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1

Let $A$ be ${\mathbb C}^2$ with the pointvise multiplication and the involution $(x,y)^*=(\overline{x},\overline{y})$, i.e., continuous functions on two points. But the norm let be $\|(x,y)\|=|x|+|y|$.


3

To avoid confusing yourself with sequences of sequences, I recommend thinking of the elements of $C_0$ as functions $f:\mathbb{N} \to\mathbb{R}$ such that $\lim_{n\to\infty}f(n)=0$. Then the claim becomes: if $(f_k)_{k=1}^\infty$ is a Cauchy sequence of functions with respect to the uniform norm, there is $f\in C_0$ such that $f_k\to f$ uniformly. The ...


0

I assume you know that a function $F$ of bounded variation on $[a,b]$ can be written as the difference of monotone functions $F=F_{+}-F_{-}$. And I'll assume you also know that $F_{\pm}$ have derivatives a.e., and these derivatives are non-negative where they exist. One pair of functions $F_{\pm}$ can be defined using the variation function $V_{a}^{x}(F)$: ...


0

The post by Nick Thompson answer you question. Let me just add that the point here is not that $Tx$ is not in $L^2$. As a matter of fact, the operator maps $L^2$ to itself. The point is that the norm of the operator is not bounded. So when you say "I tried all the obvious candidates for counter-examples and the outcomes live in $L^2$", you are not addressing ...


1

Let $\varepsilon > 0$, and $x(s):= s^{\varepsilon-1/2}$. Then $\left\| x\right\|_{2}^{2} =\frac{1}{2\varepsilon}$, and $\left\| T[x]\right\|_{2}^{2} = \frac{1}{2\varepsilon^3}$. So \begin{align} \left\|T[x]\right\|_{2} = \frac{1}{\varepsilon} \left\|x\right\|_{2}, \end{align} i.e., $\left\|T\right\|_{2} \ge \frac{1}{\varepsilon}$, for every $\varepsilon ...


1

"...from our results so far we cannot conclude that the integral is positive." As Martin pointed out in a comment, this does not mean that there are examples to the contrary, only that no proof is yet at hand.


0

I think a way to define $H^{s}$ for negative $s$ is as follows: $$ \mu\in H^{s}\leftrightarrow (1+|\xi|^{2})^{s}(F\mu)(\xi)\in L^{2} $$ and you define the norm on the domain $\Omega$ by restriction map. For detail see the Wikipedia article. But it does not seem to me that this norm is explicitly computable in most cases. For example, the Fourier transform ...


0

Hint $X$ is compact and statements about the empty set are vacuously true.


0

Hint: Use $K=X$. Then think what $K^c$ is.


0

Another place where you can find this is the book of Nigel Higson and John Roe, "Analytic K-homology". It gives a complete account of BDF theory including the Weyl-von Neumann result.


3

1) Take $T:l^p\to l^p,(x_1,x_2,x_3,\ldots)\mapsto(0,x_1,x_2,x_3,\ldots)$ (which is your idea for (4)). Then $T$ is one to one, because if $T(x)=T(y)$, then $(0,x_1,x_2,x_3,\ldots)=(0,y_1,y_2,y_3\ldots)$ and hence $x_i=y_i$ for each $i\in\mathbb N$. But $T$ is not onto, since there is no sequence $x\in l^p$ such that $T(x)=(1,0,0,0,0,\ldots)$. 2) Take ...


0

In finite (fixed) dimension all the norms are equivalent. The function $$x\longmapsto\frac{x}{\|x\|_2}$$ is continuous and bijective form $S_1$ (compact by Heine-Borel) to $S_2$, so the inverse is continuous.


1

Note that on $\Omega=\mathbb{R}^2$ we find that $H^1_0(\Omega)=H^1(\Omega)$. To see that such a constant $C$ can not exists think of the following example. Consider the functions $u_k:\mathbb{R} \to \mathbb{R}$, defined by $$ u_k(x) := \chi_{\{[-k,k]\}}(x) + (k+1 - |x|)\chi_{\{[-(k+1),-k]\cup[k,k+1]\}}(x)$$ and simply make them rotational symmetric to get ...


0

Your equation is only of first-order, only first derivatives with respect t only one variable appear there. There is no classification of such equations into elliptic, parabolic, hyperbolic equations. If you talk about systems of first-order equations, they can be classified, if they can be written as second-order equations. E.g. the system $$ ...


