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3

I would recommend practicing bounding things with inequalities, so when you encounter a problem that requires the use of the triangle inequality (it comes up fairly frequently), you can find a bound fast. Many of the definitions used in analysis deal with inequalities as well, such as convergence of a sequence, limit of a function, continuity, etc... Also ...


3

Let's fix the number $p$ and write $n = pm + k$, where $0 \leq k < p$, then we have \begin{align} \frac{x_{pm+k}}{pm+k} &\leq \frac{x_{pm}}{pm+k} + \frac{x_k}{pm+k} \\ & \leq \frac{mx_p}{pm} + \frac{px_1}{pm} \end{align} when we send $m$ to infinity, we get $\limsup\frac{x_n}{n} \leq \frac{x_p}{p}$, since here $k $ can only vary finitely from 0 ...


3

For $y\geq 2$, $0<y+2\leq 2y$ and $y-2\geq 0$ so $$(y+2)^n-(y-2)^n \leq (y+2)^n \leq (2y)^n$$ Now substitute $y=2/x$. Alternatively, show that for $x\in(0,1]$:$$(1+x)^n-(1-x)^n\leq (1+x)^n\leq 2^n$$ and then multiply both sides by $$\left(\frac 2x\right)^n$$ Note, it is true for any $n\geq 0$, not just $n$ natural.


2

We quite obviously have: $$\lim_{m\to\infty}\cos^m(n!2\pi x)=\cases{1&if $n!x\in\mathbb{Z}$\\0&if $2n!x\notin\mathbb{Z}$\\\mathrm{doesn't\ exist}&else}$$ And thus: $$\lim_{\mathbb{N}\ni n\to\infty}\left(\lim_{m\to\infty}\cos^m(n!2\pi x)\right)=\cases{1&if $x\in\mathbb{Q}$\\0&else}$$ I'll leave it to you to check the details.


2

The Heine-Borel theorem says that $[0,b]$ is compact. So, $f$ continuous on $[0,b]$ implies that it is also uniformally continuous there. By hypothesis, $f$ is uniformally continuous on $[b,\infty)$. So if you are given an $\epsilon > 0$ you can find the appropriate $\delta_1$ and $\delta_2$ given by uniform continuity on regions $[0,b]$ and ...


2

Okay: $\|Tf\|_2^2=\int_a^b|\sum_{j=1}^n\int_a^b\phi_j(t)\psi_j(s)f(s)ds|^2dt\le\int_a^b|\sum_{j=1}^n|\phi_j(t)|\int_a^b|\psi_j(s)f(s)|ds|^2dt$ $\le\|f\|_2^2\sup_j\|\psi_j\|_2^2\int_a^b|\sum_{j=1}^n \phi_j(t)|^2dt\le\|f\|_2^2\sup_j\|\psi_j\|_2^2\|\sum_{j=1}^n \phi_j(t)\|_2^2$ Thus: $\|Tf\|\le\sup_j\|\psi_j\|_2\|\sum_{j=1}^n \phi_j(t)\|_2\|f\|_2$ I.e. $T$ ...


2

By Cauchy's-Hadamard formula, with $\;R:=$ convergence radius, with the usual conventions when $\;R=0\,,\,\infty\;$ , we get: $$\frac1R=\lim_{n\to\infty}\sup\sqrt[n]{|a^n+b^n|}$$ and assuming $\;|a|\ge|b|\;$ , we get $$\sqrt[n]{|a^n+b^n|}=|a|\sqrt[n]{1+\left(\frac{|b|}{|a|}\right)^n}\xrightarrow[n\to\infty]{}|a|$$


2

Note that if $f(0)=0$ or $f(1)=1$ we are done immediately, so we can assume $f(0)\gt 0$ $f(1)\lt 1$ and let $y=\sup \left\{x\in [0,1] \land f(x)\ge x\right\}$ - this deploys the completeness of the real numbers: the set is nonempty (contains zero) and is bounded above by $1$. Then you need to use the continuity of $f$ at $y$ to show that $f(y)=y$. I put ...


2

If $u(t)=f(t,t)$ then $u'(t)=\dfrac{\partial f}{\partial x}(t,t)+\dfrac{\partial f}{\partial y}(t,t)$. Edit: This can be extended into $u=f\circ(g_1,g_2)$ with $f$ as in the question and $g_1=g_2:\mathbb R\to\mathbb R$ defined by $g_1(t)=g_2(t)=t$. Thus, $$ u'=\partial_xf(g_1,g_2)\cdot g'_1+\partial_yf(g_1,g_2)\cdot g'_2. $$


2

Yes, it is true. Define a (complex-valued) measure on $\mathbb{R}$ by $$\mu(B) := \sum_{k \in \mathbb{Z}} c_k \delta_k(B), \qquad B \in \mathcal{B}(\mathbb{R}). \tag{1}$$ Since $\sum_{k \in \mathbb{Z}} |c_k|<\infty$, $\mu$ is a finite measure on $\mathbb{R}$. From $(1)$ we see that the formula $$\int f(x)\left( \sum_{k \in \mathbb{Z}} c_k \delta_k(dx) ...


