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4

Notation: For any function $h$ let $h^{(0)}=h$, and let $h^{(n)}$ be the $n$th derivative of $h$ for $n\geq 1.$ Suppose for some $n\geq 0,$ there is a polynomial $ P_n$ in $(2 n+2)$ variables such that $$f^{(n+1)}=P_n(g^{(0)}(f),...,g^{(n)}(f),\;f^{(0))},...,f^{(n)}).$$ Then every term occurring in the polynomial on the RHS is differentiable (because $g$ is ...


4

Suppose $f \in C^k$, then $f^{(k+1)} = f'^{(k)} = (g \circ f)^{(k)}$. Since $g \in C^\infty$ and $f \in C^k$, we have that $g \circ f \in C^k$, and thus that $f^{(k+1)} = (g \circ f)^{(k)} \in C^0$, so it follows that $f \in C^{k+1}$. QED Or even more succinctly as John Ma pointed out below: Suppose $f \in C^k$, then since $g \in C^\infty$, we have that ...


3

One need not to know the whole dual space of $C[0,1]$ to see the lack of reflexivity. Simply note that $C[0,1]^*$, opposed to $C[0,1]$, is non-separable because for each $t\in [0,1]$ the map $$\langle f, \delta_t\rangle = f(t)\quad (f\in C[0,1])$$ is a norm-one linear functional on $C[0,1]$ and $\|\delta_t - \delta_s\| =2$ for distinct $s,t$. To see this, ...


2

No it is not. The dual of $C[0,1]$ is the space $\mathfrak{M}([0,1])$ of complex (or signed) regular Borel measures on $[0,1]$. The dual of $\mathfrak{M}([0,1])$ is the set $\mathcal L^\infty([0,1])$ of bounded Borel functions on $[0,1]$. For example, $\chi_{\{0\}}$, the function which vanishes everywhere, except at $x=0$, where it is equal to $1$, ...


2

Note that this is a differential equation. The equivalent Picard integral equation is $$ f(x)=f(0)+\int_0^x g(f(t))\,dt $$ From here it is trivial to observe that if $f$ is $C^n$ or better, then the composition $g\circ f$ is also at least $C^n$ and thus the anti-derivative $C^{n+1}$. Which gives that $f$ is also $C^{n+1}$ and so on.


2

Your functional assigns to every test function the value $0$. Since $$ \mathrm{e}^{-x}x^2\varphi(x)=0, $$ if $x$ is sufficiently large.


2

Consider first the functions $$h_1(x)=\frac{|\pi-x|}{2},\,\,h_2(x)=-2\ln\left|\sin\frac{x}{2}\right|.$$ It is shown here that the Fourier series of $h_1$ on $(-\pi,\pi)$ is $\sum_{j=1}^{\infty}\frac{\sin jx}{j}$, and the Fourier series for $h_2$ on $(-\pi,\pi)$ is $\sum_{j=1}^{\infty}\frac{\cos jx}{j}$. Therefore, the Fourier series of the function $\sin ...


2

Hint: for any nonnegative number $a$ and $b$ you have $$ (a+b)^2=a^2+b^2+2ab\ge a^2+b^2 $$ and $$ 2(a^2+b^2)-(a+b)^2=a^2+b^2-2ab=(a-b)^2\ge0. $$ Hence, $$ a+b\ge\sqrt{a^2+b^2}\quad\text{and}\quad \sqrt2\sqrt{a^2+b^2}\ge a+b. $$


2

Problem Let $f:(a,b)\to\mathbb R$ be a continuous function such that $$\limsup\limits_{n\to \infty}\frac{f(x_n)-f(x_0)}{|x_n-x_0|}\leq 0$$ for every $x_0\in (a,b)$ and every sequence $x_n$ converging to $x_0$ such that $x_n\not= x_0$ for all $n$. Does this imply that f is a constant on $(a,b)$? Note the absolute values in the denominator. By ...


1

Consider $f(x) = -x$. Then $\frac{f(x) - f(x_0)}{x-x_0} = -1$ for all $x \neq x_0$, and so for all sequences $x_n$ converging to $x_0$ without reaching it, $\frac{f(x_n) - f(x_0)}{x_n-x_0} = -1$, the lim sup of which is $-1 \le 0$.


1

Let $y=(y_1,...,y_n).$ $$\text { We have }\quad |y|^2-\|y\|^2=[\;\sum_{i=1}^n|y_i|\;]^2-\sum_{i=1}^n y_i^2=\sum_{1\leq i<j\leq n}2|y_i|.|y_j|\geq 0.$$ $$\text { We have }\quad 0\leq \sum_{1\leq i<j\leq n}(y_i-y_j)^2=(n-1)\sum_{i-1}^ny_i^2-2\sum_{1\leq i<j\leq n}y_iy_j=$$ $$=n\|y\|^2-\sum_{i=1}^ny_i^2-2\sum_{1\leq i<j\leq ...


1

Let us consider the family of operators $$ T_\ell ((x_n)_n) = x_\ell \cdot e_1, $$ where $e_1 = (1, 0,0,\dots) \in \ell^2$ is the first basis vector. This family satisfies the first property, since for $x = (x_n)_n \in \ell^2$ with $\| x\| = 1$, we have $$ \sum_{\ell=1}^\infty \|T_\ell x\|^2 = \sum_\ell |x_\ell|^2 = \|x\|^2 \leq 1. $$ But it is not too ...


1

The second curve is $y=1/x$ and is not $sin(1/x)$. Make sure you type the parenthesis correctly.


1

The second point is (one of) the definition(s) of the exponential function. The last one is obtained from the second by taking $z=-1$.



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