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42

This is how I used to imagine projections: If a mouse: gets run over by a steamroller: It will look like this: Now if it gets run over by a steamroller another time, it will still look like this:


6

There are basically two things to note here. First you need to understand what it means for a set to be a basis of an infinite dimensional normed space. In your context, I am all but certain that this means that it is a Schauder basis, which is a linearly independent set such that the set of all finite linear combinations of members of the set is dense in ...


5

The only thing you can say that $fg$ has the same smoothness as $g$. That is, if $g$ is in $C^k$, then $fg$ is in $C^k$ as well. If $f$ in addition has compact support, then $g\in C^k$ implies that $fg$ is in $C^k$ with uniformly continuous derivatives up to order $k$. Similarly, if $g$ is in some Sobolev space $W^{m,p}$, then $fg\in W^{m,p}$ as well.


5

I don't think you can say anything without further assumption on $g$. The constant function $f=1$ is $\mathcal{C}^\infty$, and $fg=g$ can be arbitrary.


5

Consider $H = \mathbf K^2$ and the operators $$ A = \def\p#1#2#3#4{\begin{pmatrix} #1 & #2 \\ #3 & #4 \end{pmatrix}}\p 1111, \quad B = \p 2111 $$ Then \begin{align*} \def\<#1>{\left<#1\right>}\<Ax,x> &= |x_1 + x_2|^2\\ \<Bx,x> &= (2x_1 + x_2)\bar x_1 + (x_1 + x_2)\bar x_2\\ &= |x_1|^2 + |x_1 + ...


5

Let our Banach space be $\ell^1$, which is the set of all $a \in \mathbb{R}^{\mathbb{N}}$ such that $\sum_n |a_n|<\infty$. The norm is the $L^1$ norm. For $i \in \mathbb{N}$, let $e_i \in \ell^1$ denote the $i^{th}$ standard basis vector, i.e, the sequence whose $i^{th}$ index is $1$, but all other indices are $0$. Consider the linear map $T: \ell^1 \to ...


5

It's the Hilbert space of $L^2$ functions on the circle. More explicitly, it's the space of Lebesgue measurable functions $f : \mathbb{R} \to \mathbb{R}$ which are periodic with period $2\pi$ and such that the integral $$\int_0^{2\pi} |f(x)|^2\,dx$$ converges, modulo the equivalence relation where $f \sim g$ if $\int_0^{2\pi} |f(x) - g(x)|^2\,dx = 0$.


5

Here's something to get you started. You know that $f(x)$ is a polynomial function, so it is also continuous. We also know that $$f(x)f(y)+2=f(x)+f(y)+f(xy)$$ if both $x$ and $y$ are zero. However, taking both $x$ and $y$ to approach zero, the equation is also true for either $x$ or $y$ or both being zero. Substitute $y=0$. Substituting and solving ...


5

In some sense the norm you are using is the graph norm of the Laplacian, so the operator is bounded, since $$||-\Delta u||_{L^2} \le ||u||_{2,2} $$ and since continuity implies closedness, you are done.


5

Assume that $b > a \geq 0$ (otherwise $f^{-1}((a,b))$ is empty, which is okay). We can write $$f^{-1}((a,b)) = \{ (x,y) : a < x^2+y^2 < b \} = \{ (x,y) : x^2+y^2 < b \} \cap \{ (x,y) : x^2 + y^2 \leq a \}^c$$ This is an intersection of an open ball with radius $\sqrt{b}$ and the complement of a closed ball with radius $\sqrt{a}$. The complement ...


5

Let $\left( x^{(n)}\right)_{n=1}^{\infty} \subset \ell^p$ be a Cauchy sequence. Since I see you have troubles with your notations of sequence of sequences, this is the notation that I will use for each element $x^{(n)}$ in the sequence: $$ x^{(n)} = \left( x_j^{(n)}\right)_{j=1}^{\infty} = \left( x_1^{(n)},x_2^{(n)}, \cdots \right)\in \ell^p $$ For $x= ...


