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7

From the definition, you get $$\frac{x}{1+f(x)}=f(x)$$ Thus $$x=(1+f(x))f(x)$$ Now differentiate $$1=f'(x)+2f(x)f'(x)=f'(x)(1+2f(x))$$ Or $$f'(x)=\frac{1}{1+2f(x)}$$ Now, $f(0)=0$, from definition of $f$. One remark though: I did not prove the continued fraction is differentiable or even convergent. I just assume it's true, then you can compute ...

4

One method is to write $$xy - x_0y_0 = (x-x_0)y_0 + x_0(y-y_0) + (x-x_0)(y-y_0).$$ Then you can set $$\delta = \frac{\min \{ 1,\, \varepsilon\}}{3(\lvert x_0\rvert + \lvert y_0\rvert + 1)}$$ to have $\lVert (x,y) - (x_0,y_0)\rVert_\infty < \delta \Rightarrow \lvert xy-x_0y_0\rvert < \varepsilon$.

4

In its current state, the function is not defined at zero because $0/0$ is not defined. However, if we extend the definition via $$f(x) = \left\{\begin{array}{lr} \frac{x^2}{x} & x \ne 0 \\ 0 & x = 0\end{array}\right.$$ then it's fairly easy to see that this is continuous and differentiable on the entirety of $\mathbb{R}$. Hence the discontinuity ...

3

That's not actually an upper bound, we can come as close to $\frac12$ as we want, and that's not as hard a bound to come up with ;) Choosing $m = n$, the fraction simplifies to $$\frac{2^{2n}}{2^{2n+1}+2^n} = \frac12 - \frac{2^{n-1}}{2^{2n+1}+ 2^n} > \frac12 - \frac{1}{2^{n+2}}.$$ Seeing that choosing $n\neq m$ doesn't give a value greater than ...

3

You need Hausdorff to ensure the existence of a subalgebra which separates points. Suppose $X$ is not Hausdorff. Let $x$ and $y$ be two points which cannot be separated by open sets. Let $A$ be a subalgebra of $C(X,\mathbb{R})$ and $p\in A$ be an arbitrary element. Suppose $p(x)\neq p(y)$. Then since $\mathbb{R}$ is Hausdorff we can separate $p(x)$ and ...

3

Wikipedia's statement of the theorem is Suppose $X$ is a compact Hausdorff space and $A$ is a subalgebra of $C(X,\mathbb R)$ which contains a non-zero constant function. Then $A$ is dense in $C(X,\mathbb R)$ if and only if it separates points. (Emphasis added.) What you've quoted from Rudin asserts only the "if" direction (separating points implies ...

2

In general they can't. The following would be a simple example except that we can take the open sets to be $\mathbb{R}^2$, however it illustrates the issue: $F(x,y) = (x_1-y_1,2x_2-y_2)$. To fix this, take $F(x,y) = (x_1-y_1, x_2-y_1^2)$, and use the $\|\cdot\|_2$ norm for $x$ and $y$. It is clear that $\frac{\partial f(x,y)}{\partial x} = I$ for all ...

2

When $f:\>{\mathbb R}\to{\mathbb R}$ is $C^2$ with $f(0)=f'(0)=0$ then we can write $$f(x)=f(0)+\int_0^x 1\>f'(t)\ dt=(t-x)f'(t)\biggr|_{t=0}^{t=x} -\int_0^x(t-x)f''(t)\ dt=\int_0^x (x-t)f''(t)\ dt\ ,$$ and the substitution $t:=\tau\,x$ $\>(0\leq\tau\leq1)$ gives $$f(x)=x^2\> g(x)\ ,\quad g(x):=\int_0^1(1-\tau)f''(\tau\, x)\ d\tau\ .\tag{1}$$ The ...

2

It is perfectly valid to arrive at equation $(4)$ the way you have. Doing so, however, presents difficulties in applying the Fundamental Lemma of Calculus of Variations. The Fundamental Lemma says that for some function $f(x)$ which is $k$ times differentiable on some interval $[a, b]$, if we have $$\int_{a}^{b} f(x)h(x)\,dx = 0$$ ...