1

It may require too much background for your class presentation, but you might be interested in an application of this to statistical mechanics. See my paper with R.R. Phelps: "Some convexity questions arising in statistical mechanics," Math. Scand. 54(1984), 133-156. The "pressure" is a real-valued convex function on a Banach space of "interactions"; under ...


0

let $x_n+M$ be a cauchy sequence in $X/M$ then $x_n$ is a Cauchy sequence in $X$ and as $X$ is a Banach Space and hence complete.Thus $x_n$ converges to some $x\in X$.Hence $x_n+M$ converges $x+M \in X/M$


0

Here is an elementary proof, with your notations. If one among $f,g$ is $0$, then choose $\lambda = 0$. If $f\not = 0$ and $g\not = 0$, then you have in fact equality of the kernels as both are hyperplanes, note $H$ this hyperplane. Let $e\in H$, and $D$ the line generated by $e$, which is a supplementary subspace of $H$ in $V$ as $f$ and $g$ are non zero ...


1

The heat equation is parabolic: $$ \frac{\partial f}{\partial t} = \frac{\partial^{2}f}{\partial^{2}x}+\frac{\partial^{2}f}{\partial y^{2}}+\frac{\partial^{2}f}{\partial z^{2}} $$ Laplace's equation is elliptic: $$ \frac{\partial^{2}f}{\partial^{2}x}+\frac{\partial^{2}f}{\partial y^{2}}+\frac{\partial^{2}f}{\partial z^{2}} = 0 $$ The ...


2

Yes. Note that $f$ is a function not just of dimension but of the volume of the domain.


1

Let $e_n$ denote the element of $\ell_1$ whose $m$'th coordinate is $0$ if $m\ne n$ and $1$ if $m=n$. To see more directly why the bounded linear map, referenced in my comment to the main post, $T:\ell_1\rightarrow\ell_1$ defined by $$Te_1=e_1;\ Te_n=e_1+e_n, n>1$$ is not an adjoint operator, consider the sequence $(e_n)$ in $\ell_1$. Note for any ...


3

The first condition you will want to impose is that if $N$ is a null-set, then so is $g^{-1}(N)$. To see this, note that in general $\chi_{M} \circ g = \chi_{g^{-1}(M)}$, where $\chi_{M}$ is the characteristic function/indicator function of the set $M$. Hence, if there was a null-set $N \subset [a,b]$ such that $g^{-1}(N)$ is not a null-set, then you would ...


1

HINT: Use the change of variable formula to compare $$\int_I |f( s)|^2 d s= \int_I |f( \phi(t))|^2 | \phi'(t)|\, d t \ \text{and}\\ \int_I |f( \phi(t))|^2 d t \\ $$


0

Indeed, each continuous bijection from a compact topological space into a Hausdorff space is homeomorphism(See topological books). Similarly for open functions.


0

The function $\psi:\>X\to{\mathbb R}$ defined by $$\psi(f):=\int_0^1\left|{\rm Im}\bigl(f(x)\bigr)\right|\>dx$$ is continuous on $X$: $$|\psi(f)-\psi(g)|\leq\int_0^1|f(x)-g(x)|\>dx=\|f-g\|\ .$$ Therefore $S:=\psi^{-1}(0)$ is closed in $X$.


0

No, you can't change the order of differentiation in the term $$ \frac{d}{dx} \frac{\partial J}{\partial u'}. $$ The idea here is that $u$ and $u'$ are functions of $x$, and the notation (writing $\frac{d}{dx}$ as a total derivative instead of a partial derivative) is supposed to suggest that the $\tfrac{d}{dx}$ sees that $x$-dependence. For instance, if ...


1

Take any point $y\in (x-\delta(x),x+\delta(x))$ and consider $r=min\{y-(x-\delta(x)),x+\delta(x)-y\}$ Then $r>0$ and then consider the open interval $\left(y-\frac{r}{2},y+\frac{r}{2}\right)$ and try to prove that this lies entirely in your given set.


1

Hint: Show the complement is open. Show that the nearest element of $X$ to $f$ is $\mathrm{Re}\,f$.


0

Hints: (a) $Pe_j=e_j$ for all $j$ implies $Pf=f$ for $f\in F$. (b) If $(f-g) \perp F$, then $(f-g,e_j)=0$ and $P(f-g)=0$ by definition. Also use (a) (c) $f = Pf+(f-Pf)$. Let $u=Pf$ and $v=f-Pf$. If $u,u'\in F$, $v,v' \perp F$ with $u+v=u'+v'$ then $(u-u')=(v'-v)$ is orthogonal to itself. (d) Use (c) to decompose $f=u+v$. Show $u \perp F^{\perp}$ so $f-u = ...