4

As the OP has mentioned, we have $$ \forall \lambda \in \mathbb{R}_{> 0}: \quad T_{\lambda} \stackrel{\text{df}}{=} \lambda (\lambda I - A)^{-1} - I \in B(\mathcal{H}) \quad \text{and} \quad \| T_{\lambda} \|_{B(\mathcal{H})} \leq \rho. $$ This is a consequence of the following theorem, whose statement is found in Theorem $ 13.35 $ of Walter Rudin’s ...


4

In case that $W$ is closed, there is indeed a canonical choice. It is called the quotient norm and defined by \begin{equation*} \| [\hat v] \|_{V / W} := \inf_{v \in [\hat v]} \| v \|_V . \end{equation*} This also leads to $\|\pi\| = 1$ (in case $W \ne V$). As already mentioned by matt biesecker, you always have $\operatorname{ker}(\pi) = W$, since this ...


4

Write $J(y) = \int_0^1 L(x,y,y') \ dx$ where $L(x,y,y') = y'.\sin(\pi y) - (t + y)^2$. Then by the E-L equation, $J$ has extrema when $$\cfrac{\partial L}{\partial y} - \cfrac{d \ }{dt}\left(\cfrac{\partial L}{\partial y'}\right) = 0$$ I.e., $$\left( \pi.y'\cos(\pi y) - 2(t+y)\right) - \cfrac{d \ }{dt}(\sin(\pi y)) = 0 $$ equivalently $$\pi.y'\cos(\pi y) ...


4

$T$ with the condition you gave is clearly one-to-one, as $Tx = 0$ implies $$ \def\norm#1{\left\|#1\right\|}\norm{x}_X \le \frac 1C \norm{Tx}_Y = 0. $$ But $T$ need not be onto $Y$, as the example $T(x_1, x_2, \ldots) = (0, x_1, x_2, \ldots)$ on $\ell^2$ shows (note that $\norm{Tx}_2 = \norm x_2$ in this case). But, $T$ is onto $TX$ (obviously) and hence ...


4

Consider a sequene $(y_n)$ of elements of $X$ with unit norm and such that $f(y_n)\leqslant -n^2$ (such a sequence exists since $f$ is not continuous). If $x$ is an element of $X$, define $x_n:=x+n^{-1}y_n$; then $\lVert x-x_n\rVert\to 0$ and $$f(x_n)=f(x) +\frac 1nf(y_n) \leqslant f(x)-n,$$ hence $x_n$ belongs to $C_1$ for $n$ large enough.


4

If $\{ x_{n} \}$ converges to $x$ in $\|\cdot\|_1$, then it converges in $\|\cdot\|$ because $$ \|x-x_n\| \le \|x-x_n\|_1. $$ Conversely, if $\{ x_n \}$ converges to $x$ in $\|\cdot\|$, then $\{f(x_n)\}$ converges to $f(x)$ because $f$ is continuous; therefore, $\{ x_{n} \}$ converges to $x$ in $\|\cdot\|_1$.


4

Make a substution $y=\alpha x$ and examine $g(y)=\frac{e^y - 1}{y}.$ Then $$ g'(y)= \frac{ye^y - e^y + 1}{y^2}. $$ To show $g'(y) > 0$ for $y>0,$ it suffices to show that $(y-1)e^y > -1.$ Let $h(y) = (y-1)e^y.$ Then $h'(y)=y e^y,$ we have $h(0)=-1,$ $h'(0)= 0,$ and $h'(y)>0$ for $y>0,$ implying that $h(y)>-1$ for $y>0.$ Thus ...


4

Rather, each stage of the Cantor set's construction is comprised of a finite union of closed intervals, so is closed. The Cantor set, itself, is the intersection of these (countably-many) stages. As an intersection of closed sets, then, the Cantor set is closed.


3

Asking questions about the intersection $H_1 \cap H_2$ of two arbitrary Hilbert spaces $H_1$ and $H_2$ is "evil" in the mathematical sense of the word. Unless the two Hilbert spaces are linear subspaces of some larger Hilbert space, then this question is ill-posed. In general, it is "evil" to ask questions about structured objects such as groups, ...


