New answers tagged real-analysis
0
For the second one: Assume there is a continues $g(x)$ in $[0,2]$
that satisfy $${x\leq g(x)\leq x^{2}-x+1}$$ note that this
imply $$0.5\leq g(0.5)\leq0.25-0.5+1=0.75$$
Consider $\widehat{g}:[0,2]\to\mathbb{R}$ s.t $\widehat{g}(x)=g(x)$
for all $x\neq0.5$ and such that $$\widehat{g}(0.5)=g(x)+\delta$$
where $\delta\neq0$ is some number such that the above ...
0
If $I$ is an intervall and $f,h\colon I\to\mathbb R$ are continuous functions with $f(x)\le h(x)$ for all $x\in I$. Then
If $f(a)=h(a)$ for some $a\in I$, then any function $g\colon I\to\mathbb R$ with $f(x)\le g(x)\le h(x)$ for all $x\in I$ is continuous at $a$.
If $f(a)\ne h(a)$, there exists $g\colon I\to \mathbb R$ with $f(x)\le g(x)\le h(x)$ for all ...
0
Hint: For the first part, you know that your function can't be "further away" from 1 than either of the bounding functions are. For the second part, you should be able to just draw a counterexample. Start by sketching the two bounding functions.
1
Hint: As @Ted noted you will have $$C: \mathbf{r(t)}=\big(\cos(\theta),~\sin(\theta),~~2-2\cos(\theta)-\sin(\theta)\big),~~~~0\le\theta\le2\pi$$ Now evaluate $$\oint_C F\cdot dr$$
0
This answer assumes $a,b>0$. (If it were otherwise, the presence of the minus sign in the $-2bx^2$ term seems unnecessary). If this is not a valid assumption, my result at least treats one of the four possibilities; note that if $a>0,b<0$ the equation has no positive roots at all, so really only two cases remain.
Anyway let ...
1
As $F,G$ are given to be differentiable so they are continuous. And all continuous functions in a closed inteval is integrable so $F,G$ are integrable. Integrability of $Fg,Gf$ follows from this.
3
Alas, the answer is no.
$$f(x)=\begin{cases}-4& x\in[0,2]\\ -2& x\in [2,4]\\0& x\in[4,6]\end{cases}$$
$$g(x)=\begin{cases}5 & x\in [0,1]\\3& x\in[1,3]\\1& x\in[3,5]\\ 0 & x\in[5,6]\end{cases}$$
$$q(x)=\begin{cases} 1 & x\in [0,1]\\ -1& x\in[1,2]\\ 1 & x\in[2,3]\\ -1 & x\in[3,4]\\ 1& x\in[4,5]\\ 0 & ...
0
If $x \in N_0$, then $T(x) \in N$ (as "$N_0$ is mapped into $N$") so $T(x)$ is in $S$ (as we assume that $N \subset S$). So by definition, $T(x) \in S_0$, as $S_0$ is defined the be the inverse image of $S$ ($S_0 = T^{-1}[S]$), so exactly the set of all $p$ with $T(p) \in S$, and the $x$ we start with is such a point.
0
If $p>1$, use Hölder's inequality. Then write
$$\int_{\Bbb R}\left|\frac{t}{t^2+y^2}\right|^qdt\leqslant \int_{\{|t|\geqslant R\}}\frac 1{|t|^q}dt+(2R)^{Q+1}y^{-2q}.$$
If $p=1$, use dominated convergence theorem, where $|f(x-t)|\frac{|t|}{t^2+1}$ is a dominating function ($y>1$).
1
No, there are functions such that all your nested partial sums converge to $0$, but that are nonzero :
Suppose you have constructed the first $k(n)$ terms for $S_0$, such that $S_1(k(n)) = S_2(k(n)) = \ldots = S_n(k(n)) = 0$, and let $\varepsilon > 0$.
You want to pick the next $k(n+1)-k(n)$ terms so that the next partial sums stay bounded by ...
0
Hmm, it might be that I did not understand the question correctly, but I find the problem interesting so I'll give it a try and please correct me if I'm wrong.