2

I suppose if you can see that $\liminf a_n = -1$ and $\limsup a_n = +1$, then trivially $\liminf a_n \leq \limsup a_n$. Indeed for "any" sequence that is the case, the definition forces that, so it is among properties of limsup and liminf. From Wikipedia http://en.wikipedia.org/wiki/Limit_superior_and_limit_inferior Whenever $\liminf x_n$ and $\limsup x_n ... 2 I'll do part (a). Set$a_n = \frac{x^n}{2^n \sqrt{n}}$. Then$\frac{a_{n+1}}{a_n} = \frac{\sqrt{n}}{\sqrt{n+1}} \frac{x}{2} \to \frac{x}{2}$as$n \to \infty$. Thus, the series converges absolutely for$|x| < 2$. We see that the series diverges at$x = 2$by the$p$-series test. For$x = -2$, the series converges (not absolutely) by the alternating ... 2 It's not valid in its current state: All you can conclude is that$M$is an upper bound for$S$. What happens, say, if we choose$S = [0, 1]$and a sequence $$K_n = 2 + \frac 1 n$$ If our initial choice of$s$is$0$, then our intervals are $$I_n = \left[0, 2 + \frac 1 n\right] \implies \bigcap_n I_n = [0, 2]$$ So$M = 2$. The gap starts in the line ... 2 You need to make the gap between the upper bounds and$S$shrink to$0$. HINT: Given$s_n\in S$and an upper bound$K_n$of$S$, let$x_n=\frac12(s_n+K_n)$. If$x_n$is an upper bound for$S$, let$s_{n+1}=s_n$and$K_{n+1}=x_n$. If not, choose$s_{n+1}\in[x_n,K_n]$, and let$K_{n+1}=K_n$. Now consider the intervals$I_n=[s_n,K_n]$. 2 Hint: Use Fubini-Tonelli Theorem instead of Fubini's Theorem, which only requires that your integrant is non-negative, and has the same conclusion as ordinary Fubini's Theorem. The statement of Fubini-Tonelli Theorem is: Let$f:\mathbb{R}^d\to [0,\infty]$be measurable, and$x\mapsto \int f_x\,d\lambda_2$and$y\mapsto int f^y\,d\lambda_1$are Borel ... 2 It is a removable singularity. Your$f(x)$is not defined at$x=0$, but$\lim_{x \to 0}f(x)=0$It is reasonable to define a new function$g(x)=\begin {cases} f(x) & x \neq 0\\0&x=0 \end {cases}$and note that$g(x)=x$2 Already answered by Jonathan Y. in a comment, but here is another approach, based on scaling. For every$\lambda>1$the dilated cube$\lambda \cdot Q$(same center, sidelength multiplied by$\lambda$) has volume$|\lambda\cdot Q|=\lambda^n |Q|$. Since$\overline{Q}\subset \lambda\cdot Q$, we have $$|\overline{Q}|\le \lambda^n |Q|$$ Since$\lambda>1$... 1 If$X$is an uncountable null set then it cannot have a minimal cover. A fat Cantor set cannot have a minimal cover either since every non-trivial closed cube will leak out of it. If$X \subseteq \mathbb{R}^d$has finite outer measure and$\{Q^{\star}_n : n \geq 1\}$is a minimal cover of$X$, then$X$is a full outer measure subset of$Y$where$Y = ...

1

I am giving a complete solution, but please for you to learn try to prove that $\sup (A +B) \leq \sup A + \sup B$ Solution Notice $A + B = \{ a + b : a \in A \; \; and \; \; b \in B \}$. By definition, then we have $a + b \leq \sup (A + B)$. But this implies $$a \leq \sup(A+B) - b \implies \sup(A+B) - b \; \text{is upper bound for set A}$$ $$... 1 If x\in A+B then there are elements a\in A and b \in B with x=a+b. By definition a\leq\sup A and b\leq\sup B leading to x=a+b\leq\sup A+\sup B. This is true for every x\in A+B and consequently \sup\left(A+B\right)\leq\sup A+\sup B Slightly different: Assume that \sup\left(A+B\right)>\sup A+\sup B. Then a\in B and b\in B exist ... 1 You want to show that lim_{p \to \infty} \|f\|_p \leq \|f\|_\infty and \|f\|_\infty \leq lim_{p\to\infty}\|f\|_p. There are two relevent estimates. The first is \|f\|_p \leq \|f\|_\infty \mu(X)^{1/p}. For the second, let \epsilon > 0, and pick E \subset X with \mu(E) > 0 such that |f(x)| > \|f\|_\infty - \epsilon for all x \in E. ... 1 Let C be the usual Cantor set, and \nu_C its probability measure. Clearly \nu_C(A)=0, where A is the subset of [0,1], consisted of the numbers in which the digit 1 appears in their ternary expansion, while the Lebesgue measure of A is 1. The support of \nu_C is the set of the numbers in which the digit 1 does not appears in their ternary ... 1 Consider the probability measure \mu_p on the Borel sigma-algebra on [0,1], which is the image of the product of the Bernoulli measures p\delta_1+(1-p)\delta_0 on \{0,1\} by the function (x_n)_{n\geqslant1}\mapsto\sum\limits_{n=1}^\infty x_n/2^n. Then for every p\ne\frac12, the measure \mu_p is purely singular, \mu_p has no atom, and every ... 1 0 < x < u for all x \in S, which implies 0 < x^2 < u^2 for all x \in S, so u^2 is an upper bound on T. Now we must prove it is the least upper bound. Suppose that it is not the least upper bound, so for some number \alpha < u^2 we have 0 < x^2 < \alpha for all x \in S. Then, we have that 0 < x < \sqrt{\alpha} ... 1 The standard proof of (a) goes by noticing that since f is monotone, given any partition P, if [x_{i-1},x_i] is one of the intervals in the partition, the largest value of f on the interval is one of f(x_{i-1}) and f(x_i), and the least value is the other one: If f is increasing, the largest value is f(x_i). If f is decreasing, it is the ... 1 Here is how to prove \liminf \{ x_n \} \leq \limsup \{ x_n \} for a general real sequence \{ x_n \}. Set L = \liminf \{ x_n \} and S = \limsup \{ x_n \}, and suppose for the sake of contradiction that L > S. Say L = S + h for some h > 0. Then there are infinitely many n for which x_n \in (L - h, L + h). Therefore, there are ... 1 Using the definition of the derivative,$$E'(x) = \lim_{h\to 0}\frac{E(x+h)-E(x)}{h} = \ldots =\left(\lim_{h\to 0}\frac{E(h)-1}{h} \right)E(x).  By the functional equation, $E(0)=E(0)^2$, so $E(0)=0$ or $1$. If $E(0)=0$, the functional equation gives $E(x)=0$ for all $x$, and you wrote in a comment that you didn't want the zero solution. So $E(0)=1$. ...

1

Following up on Dan's suggestion we have (d/dx)E(x+y) = E'(x)E(y). Setting x = 0 we get E'(y) = E'(0)E(y). The unique solution of this differential equation is E(y) = $Ce^{ay}$ where a = E'(0). But if E(0) = 1, C = 1. The case of E $\equiv$ 1 comes when a = 0; E $\equiv$ 0 is from C = 0.

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