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5

The original statement is trivial if $f(a)=a$ or $f(b)=b$. Otherwise, $f(a)>a$ and $f(b)<b$, so $g(a)>0$ and $g(b)<0$. Now you can use the IVT as you suggested (but you want to replace $h$ with $g$ or the other way around). Also, $g$ as you defined maps to $[a-b,b-a]$, not $[a-b,b]$.


4

It is true. If the sequences $\{x_n\}$ and $\{y_n\}$ are Cauchy, then they are both bounded. Thus there is an $M\gt 0$ such that $|x_n|\lt M$ for all $n$, and $|y_n|\lt M$ for all $n$. It follows that $|x_n+y_n|\lt 2M$ for all $n$. Thus the sequence $\{\frac{1}{x_n+y_n}\}$ cannot have limit $0$. Remark: We did not need the condition $x_n+y_n\gt 0$. We ...


4

The limit is in fact $\pi$. Hint: Let $a_n$ and $b_n$ be the half the perimeter of the of the inscribed and circumscribed $n$-gons of the unit circle. Then $$b_{2n}=\frac{2a_nb_n}{a_n+b_n},$$ $$a_{2n}=\sqrt{b_{n+1}a_n}.$$ Further, $b_3=2\sqrt3$. Historical note: Gauss used a similar process to compute (approximate) values of elliptic integrals.


3

The fact that $(C[0,1],\| \cdot \|_{L^\infty})$ is complete follows from the triangle inequality estimate $|f(x)-f(y)| \leq |f(x)-f_n(x)| + |f_n(x)-f_n(y)| + |f_n(y)-f(y)|$. The fact that $(C[0,1],\| \cdot \|_{L^\infty})$ does not have a compatible inner product can be checked by demonstrating that the parallelogram law is not satisfied. One way to do this ...


3

Ms. Chris's sis asked me exactly same question a few days ago in chatroom & I could answer it. Here is my answer. Let $I$ be the integral. Using magic substitution $2t=1+x$ we get \begin{align} I&=\int_{\frac{1}{2}}^1 \frac{\log(2t)\log(2-2t)}{t}dt\\ &=\int_{\frac{1}{2}}^1 \frac{\log t\log(1-t)}{t}dt+\ln2\int_{\frac{1}{2}}^1 ...


2

This stems from the more general fact (see this question) that given $A \subseteq X$ $$\operatorname{Int} (A) = X \setminus \overline{ X \setminus A },$$ or, equivalently, $\operatorname{Int}( X \setminus A ) = X \setminus \overline{A}$. (Note that this is valid in all topological spaces, not just metric spaces.) So $A$ is dense if and only if $X = ...


2

If $G$ is bounded, then for $w\in H^1(\Omega)$, $$ |\langle g(v),w\rangle| \le \int_\Omega |g(v)|\cdot|w| dx \le M \|w\|_{L^1(\Omega)}, $$ and you are done. As you can see, there is a lot of room to relax the assumptions on $g$. In fact, one can allow a certain growth of $g(x)$ for $|x|\to\infty$. Then $g$ is well-defined. Since $g \in W^{2,\infty}$ it is ...


2

It suffices to show that, for every $a,b> 0$, $$ (a+b)^p\le a^p+b^p, $$ or equivalently $$ 1\le \left(\frac{a}{a+b}\right)^p+\left(\frac{b}{a+b}\right)^p. $$ But $$ \left(\frac{a}{a+b}\right)^p\ge \frac{a}{a+b} \quad\text{and}\quad \left(\frac{b}{a+b}\right)^p\ge \frac{b}{a+b}, $$ as $0<p<1$. Thus $$ 1= \frac{a}{a+b}+\frac{b}{a+b} \le ...


2

Topology is weird sometimes. It ends up that the statement you are trying to prove is true in $\Bbb R^n$ (in which we can use the Heine Borel theorem as in this proof), but is not true for bigger metric spaces. Note, however, that as long as bounded and closed sets are compact, this statement holds. Another counterexample: we can consider the metric space ...


1

Let $$ h(x)=\left\{\begin{array}{lll}\mathrm{e}^{-1/x^2} & \text{if} & x>0,\\ 0 & \text{if} & x\le 0.\end{array}\right. $$ Then $h\in C^\infty(\mathbb R)$. Then set $$ j(\boldsymbol{x})=c\,h\big(1-\|\boldsymbol{x}\|^2\big), $$ where $\boldsymbol{x}\in\mathbb R^n$, and $c>0$, so that $\int_{\mathbb ...


1

Rewrite $f_{\epsilon}(x)$ by substituting $y=y'-\frac{1}{\epsilon}x$: $$ \begin{align} f_{\epsilon}(x) & = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}g(\epsilon y') e^{-(y'-\frac{1}{\epsilon}x)^{2}/2}\,dy' \\ & = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}g(\epsilon y')e^{-(\epsilon y'-x)^{2}/2\epsilon^{2}}\,dy' \end{align} ...


1

If by "divergent" you mean precisely "not convergent" (as opposed to "diverges to $+\infty$" or something), then the answer is indeed no. Set $a_1=a_2=a_3=a_4=0$. For each $m\ge2$, define $a_{2^m+1}$ through $a_{2^{m+1}}$ as follows: $$ a_n = \begin{cases} 2^{-m}, &\text{if } 2^m < n \le 3\cdot 2^{m-1} \text{ and $n$ is odd}, \\ 0, &\text{if } n ...


1

I would ordinarily leave a comment in this case, because I don't have any real answers. But it's too long to leave in a comment. For ODEs $-ay''+by'+cy=\lambda w y$, the common techniques are Multiply on the left by $\rho$ to get $\rho(ay''+by')= ((\rho a) y')'$ which (essentially) uniquely determines $\rho$. This requires minimal smoothness of ...


1

Fix a point $c\in (0, 1)$. Define $I := \{ f \in C[0,1] : Z(f)$ contains an open nbd of $c \}$ where $Z(f)$ denotes the zero set of $f$. It can be easily checked that $I$ is a radical ideal and $I \subsetneq M_c := \{ f \in C[0,1] : f(c) = 0 \}$. Also note that $I \nsubseteq M_b$ for any $b \in [0, 1], b\neq c$. Now use the following fact: In a commutative ...


1

Any set that is Jordan measurable has Jordan measure that agrees with its Lebesgue measure. However, if you mean a set that has outer Jordan measure $1$ and Lebesgue measure $0$, consider $\Bbb Q \cap [0,1]$.


1

If $f$ is continuous at $b$, then for any $\epsilon > 0$ there exists $\hat{\delta}> 0 $, which may depend upon $b$, such that if $||x-b|| < \hat{\delta}$, then $||f(x) - f(b)|| < \epsilon/2$. Suppose $x,y \in D \cap B(b;\hat{\delta}).$ Then $$f(x) < f(b) + \epsilon/2 \\ \sup \{f(x):x \in D , ||x-b||< \hat{\delta}\} \leq f(b) + ...


1

A similar construction for a uncountable set is not possible since the sum of uncountable many positive numbers doesn't converge. See The sum of an uncountable number of positive numbers.



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