# Tag Info

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 $$\sum_{n=1}^N \alpha_n X_n\quad\text{ and }\quad\Big(\sum_{n=1}^N |\alpha_n|^p\Big)^{1/p}X$$ ...

Only top voted, non community-wiki answers of a minimum length are eligible