Here is, what I understand from the definitions of the $S_m$ with an example at $n=4$:
$$ \begin{eqnarray}
S^1(4) &=& S^0(1)+ S^0(2)+ S^0(3)+ S^0(4) \\
&=& 1 S^0(1)+ 1 ...
0
Inverse images preserve set inclusion. More explicitly:
$x\in N_0 \Rightarrow Tx \in N \Rightarrow Tx \in S \Rightarrow x \in S_0.$
1
Use limit comparison test with harmonic series. We have
$$\dfrac{1/n}{1/n^{1+{1/\log(1+n)}}} = n^{1/\log(1+n)} \to e$$
1
$$\int \frac{1}{x^{1+1/\ln(x)}}dx=\int\frac{dx}{x}(1/x)^{(1/\ln x)}$$
Set $u=\ln x$, then $x=e^u$ and $1/x=e^{-u}$ so the integral becomes $$\int du (e^{-u})^{1/u}=\int e^{-1}du$$
which is divergent. Replacing $1/\ln x$ with $(1/\ln x)^s$ will be divergent (resp. convergent) if $s\ge 1$ (resp. $s<1$).
1
From the mean value theorem, $\forall x,y\in\mathbb R$ we have that there exists $\xi$ between $x$ and $y$ such that
$$|f(x)-f(y)|=|f'(\xi)|\cdot |x-y|\leq a|x-y|\Rightarrow$$
$$|f(x)-f(y)|\leq a|x-y|,\forall x,y\in\mathbb R$$
Now take $$|s_{n+1}-s_n|=|f(s_n)-f(s_{n-1})|\leq a|s_n-s_{n-1}|=$$
$$a|f(s_{n-1})-f(s_{n-2})|\leq a^2|s_{n-1}-s_{n-2}|=\dots$$
...
0
By the mean value theorem, $\forall x, y \in \mathbb{R}$, $|f(x)-f(y)|\leq|x-y||f'(\theta)|\leq a|x-y|$ (where $\theta$ is some value in the open interval bounded by $x$ and $y$.) We call such an $f$ a contraction. The proof of this theorem is identical to the proof of the more general "contraction mapping theorem" in a complete metric space, the statement ...
0
Hint: use the fact that $|f(x)-f(y)|=|f'(c)(x-y)|\le a|x-y|$ to prove that your sequence is Cauchy.
1
The function $f$ is integrable. The step functions are dense in $L^1(\mathbb{R})$ (corollary 18 here). A fortiori in $L^1([0,1])$. So there exists a sequence $(s_n)$ of step functions converging to $f$ in $L^1$. Finally, there exists a subsequence of the latter converging a.e. pointwise to $f$. As it is the case in general for convergent sequences in an ...
0
Let $LI^m$ and $L^m$ be the set of invertible linear transformations of $\mathbb{R}^m$ in itself and the set of linear transformations of $\mathbb{R}^m$ in itself respectively. Define $\operatorname{Inv}: LI^m\to L^m$ by $\operatorname{Inv}(A)$ is the inverse matrix of $A$. First you can note that if $f$ is a function satisyfing your conditions, then ...
2
There is another way which is based on Caparison Test again. In fact, the following limit says that the improper integral is convergent: $$\lim_{x\to 0^+}~~x^{1/2}\times\frac{1}{x^x}=0<\infty$$
3
$\det:M_n(\mathbb R)\to\mathbb R$ is continuous, since it is a polynomial map, if you regard it as a map $\det:\mathbb R^n\to\mathbb R$:
$$\det A=\sum\limits_{i+j=n}(-1)^{i+j}a_{ij}A_{ij}$$
where $A=(a_{ij}),$ and $A_{ij}$ is the determinant of the $(n-1)\times(n-1)$ matrix if we remove row $i$ and column $j$ from $A$.
Now, $\dim GL_n(\mathbb R)=\dim ...
0
Hint: Use the limit comparison test.
3
It suffices to show that
$$\lim_{x \to 0^+} x^{-x} =1,$$
because then we could consider the integral
$$\int_0^1 x^{-x} dx$$
as an integral of a continuous function on a compact set (which has convergent value).
Taking logs, this is the equivalent to
$$\lim_{x \to 0^+} -x \log x =0.$$
This is a standard result, which can be shown using l'Hopital's rule.
