2$ is the smallest number which is an upper bound of $S$.


2

We do (sort of) need the definition of derivative to deal with differentiability at at $x=0$. So we want to show that $$\lim_{h\to 0} \frac{f(h)-f(0)}{h}$$ exists. It is clear that the limit above exists and is equal to $0$, if $h$ approaches $0$ from the right So it remains to show that $$\frac{\sqrt{4-h^2}-2}{h}\to 0$$ as $h$ approaches $0$ from the ...


2

You get $U(|f|,P) -L(|f|,P) \le U(f,P) - L(f,P)$, and since $f$ is integrable, you can choose a partition so the right hand side is as small as you want. (Also, we have $0 \le U(|f|,P) -L(|f|,P)$.) Addendum: You have $\sup_{x\in I}|f(x)|-\inf_{x\in I}|f(x)|\le \sup_{x\in I}f(x)-\inf_{x\in I}f(x)$ for all intervals in a partition $P$. Hence $\sup_{x\in ...


1

The fact that $Q$ is a Borel set is irrelevant here. If $f : \mathbb{R}^d \to \mathbb{R}$ is differentiable, then $\operatorname{supp}(f') \subseteq \operatorname{supp}(f)$. To see this, suppose $x \not\in \operatorname{supp}(f)$, then $f(x) = 0$. Furthermore, as $\operatorname{supp}(f)$ is closed, there is an open neighbourhood $U$ of $x$ such that $f|_U ...


1

This is an application of Fatou's Lemma. Let $E \subset X$ be $\mu$-measurable. By using Fatou's Lemma twice we get $$\begin{align*} \int_E f \, \mathrm d\mu &= \int_X f \, \mathrm d\mu - \int_{E^{\mathrm c}} f \, \mathrm d\mu \\ &\geq \lim_{n \to \infty} \int_X f_n \, \mathrm d\mu - \liminf_{n \to \infty} \int_{E^{\mathrm c}} f_n \, \mathrm d\mu ...


1

My concern is that I said $a<b$. Am I allowed to do that? Yes, because $a,b \in (1/2,1)$ are arbitrary and $$|\sin^{-1}(a)-\sin^{-1}(b)| = |\sin^{-1}(b)-\sin^{-1}(a)|, \quad \text{and} \quad |a-b| = |b-a|.$$ And more importantly I let $c∈(−1,1)$ and not $(a,b)$. Is that wrong? Yes, you must have $c \in (a,b)$. However, this does not changes the ...


1

Basic is that $\nu\left(\bigcup_{n=1}^{\infty}A_{n}\right)=\sum_{n=1}^{\infty}vA_{n}$ if the $A_{n}$ are measurable and disjoint. On this it can be proved that $A_{n}\uparrow A$ implies $\nu A_{n}\uparrow vA$. This by setting $B_{1}:=A_{1}$ and $B_{n}:=A_{n}-A_{n-1}$ if $n>1$. Then: ...


1

We are of course assuming that $S$ and hence $T$ is nonempty. Since $\sup T\ge x$ for all $x\in T$, we must have $\sup T\ge x$ for all $x\in S$. So, $\sup T$ an upper bound of $S$, hence $\sup S\le\sup T$, since $\sup S$ is the smallest upper bound of $S$.


1

(a) In $\mathbb R$, let $\mu$ be Lebesgue measure restricted to $[0,1]$. Then the density $$ h(x) = \begin{cases} 1,\qquad 0 \le x \le 1\\ 0,\qquad \text{otherwise} \end{cases} $$ is not continuous. (b) In $\mathbb R^2$, let $\mu$ be "arc length" measure on a circle. It is nonatomic. But it has no density (continuous or not) with respect to Lebesgue ...


1

You want to apply the sleaky AM-GM inequality...: $\sqrt[n^2]{n} = \sqrt[n^2]{1\cdot 1\cdot 1\cdots 1\cdot \sqrt{n}\cdot \sqrt{n}} \leq \dfrac{(n^2-2)\cdot 1+ 2\sqrt{n}}{n^2} = 1+\dfrac{2\sqrt{n}-2}{n^2} < 1+\dfrac{1}{n}$. This sounds better than yours....


