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19

Consider $g(x) = f(x)e^{-x}$. Since $g(0), g(1) = 0$, the derivative $g'(x) = \left(f'(x) - f(x)\right)e^{-x}$ must vanish somewhere on $(0, 1)$. Thus there exists some $x_0\in (0, 1)$ with $f(x_0) = f'(x_0)$.


12

Define $g(x) := \int_a^x |f(t)| dt$ for $x \in [a, b]$. Then $g$ is differentiable, non-negative, $g(a) = 0$ and $$ g'(x) = |f(x)| \leq M \int_a^x |f(t)| dt = M g(x) \, . $$ Now let $h(x) := g(x)e^{-Mx}$. Then $h$ is non-negative, $h(a) = 0$ and $$ h'(x) = g'(x) e^{-Mx} - Mg(x) e^{-Mx} \le 0 \, . $$ So $h$ is decreasing on $[a, b]$ and therefore identical ...


5

I don't think this is possible. To see this, let \begin{align*} f(x)\equiv&\,\frac{1}{x},\\ g(x)\equiv&\,\frac{1}{x^2},\\ \end{align*} for each $x\in[1,\infty)$. Clearly, $f>g>0$. I claim that there exists no simple measurable function $\varphi:[1,\infty)\to\mathbb R$ such that $f\geq\varphi\geq g$ pointwise (let alone a sequence of such ...


5

If you mean a function that is nowhere continuous on $\mathbb{R}$, you may consider the indicator function of $\mathbb{Q}$.


5

Hint: Consider the functions $f_n = g\cdot \chi_{\{g\le n\}}.$


5

Hint: Is the complementary set open in the extended line?


4

Consider bounded cutoffs of $g$, i.e. $g_n(x)=\begin{cases} g(x) & \text{if } g(x) \in [0,n] \\ 0 & \text{otherwise} \end{cases}$. These are all in $K$, and converge pointwise to $g$. By compactness there is an $L^1$-convergent subsequence whose limit is in $K$. Can you use this to argue that in fact $g \in K$, hence $g \in L^1$?


4

By setting $\sigma=\frac{-1+\sqrt{5}}{2},\overline{\sigma}=\frac{-1-\sqrt{5}}{2}$ we have: $$ \sum_{k\geq 1}F_k\, x^k = \frac{x}{1-x-x^2} = \frac{-\sigma}{x-\sigma}+\frac{\overline{\sigma}}{x-\overline{\sigma}}\tag{1} $$ hence: $$ \sum_{k\geq 1}\frac{F_k}{k}\,x^k = \frac{1}{\sqrt{5}}\,\log\left(\frac{1+\sigma x}{1+\overline{\sigma}x}\right)\tag{2}$$ and: $$ ...


3

Consider $\mathbb R^2$ with standard metric and let $E=\{\,(x,y)\mid xy=1\,\}$, $F=\{\,(x,y)\mid xy=0\,\}$. Your argument about $e,f$ being accumulation points is false. Instead of $$ \exists e\in E\exists f\in F\forall r>0\colon d(e,f)<r$$ we only have $$\forall r>0\exists e\in E\exists f\in F\colon d(e,f)<r $$


2

This is a special case of Grönwall's inequality, an essential tool for any analyst. See the "Integral form for continuous functions" section at the above Wikipedia link, and apply it with $\alpha(t) = 0$ and $\beta(t) = M$. You conclude $f(x) \le 0$ everywhere. Then apply it again to $-f$. The Wikipedia page also gives the proof, which is pretty much the ...


2

Fix $v \in \mathbb{R}$. Since $f(x) \to -\infty$ as $x \to -\infty$, there exists $a \in \mathbb{R}$ such that $f(a) < v$. Since $f(x) \to \infty$ as $x \to \infty$, there exists $b \in \mathbb{R}$ such that $f(b) > v$. Now apply the intermediate value theorem.


2

In a sense, a postulate means here is where I want to start. If you start at a more elementary level, then your postulates may become theorems. The field postulates for real numbers are a good example of that. In (usually) Foundation of Mathematics, the basic properties of the integers, rationals, and reals are all derived from basic set theory.


2

Suppose that $\int_0^1 g(t)\,{\rm d}t = \infty$. Now decompose $[0,1]$ into sets where $g\leqslant n$ What can you say about the the sequence of restrictions of $g$ to these sets?


2

The p-adic numbers are an example of this phenomenon and there are a lot of similar examples, since any discrete valuation ring or its quotient field have the property that open disks are closed and vice versa. This includes finite extensions of the p-adics as well as rings of the form $K((X))$ (Laurent series with finite principal part over a field). Note ...


2

The complement of $(-\infty,\infty)$ is $\{-\infty,\infty\}$. A finite set is closed (but not open) in the standard topology on $\mathbb{R}$ (and the extended real line inherits topological properties from $\mathbb{R}$). Thefore $(-\infty,\infty)$ is not closed, but it is open. EDIT: More directly, a set in $\mathbb{R} \cup \{ \pm \infty\}$ is closed if and ...


2

Let $f(x)=\sqrt{1+4x}$. Choose any $x \in (0,\infty)$, then by IVT: $$\frac{f(x)-f(0)}{x}=f'(\xi)$$ where $\xi \in (0,x)$.But: $$f'(\xi)=\frac{2}{\sqrt{4\xi+1}}$$ So (because $\sqrt{4\xi+1}>1$): $$f'(\xi) < 2 $$ Therefore: $$\frac{f(x)-f(0)}{x}=\frac{\sqrt{4x+1}-1}{x}<2$$ Finally (because $x>0$): $$\sqrt{4x+1}<2x+1$$


1

The answer is, of course, yes. Lets consider the case with "classical" induction. The argument basically goes like this: Let $P(n)$ be statements indexed by $n\in\mathbb{N}$. Suppose $P(1)$ is true, and $P(i)\Longrightarrow P(i+1)$, then $P(n)$ is true for all $n\in\mathbb{N}$. Fundamentally, this is just saying that if you have a situation where The ...