...
1
Well first you have to decide what exactly the "topology" on matrices is. Suppose we considered matrices just as vectors in $\mathbb{R}^{n^2}$, with the usual metric topology. Matrix multiplication by say $A$ take a matrix $B$ to another $\mathbb{R}^{n^2}$ vector where the entries are polynomials in the components of $B$.
The other bit is similar, if you ...
2
$(0,1)$ is topologically complete (also referred to as completely metrizable): it’s homeomorphic to $\Bbb R$ and therefore has a complete metric that generates its topology.
0
I'll go on from where you stopped at $ a^3|e^{-xy_1^2}-e^{-xy_2^2}|$. Let's note that $ 0 \leq e^{-xy_1^2}\leq 1$ because this is a monotone decreasing function of $x$, and therefore, it receives a maximum when $x=0$, such that:
$ a^3|e^{-xy_1^2}-e^{-xy_2^2}|\leq a^3|1-e^{-xy_2^2}|.$
Let's use the fact that $ e^{-xy_2^2}\geq 0$, to conclude that:
$ ...
0
Try to find a Lipschitz constant of $x \mapsto \mathrm{e}^{-x}$, $x \ge 0$. Hint: Fundamental Theorem of Calculus.
Then, you can use this in your Lipschitz estimate.
4
Yes.
It follows easily from induction that $f(nx)=nf(x) \, \forall n \in \mathbb{Z}^+$. Thus $f(\frac{m}{n})=\frac{1}{n}f(m)=\frac{m}{n}f(1)$.
If $f(y)<0$ for some $y \in[0, +\infty)$, then $f(ny)=nf(y)<m$ for sufficiently large $n$. Thus $f(y) \geq 0 \forall y \in [0, +\infty)$. Now if $a \geq b$ then $f(b+(a-b))=f(b)+f(a-b) \geq f(b)$, so $f(x)$ ...
2
No, your argument is not correct. I’ll comment on it in detail.
Consider the interval $E=[0,1]$ which is closed. By definition, every point of $E$ is a limit point.
While it is true that every point of $[0,1]$ is a limit point of $[0,1]$, this is not a matter of definition: there is nothing in the definition of closed set that requires every point of ...
1
The answer is no. Take a subspace of $\mathbb{R}^2$ consisting of a single point $P$. Then $\{P\}$ is closed, since its complement $\mathbb{R}^2\setminus\{P\}$ is open in the usual topology of $\mathbb{R}^2$. Now, $P$ is a point of $\{P\}$ but not a limit point for $\{P\}$ since you can't define in any way $P$ as the limit of a sequence of points of $\{P\}$ ...
0
Ron:
I am not trying to be pretentious, and I realize this is an old post, but if I may offer a little input on the evaluation of your zeta sum. If I can assume you're still interested at this point.
I believe it was Choi who has done research on these series.
Using the Barnes G function, many series involving zeta have been evaluated.
Here is a ...
2
For finite-dimensional spaces, matrix norms are equivalent, so the choice does not matter.
1
First let $\Delta\Omega$ be the boundary of $\Omega$ as $\Omega$ is open and bounded $\Delta\Omega$ is compact and disjoint from $\Omega$ (hence does not intersect $K$).
Now as $K$ and $\Delta\Omega$ are non intersecting compact sets there exists some $\varepsilon>0$ such that no pair of points $x\in K, y\in\Delta\Omega$ have $\|x-y\|<2\varepsilon$ ...
2
For each $p\in K$, let $B(\delta_p,p)\subset\Omega$ where $B(\delta_p,p)$ is an ball of ratio $\delta_p$ and ceter $p$. Because $K$ is compact, you can find a finite number of balls $B(\delta_{p_i},p)$ such that $$K\subset\bigcup_{i=1}^mB(\delta_{p_i},p)\subset\Omega$$
To finish in we have to consider the points where any two balls intersect. Denote by ...