1

DIFFERENTIATING AT THE ORIGIN. Take the vector space to be the ordinary $(x,y)$ plane. In polar coordinates, restricting to $r \geq 0,$ let $$ f(r, \theta) = r \sin 3 \theta . $$ A line through the origin is given by fixing a value of $\theta;$ doing so, the Gateaux ratio is just the constant $\sin 3 \theta ,$ so that is also the limit. The Gateaux ratio ...


1

Let $A$ be an element of $F\setminus F_n$ (i.e., $A$ has at least $n+1$ elements and may even be infinite). Pick $n+1$ distinct points $a_0,\ldots, a_{n}\in A$ and let $r=\min_{0\le i<j\le n}p(a_i,a_j)>0$. Now let $B$ be any element of $F_n$. Then for any $b\in B$ there exits at most one $i$ with $p(a_i,b)<\frac r2$, hence by pigeon-hole there ...


0

To prove that $F_n$ is closed. Take an element $K$ not in $F_n$, then we can take $n+1$ points $x_1,x_2,...,x_{n+1}$ in $K$. Let $2r$ be the minimum distance between any two of these $n+1$ points. Then $$B(K,r)$$ is disjoint from $F_n$. The reason is that any element $J$ of $B(K,r)$ ought to have a point in the $n+1$ balls (in $X$) with centers ...


1

Since $$ p=p^2=(p_1+\cdots+p_n)^2=\sum_{i=1}^{n}\sum_{j=1}^{n}p_i p_j= $$ $$=p_{1}^{2}+\cdots+p_{n}^{2}+\sum_{1\leq i<j\leq n}(p_ip_j+p_j p_i)=p+\sum_{1\leq i<j\leq n}(p_ip_j+p_j p_i)$$ we have $$ \sum_{1\leq i<j\leq n}(p_ip_j+p_j p_i)=0. \tag5 $$ Without loss of generality we may assume that $A$ is a $C^*$-subalgebra of $B(H)$, where $H$ is a ...


0

let $f(x) = \dfrac{a_1}{x - \lambda_1} + \dfrac{a_2}{x - \lambda_2} + \dfrac{a_3}{x - \lambda_3}.$ you can verify that $f$ is continuous everywhere except at $\lambda_1, \lambda_2$ and $\lambda_3.$ at these points $f$ has a vertical asymptote. then $$\lim_{x \to \lambda_1 + }f(x) = \infty, \lim_{x \to \lambda_2 - }f(x) = -\infty.$$ therefore by ...


0

Suppose that $x\in V\cap\operatorname{cl}Y$, and let $U$ be an open nbhd of $x$. $V$ is an open nbhd of $x$, so without loss of generality we may assume that $U\subseteq V$. Then $U\cap Y\ne\varnothing$, since $x\in\operatorname{cl}Y$, so $$U\cap(V\cap Y)=U\cap Y\ne\varnothing\;,$$ and $x\in\operatorname{cl}(V\cap Y)$.


2

Just as an example, consider the operator $\frac{d^{2}}{dx^{2}}$ on the set $\mathscr{D}$ of all twice absolutely continuous functions $f \in L^{2}[0,2\pi]$ with $f'' \in L^{2}[0,2\pi]$. Then $T_{\alpha,\beta}=\frac{d^{2}}{dx^{2}}$ is selfadjoint on the domain $\mathcal{D}(T_{\alpha,\beta})$, $0 \le \alpha,\beta < \pi$ consisting of all $f \in ...


1

No, it is not. The reason why is that, if $\mathbb{K}$ is complex, then the inner product is conjugate linear and therefore only induces an anti-isomorphism, not an isomorphism, between $H$ and its dual. We find that $y$ is identified with the conjugate of the gradient in this case. To be more concrete, pick $H = \mathbb{C}$ (over $\mathbb{C}$). Then pick ...


1

So does the author just mean that the basis is not a Hamel basis? Yes, precisely that; "a basis, in the sense of algebra" is a Hamel basis. I think that a total orthonormal sequence must be a Schauder basis Yes, that is correct, but a total orthonormal family need not be countable, in contrast to a sequence, and if you have an uncountable ...


1

The standard definition requires $A$ to be bounded. There are many text on nonlinear analysis, like this one.