1

Also $H_2$ is NOT satisfied, I think it is a typo and is should be: $f(t,u)=81u^2+4015\sin u +tu$ (or atleast along those lines) since $H_1$ is then satisfied and: $\int_0^u81v^2+4015\sin v +tvdv=(81/2)u^2-4015\cos u +(1/2)u^2t\le (82/2)u^2+4015\lt (\pi^4/2)+4015$ So $H_2$ is satisfied but it wasn't before.


1

We need two ingredients: If a sequence $(f_n)_n$ converges in some $L^p$ then there exists a subsequence $(f_{n_k})_k$ that converges almost everywhere. A subsequence of a convergent sequence converges to the same limit. Now, since $(f_n)_n$ converges to $g$ in $L^1$ then, (by 1.), there there exists a subsequence $(f_{n_k})_k$ that is simply ...


1

In fact if $T$ is non-expansive then the proposed inequality follow from the Cauchy-Schwartz inequality, since with notation of the proposer we have $$ \langle T(x)-T(y),x-y\rangle= \langle A(x-y),x-y\rangle\leq\Vert A(x-y)\Vert\,\vert x-y\Vert\leq\Vert x-y\Vert^2. $$ Now, the converse is not correct in general without supplementary assumptions, for ...


1

The argument looks fine, although the notation is somewhat sloppy: for instance I would write the first step as $$\int_0^{\infty} f(x)\cos(nx)\,dx = \cos(nx)\int_0^x f(y)\,dy\ \biggr\vert_{x=0}^{\infty} + n\int_0^{\infty} f(x)\sin(nx)\,dx.$$ The claim becomes false if $f(x)$ is locally integrable but not $L^1$, i.e. if $\lim_{x\to \infty} \int_0^x ...


1

I think it is that straightforward. The reverse implication is trivially true. If there exists an $n \in \mathbb{N}$ such that $1 \leq x \leq 1+1/n$, then $x$ must necessarily lie in $[1,2]$, since $\sup_{n \in \mathbb{N}}{(1+1/n)}=2$. Note that for any $n \neq 1$, $x$ must still lie in $[1,2]$ even though $x\neq 2$.


1

Yes, the $Max$ function of two Riemann -integrable functions is Riemann -integrable. This is because the $Max$ of two a.e -continuous bounded functions is also a.e -continuous and bounded. For $f,g$ continuous, the function $Max${f,g} is continuous. This implies that (since a Riemann-integral must be a.e -continuous) , that $Max${f,g} is a.e -continuous; ...


1

Yes, the convergence/divergence depends on $r$. In your case, the series converges when $\dfrac1{2^r} < 1$, i.e., for $r>0$. Specifically for $r=0$, note that the sum is $\displaystyle \sum_{n=0}^{\infty} n^2$, which clearly diverges. For $r>0$, you can infact simplify and obtain a closed form as shown below. First note that for $\vert x \vert ...


1

Once you get to $$(1+x)^n=(1+x)^{n-1}(1+x)$$ you have rearranged the expression in a form in which you can use the inductive hypothesis, because you can assume that $$(1+x)^{n-1}\ge 1+(n-1)x$$ You should substitute that information into the expression you have for $(1+x)^n$ and see where it gets you when you follow it through logically. So in comments ...


1

Since $a_n = a^n + b^n$, then $$\sum_{n=1}^\infty a_n x^n =\sum_{n=1}^\infty (ax)^n + \sum_{n=1}^\infty (bx)^n .$$ For this to be convergent, both series must be convergent, but these are regular geometric progressions, so the conditions for their convergence are that $|ax|<1$, and $|bx|<1$. So we must have simultaneously $$|x|< \frac{1}{|a|}$$ ...


1

A continuous function on a compact set is uniformly continuous. Since $f$ is uniformly continuous on $[0, b]$ and uniformly continuous on $[b, \infty)$, it is uniformly continuous on $[0, \infty)$. To see that it is uniformly continuous, let $\epsilon > 0$. We can find $\delta_1, \delta_2 > 0$ such that $|x - y| < \delta_1, \delta_2$ implies ...


1

It means, along the tangent vector to the curve. There is a very simple way to do this: substitute $x,y,z$ with the coordinates of the curve, getting: $$ F(t) = f(x(t),y(t),z(t)) = f(e^{-t}, 1+2\sin t, t-\cos t) = ... $$ Now simply take the derivative with respect to $t$! EDIT Otherwise, take the scalar product between the gradient of $f$, and the ...


1

If $x\in\mathbb{Q}$ then $x=\frac{a}{b}$ for some $a,b\in\mathbb{Z}$ so we have for $$n>b$$ $n!x\in\mathbb{Z}$ and therefore $$\lim_{m\to \infty}{\cos^m{(n!2\pi x)}}=\lim_{m\to \infty}{1^m}=1$$ Thus $\forall \epsilon>0$ we can take $n>N=b$ to get $$\left|\lim_{m\to \infty}{\cos^m{(n!2\pi x)}}-1\right|=|1-1|=0<\epsilon$$ so the limit is $1$. If ...



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