2
I think you can consider the following function instaed: $$\pi:\mathbb R^2\to\mathbb R\\\\ \pi\left((x,y)\right)= x$$
0
If you consider $\def\R{\mathbb{R}}\def\exp{\mathrm{exp}}\exp\colon\R\to\R$, then the example is correct: the map is not closed, because
$$\exp(\R)=\R_{>0}$$
which is not closed in $\R$. On the other hand, the image of an open set in $\R$ is open in $\R_{>0}$, so also open in $\R$. Hence the function is open but not closed.
If you consider ...
4
Your tutor is wrong. Every homeomorphism (such as $\exp:\mathbb R\to\mathbb R_{>0}$) is closed. The set $\mathbb R_{>0}$ is a closed set in the standard topology on $\mathbb R_{>0}$.
1
The term "neighborhood" of a point comes from topology, where it is taken to mean any open set containing that point. This convention carries over into complex analysis.
Often, a circular neighborhood of a point and a more general neighborhood of a point are interchangeable in practice. If you need a circular neighborhood of a point, it suffices to find a ...
1
It probably depends on the textbook you're using. Some authors write "neighbourhood" in the sense "open neighbourhood". Some do not.
Some authors might only consider open discs, but the most common convention is that an (open) neighboorhood of a point is any open set containing the point. In practice, you can almost always work with disc shaped ones though. ...
2
Certainly non-circular neighborhoods are allowed. Using Cauchy's Theorem it is often possible to replace an integral along a curve by an integral along a circle. But there is no need, and little sense, in limiting the discussion only to function defined on circular neighborhoods. Moreover, the concept of analytic continuation is extremely important and one ...
1
Norm $1$ and $3$ are equivalent, so $C[0,1]$ is also complete in the norm $3$.
$||f||_{\infty}^{0,1}=||f||_{\infty}+|f(0)|+|f(1)|\le 3||f||_{\infty}$
4
Besides to @Dominic's neat answer, I am noting a good Theorem below. However it is about $\mathbb R$, you can get the points in another dramatized answer better:
Theorem: Let $E\subset\mathbb R$ including at least two points. Then $E$ is connected if and only if it is an interval.
In fact, if $E$ is not an interval,so $$\exists~p\notin E, a,b\in E~~ ...
1
Polygonal connectedness (which is a stronger condition than topological connectedness) can be defined for any subset of the complex plane. The definition will be, as you'd expect, that a set $A\subseteq \mathbb C$ is polygonally connected if any two points in it can be connected by a polygonal path inside $A$.
The authors simply did not bother with the ...
2
The connectedness at here is path connectedness, which means any two points can be connected by a path traveling across the domain. For open sets it is not difficult to show polygon path connectedness and path connected is the same, because any polgyon path is a path, and any path can be approximated by a polygon path given the domain is open. Using some ...
6
Because there is a difference between path connected and conected. And there is even a difference between polygonal connected and path connected. For example taking the set
\[ \{ z \in \mathbb{C}: |z|=1\}\]
has no polygonal paths connecting two points. But you intuitively says that this set should be connected. Connected in the pure topologic version sound ...
2
Let $U$ be an open annulus in $X$. Then the statement is not true.
2
A usual example is
$$
f(x)=\sum_{k\in A} e^{-\sqrt{k}}\cos kx
$$
where $A=\{2^n:n\in\mathbb{N}\}$.
1
$$f(x)=\begin{cases} e^{-1/x^2} & x>0\\ 0 & x\le 0\end{cases}$$ is the usual example.
2
Let $\{I_{k,j}:j=1,\ldots,2^k\}$ be the set of open intervals we are throwing away on $k$-th step of our construction. Note that $m(I_{k,j})=5^{-k-1}$ for all $k\in\mathbb{N}$ and $j=1,\ldots,2^k$.
Let
$$
C_0:=[0,1]\qquad\qquad
C_{k+1}:=C_k\setminus\bigcup\limits_{j=1}^{2^k}I_{k,j}
$$
Now we proceed to the proof of inequality. Define $f_n=\chi_{C_n}$, ...
0
The idea is to consider sequence of continuous functions that converge to the function of class $C([a,b])\setminus C^1([a,b])$. Then one needs to integrate this functions $n$ times and prove that the result is Cauchy sequence without limits in $C^n([a,b])$. It will not have the limit because $n$-th derivatives will converge to non-$C^1([a,b])$ function.
...
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