1

If $\phi$ is constant on a subset of positive measure, then a function whose support is only there will also be an eigenvector. For instance if the space is $[0,2]$ with the Lebesgue measure, $\phi=\chi_{[0,1]}$, and $f(x) = \max \{ 0,1-x \}$, then $f$ is an eigenvector of $M_\phi$.


1

Suppose $y(0) = y_0; \tag{1}$ then the unique solution to the equation $y' = a(x)y + b(x) \tag{2}$ is $y(x) = e^{\int_0^x a(s)ds}(y_0 + \int_0^x e^{-\int_0^s a(u) du} b(s) ds); \tag{3}$ formula (3) is very well known; a derivation may be found here; for $x = T$ we thus have $y(T) = e^{\int_0^T a(s) ds}(y_0 + \int_0^T e^{-\int_0^s a(u) du} b(s) ds). ...


1

http://mathworld.wolfram.com/IntegratingFactor.html I am suppressing $(x)$ in $a(x)$ etc. $$y=e^{\int a\, dx}\big(\int b\,e^{-\int a\, dx}dx +C \big)$$ Let $\int_0^T a\, dx = M$ and $\int_0^T b\, dx = N$. Then we can write $\int a\, dx=A(x)+\frac MTx$ and $\int b\, dx=B(x)+\frac NTx$ where $A$ and $B$ are zero mean AND periodic. FACT: Integral of a zero ...


1

you are defining a linear transformation h on C[0,1] defined by h(f)=f(g) and you are trying to find eigen vectors and values. (a) which you are considering is an eigen value and f is an eigen function. Why not consider analytic functions and then we see that g(x)=cx+dx^2+... while f(x)=A+Bx+Cx^2+.... Then we see that A=0. It seems the problem is solvable ...


1

The issue is that if you take finite number of points $\{x_1,\dots,x_k\dots,x_K\}$ and write the metric space as union of these points. Each singleton is not nowhere dense anymore because each singleton itself is a sphere so it is dense in itself.


1

Yes. There could be for some manifold. For example if you have a Killing vector fields $X$ on $M$, then the diffeomorphism group generated by $X$ are isometries. Thus it sends geodesics to geodesics. It is known that such a vector fields are Jacobi fields (along any geodesics). For example, any constant vector fields on $\mathbb R^n$ and flat torus are ...


1

What if you define $x=e^{-a}$ ? This would make $$S(a)=\sum_{n=1}^\infty n e^{-n a}=\sum_{n=1}^\infty n x^n=x \sum_{n=1}^\infty n x^{n-1}=x \frac{d}{dx}\Big( \sum_{n=1}^\infty x^n\Big)$$ You have then a classical summation; differentiate it to get $$S(a)=\frac{x}{(1-x)^2}=\frac{e^a}{\left(1-e^a\right)^2}$$ which undefined only for $a=0$; you could also note ...


1

we know that sigma n x^n converges to x/(1-x)^2 for abs(x)<1. This is easy to show by ratio or root test. our series is the same as sigma n(1/e^a)^n,0<1/e^a<1, since a>0. In fact we can fond the sum.


1

You can also use a comparison test by noticing that $ne^{-na}=O\left(\frac{1}{n^2}\right)$ for $a>0$.


1

In order to compute both integrals you may exploit: $$ \forall a,b>0,\quad \int_{0}^{1} x^{a-1}(1-x)^{b-1}\,dx = B(a,b) = \frac{\Gamma(a)\,\Gamma(b)}{\Gamma(a+b)}, $$ where: $$ \Gamma\left(\frac{1}{2}\right)=\sqrt{\pi},\qquad \Gamma(z+1)=z\,\Gamma(z).$$


1

If $\lim_{n \to \infty} f(n) = 0$, then your function is $O(1)$, i.e. bounded above by a constant function. So there's no need to use positive powers on $n$, you can come up with better big-O bounds. However if you really want to use (positive) powers on $n$ for your big-O bounds, then what you say is certainly true, for any positive power, or, for that ...



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