0

In you post, I read $C$ is bounded in $BV$ norm, i.e., the function of bounded variation. If you know Sobolev space, you could just think $BV$ is the relaxation of $W^{1,1}$ and hence similar to $W^{1,1}$ space, bounded sequence in $BV$ will compact in $L^1$, and even in $L^q$ for any $q<N/(N-1)$ where $N$ is the dimension of space. For reference, ...


0

For some reasons I read that you wanted to have $T$ self-adjoint on one domain and not-selfadjoint on the other. If $T$ is densely defined (its domain is dense in $H$), then its adjoint exists. If moreover the adjoint is densely defined then $T$ is closable (there exists a closed operator which graph contains the graph of $T$). It may happen that its ...


2

As a first approach you can do this component wise. The trouble with this is that usually you want to have a definition which does not depend on the choice of a basis, so this is what you would have to check in this case. A (very) general definition can be found in several places, e.g. in Rudin's functional analysis, chapter 3, section 'Vector valued ...


4

For $K \subset \mathbb{R}^n$ compact, consider the space $$\mathcal{D}_K := \{ \varphi \in \mathcal{D} : \operatorname{supp} \varphi \subset K\}$$ endowed with the seminorms $$\lVert\varphi\rVert_k = \sup \{ \lvert D^k\varphi(x)\rvert : x \in \mathbb{R}^n\}$$ for $k \in \mathbb{N}^n$. It is straightforward to show that $\mathcal{D}_K$ is then a Fr├ęchet ...


2

The problem here is that $\delta$ isn't really a function--rigorously, it's a functional. The idea that $\int_{-\infty}^\infty f(t) \delta(t) dt=f(0)$ for any real function is perfectly fine, but the notion that $\delta$ is some bizarre function that assumes an infinitely tall value in an infinitesimally small region to have unit area can't be made rigorous. ...


0

If I set $x_i=|T(e_i)|^{q-1}$ when $T(e_i)>0$ and $x_i=-|T(e_i)|^{q-1}$ when $T(e_i)<0$ then $x_iT(e_i)=|T(e_i)|^q$ and $\frac{|T(\mathbf{x})|}{\|\mathbf{x}\|_p}=(\sum_{i=1}^{k}|T(e_i)|^q)^\frac{1}{q}$, for $\mathbf{x}=(x_1,...,x_k)$.


2

The result perhaps can be proved inductively showing that $\text{index}\,( T- \delta) = \text{index}\,T$ for $\delta$ of rank $1$. A useful particular case is showing by hand that $\text{index}( I - \delta) = 0$, starting with rank of $\delta=1$. The different cases presented suggests that there could be a uniform proof. We sketch one below. Break it into ...


1

First note that given any $x,n\in L^2(-1,1) $ with $n\neq 0$ there is a unique scalar $\lambda$ and $y \in L^2(-1,1)$ such that $x = \lambda n + y$ and $y \bot n$. For existence, let $\lambda = { 1\over \|n\|^2}\langle n , x \rangle $ and $y = x-\lambda n$. It is easy to check that $y \bot n$. For uniqueness, suppose $\lambda_1 n + y_1 = \lambda_2 n + ...


2

It is a consequence of Poisson summation formula. You just have to prove that, if $$ f(x) = \exp\left(-\pi a x^2+2\pi i b x\right), $$ then its Fourier transform is: $$ \widehat{f}(s) = \frac{1}{\sqrt{2\pi a}}\exp\left(-\frac{(2b\pi+s)^2}{4a\pi}\right).$$


3

Try something like $f_n(x) = ne^{-nx}$. Then $$\|f_n\|_1 = \int_0^1 ne^{-nx} \, dx = 1 - e^{-n}$$ for all $n$ but $\|f_n\|_\infty = n$.


6

This is not true. Simply take $f = \chi_{[-1,1]}$ (the indicator function of the interval $[-1,1]$). Then $f \in L^p$ for all $p \in (0,\infty]$, but if $\widehat{f} \in L^1$ was true, then Fourier inversion would imply that $$ f = \mathcal{F}^{-1} \widehat{f} \in C_0 $$ would be (almost everywhere equal to) a continuous function. This is clearly not the ...


1

Your professor is wrong and Umberto P. is right. If we take $$ f(x)= K_0(|x|) $$ where $K_0$ is a modified Bessel function of the second kind, we have $f\geq 0$ and: $$ \int_{-\infty}^{+\infty}K_0(|x|)\,dx = \pi,\qquad \int_{0}^{+\infty}K_0(|x|)^2\,dx = \frac{\pi^2}{2},$$ so $f\in L^1\cap L^2(\mathbb{R})$, but: $$ \widehat{f}(s) = ...



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