3

Consider $$ h(x)=f(x)-\frac{1}{b-a}\int_a^b f(t)\,dt. $$ It suffices to show that $h\equiv 0$. In turn, it suffices to show that $$ \int_a^b h(x)w(x)\,dx=0, \quad\text{for all $w:[a,b]\to\mathbb R$ continuous}. $$ Take one such $w$, and let $\tilde w(x)=w(x)-\dfrac{1}{b-a}\int_a^b w(t)\,dt$. Then $\int_a^b \tilde w(x)\,dx=0$, and hence we know that $$ ...


3

The idea of compact came from sequences. You can see how one might come up with the name 'compact' to describe a set where you can't have an infinite set of points that are a minimum fixed positive distance from each other; that would not be a 'compact' set. Finding a cluster point gives you way to find a limit, and that's the importance of a 'compact' set. ...


3

My favourite proof is due to Hartig: D. G. Hartig, The Riesz representation theorem revisited, American Mathematical Monthly, 90(4), 277–280. Hartig claims that his proof is ...category-theoretic but having unwrapped the details I must say it is really functional-analytic. The idea is as follows. Step 1. We can easily prove this theorem for extremely ...


3

Suppose that $T$ is bounded. Then $$ \|x\|^{2}+\|Tx\|^{2}=((I+T^{\star}T)x,x) \le \|(I+T^{\star}T)x\|\|x\| \\ \|x\|^{2} \le \|(I+T^{\star}T)x\|\|x\| \\ \|x\| \le \|(I+T^{\star}T)x\|. $$ The last inequality can be use to show that the range of $I+T^{\star}T$ is closed, and $I+T^{\star}T$ has a bounded inverse on ...


3

Let $k(x,y) = -x^2+e^{-x^2+y}$, then $k \in C([a,b]^2)$. It is uniformly continuous on $[a,b]^2$, so for any $\epsilon > 0, \exists \delta > 0$ such that $$ |(s,t) - (x,y)| < \delta \Rightarrow |k(s,t) - k(x,y)| < \epsilon $$ Thus, $$ |T(f)(s) - T(f)(x)| = |\int_a^b k(s,t)f(t)dt - \int_a^b k(x,t)f(t)dt| \leq \int_a^b |k(s,t)-k(x,t)||f(t)|dt $$ ...


3

Yes, a linear transformation is not required to be invertible. All that is required is that it is a mapping from one vector space $V$ to another $W$ that satisfies $$L(\alpha_1 v_1 + \alpha_2 v_2) = \alpha_1 L(v_1) + \alpha_2 L(v_2)$$ where $\alpha_i$ are elements of the field over which the V is defined and $v_i$ are elements of $V$. A matrix is such a ...


3

It is well-known (and easy to see) that $c/Y$ is isomorphic to $c_0$. So we just need to renorm $c_0$ so that the natural map between them is an isometry. In fact, the norm $|||\cdot|||$ will do, where \begin{equation*}|||(x_n)|||=\frac{1}{2}\sup_{m,n}|x_n-x_m|.\end{equation*} Notice that for any $(x_n)\in c_0$ we have ...


3

For $a\in A$, $$\phi(a^*a) = \langle \pi(a^*a)h,h\rangle = ||\pi(a)h||^2\geq 0$$ Which shows that $\phi$ is a positive linear functional. By theorem 3.3.3 of Murphy's C*-algebras and operator theory, if $\{u_i\}$ is approximate unit of C*-algebra $A$, then $\|\phi\|=\lim \phi(u_i)$. Using this we have $$\|\phi\|=\lim \phi(u_i)=\langle \pi(u_i)h,h\rangle = ...


3

I presume $\text{aff} A$ means the affine hull of $A$ (in some normed linear space), and $\text{ri} A$ is the relative interior of $A$ with respect to this affine hull. No, it's not true. In $\mathbb R^2$, let $A$ be a line segment, and let $B$ be the union of $A$ with some point not on $A$. Then $\text{ri}(A)$ is nonempty, but $\text{ri}(B)$ is empty.


3

What if you have eignnvectors of different eigenvalues? If $T(x) = \lambda x$ and $T(y) = \mu y$, then $T(x+y) = \lambda x + \mu y$, which is not a scalar multiple of $x + y$, unless $\lambda = \mu$.



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