1

The extended line is first-countable, so a set is closed if and only if it contains all the limit of convergent sequences in it. What about $x_n=n$? It is a sequence of elements from $(-\infty,\infty)$, and it is convergent in the extended line. Where does the limit lie?


1

Consider: $\displaystyle f_n(x) = \begin{cases} -\frac{1}{n} &\mbox{if } x < \frac{1}{2}-\frac{1}{n} \\ x -\frac{1}{2}& \mbox{if } \frac{1}{2}-\frac{1}{n} \leq x \leq \frac{1}{2}+\frac{1}{n}\\ \frac{1}{n} & \mbox{if }x > \frac{1}{2}+\frac{1}{n}\end{cases} $ $f_n \rightarrow 0$, but $\phi(f_n) \rightarrow 1$


1

First we prove that for $z > x$, $$\lim_{t \to \infty} \int_z^\infty \frac{1}{\sqrt {4\pi t}}e^{-\frac{|x-y|^2}{4t}}dy = \frac{1}{2}$$ To see this, let $u = y - x$, then by change of variables $$\int_x^\infty \frac{1}{\sqrt {4\pi t}}e^{-\frac{|x-y|^2}{4t}}dy = \int_{z - x}^\infty \frac{1}{\sqrt {4\pi t}}e^{-\frac{u^2}{4t}}du$$ Taking limits: $$\lim_{t \to ...


1

No, it is not. We may just consider a positive function $f(t)=\frac{g(t)}{t}$ that belongs to $L^1\setminus L^{\infty}(\mathbb{R}^+)$ and is unbounded on every interval $(r,+\infty)$, like: $$ f(t) = \sum_{n\geq 0}\frac{\mathbb{1}_{(n,n+1)}(t)}{2^n\sqrt{t-n}}.$$ We have: $$ \| f \|_1 = \sum_{n\geq 0}\frac{2}{2^n} = 4 $$ but for every $r\in\mathbb{R}^+$ we ...


1

I'm going to assume that you know basic things about compactness. And give you two steps to begin with. If $A$ is a closed and bounded set, then it is compact. Prove, even in general, that a compact metric space is always the closure of a countable set (or in other words, a compact metric space is always countable). You can do that by finding a particular ...


1

There does not have to be a fixed point $x$ such that $f(x) = x$. Here's a counterexample: the constant function $f(x) = 2$. Obviously, there is no $x \in [0, 1]$ for which $x = 2$. Given a function $f(x)$ as described in the problem, consider the function $g(x) = f(x)-2x$. We have $g(0) \in [0, 2]$, and $g(1) \in [-2, 0]$. Since $g(x)$, like $f(x)$, is ...


1

Why is there not a fixed point $f(x)=x$? Nobody said there wasn't. To show there does exist an $x$ with $f(x)=2x$, let $g(x)=f(x)-2x$. Then $g(0)=f(0)\ge 0$, while $g(1)=f(1)-2\le0$. So the intermediate value theorem shows there exists $x$ with $g(x)=0$. (Or, if you have a theorem saying any map from $[0,1]$ to itself has a fixed point, consider $f/2$.)


1

Note that \begin{align*} \underbrace{\color{red}{|a_1-c_1|}+\color{red}{|c_1-b_1|}}_{\clubsuit}\leq&\,\underbrace{\max\{\color{red}{|a_1-c_1|},|a_2-c_2|\}+\max\{\color{red}{|c_1-b_1|},|c_2-b_2|\}}_{\star},\\ ...


1

No, it does not follow. Let $b=2$, $r=1.1$, and $m=1$. Let's assume that $f(n) = 1+\lfloor \log_2(n) \rfloor$. Then $n$ must satisfy $$ 1 \leq n < r+1 = 2.1.$$ So let $n=2$. Then $f(2) = 1 + \log_2(2) = 2$, but the proposed upper bound is $\lceil m \cdot \log_b r \rceil = \lceil \log_2(1.1) \rceil = 1$.


1

Hint. Slice the image of $f(x)=\frac{2}{1+x^2}$, i.e. $(0,+\infty$ in equal slices of thickness $\frac{1}{n}$. Compute the reverse image of each slice which is for each a union of a pair of disjoint intervals. Take the measure of each reverse slices and make the Lebesgue sum. Then prove that this is the $\sup$ of all $\{\psi \le f, \psi\text{ simple ...


1

To speak of density of a subspace $Y$ in a space $X$, $X$ and $Y$ have to be topological spaces. So, the answer to your question is: it depends on the topology of $X$. In your example, you may equip $X$ either with the topology of uniform convergence or the topology of pointwise convergence.


1

No. In fact, the conclusion $\exists~e \in E,f \in F$ such that $~\forall ~r>0,~d(e,f)<r$ implies that $d(e, f) = 0$, and since $d$ is a metric, this is forces $e = f$ (and so that $X$ is a singleton). The definition of distance between sets $E, F$ in a metric space $(X, d)$ is $$d(E, F) := \inf \{d(e, f) : e \in F, f \in F \},$$ so $d(E, F) = ...


1

No, thay cannot be naturally generalised to exponents $p>2$. A random variable $X$ is $p$-stable if whenever we have a sequence $(X_n)_{n=1}^\infty$ of independent copies of $X$ then for any finite sequence of scalars $(\alpha_n)_{n=1}^N$ the random variables $$\sum_{n=1}^N \alpha_n X_n\quad\text{ and }\quad\Big(\sum_{n=1}^N |\alpha_n|^p\Big)^{1/p}X$$ ...



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