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5

All normed linear spaces are metric spaces, and metric spaces are Hausdorff.


4

For $0 < p < 1$, let $$\ell^p(\mathbb{N}) = \Biggl\{ x \colon \mathbb{N}\to \mathbb{C} : \sum_{n = 0}^\infty \lvert x(n)\rvert^p < +\infty\Biggr\}.$$ We endow it with the $p$-seminorm $$s_p(x) = \sum_{n = 0}^\infty \lvert x(n)\rvert^p$$ and the metric $d_p(x,y) = s_p(x-y)$ derived from it. Note the difference from the case $p \geqslant 1$. If ...


4

Scalar multiplication will not be continuous unless the topology on the scalar field $K$ is also discrete (or unless $X=0$). Indeed, if $x\in X$ is any nonzero vector, $a\mapsto a\cdot x$ is a continuous injection $K\to X$, as the restriction of the scalar multiplication map $K\times X\to X$ to the subspace $K\times\{x\}\cong K$. If $X$ has the discrete ...


3

There are a few ways to define the span of $M$, what you wrote in your first paragraph is one of them. Taking all scalar multiples of elements of $M$ does not give the correct definition of the span of $M$: in $\mathbb{R}^2$, the set of all scalar multiples of $M=\{(1,0),(0,1)\}$ is the union of the $x$-axis with the $y$-axis which is not even a subspace. ...


3

The Cauchy-Kowalevski Theorem, together with Lewy's example, provides some food for thought. The CKT is the main result about local existence of a solution of an analytic first order system of PDEs. Roughly speaking, if all the coefficients, the force and the boundary datum are analytic at some point the PDE admits a local solution which is $C^{\infty}$ and ...


3

zhw points out a nice property of analytic functions: If $f,g$ are analytic on $(a,b)$ and $f(x_n)=g(x_n)$ for a sequence of distinct points converging to some some $x_0\in(a,b)$, then $f(x)=g(x)$ for all $x\in(a,b)$. This becomes false if we loosen the restriction of analyticity, as can be seen by considering the functions ...


3

Daniel Fischer's comment says yes, since $\ell^1=c_0^*$. One can give an explicit projection from $(\ell^\infty)^*$ onto $\ell^1$ by saying $$P\Lambda=(\Lambda e_1,\Lambda e_2,,\dots).$$


2

It is enough to show that when one of the norms is 1, the other is bounded between two positive constants, $m$ and $M$. It is easiest to look at the case when $||(x,y)||_\infty=1$. What does that equation tell you about $x$ and $y$? You can use this to bound $||(x,y)||_3$. In terms of the shape of $||.||_3$, you really just want to sketch the curve ...


2

You proved that $T(t)x$ is a solution. So, it remains to show that the problem has only one solution. Let $u$ and $v$ be two solutions. Notice that the function $w=u-v$ satisfies $$\left\{\begin{align} &\frac{dw}{dt}=Aw, & 0\le t \le t_{e} \\ &w(0)=0. \end{align}\right.$$ Pick $s\in[0,t_e]$ and define $\phi:[0,s]\to X$ be setting ...


2

The precise statement is the following: a set $A\subseteq V$ is in the weak topology iff $$\forall a\in A \exists \epsilon\in \mathbb{R}_{>0}\exists F\subseteq V^*: F\text{ is finite and }\{y\in V\mid\forall\phi\in F:|\phi(a-y)|<\epsilon\}\subseteq A.$$ It sounds like the key point you are missing is the implicit universal quantifier in the definition ...


2

Let $f_n$ be a square function with width $1/n$ and height $\sqrt{n}$. Then $f_n\rightarrow 0$ in $L^2$. Let $g_n=f_n$...


2

HINT: Note that we have $$\frac{\partial g}{\partial \nu}=\frac{2}{\epsilon}$$ on $\Gamma_{\epsilon}$. Therefore, $$\lim_{\epsilon\to 0}\int_{\Gamma_{\epsilon}}\left(g\frac{\partial\phi}{\partial\nu}-\phi\frac{\partial g}{\partial\nu}\right)\,ds=-4\pi\phi(0)$$


2

This looks suspiciously like one of the questions in my differential equations workshops this year. Anyway, let $z=||x-y||^2.$ Then $\frac{d}{dt}z=\frac{d}{dt}||x-y||^2$ which is equal to $$ =2 \begin{pmatrix} x_1-y_1 \\ x_2-y_2 \end{pmatrix} \begin{pmatrix} -x_1+2x_2+y_1-2y_2\\ -2x_1-x_2+2y_1+y_2 \end{pmatrix} \\ ...


2

Of course every finite dimensional Banach space has a finite basis. As a consequence of Baires category theorem every infintedimensional Banach space cannot have a countable Hamel Basis, but very often a Schauder basis.There exist separable Banach spaces that don´t even have a Schauder basis so as was shown by Per Enflo. So You are right, the word ...


2

The limit always exists and this is called Gelfand's formula for the spectral radius. Any textbook on spectral theory has a proof of it, for example N. Burbaki, Theories spectrales. This one is also very good: B. Aupetit, A primer on spectral theory, Springer-Verlag, New York, 1991. For more detail about such formulas you may look at this paper, for example: ...


2

The derivative of the Dirac delta will give you some trouble...


2

The set $S$ is convex and symmetric, i.e., $-x \in S$ for all $x\in S$. If $x$ is an interior point of $S$ and $B(x,\varepsilon)\subseteq S$ (where $B(x,\varepsilon)$ is the ball around $x$ with radius $\varepsilon$) you get $B(0,\varepsilon/2)\subseteq S$ since for $\|y\| <\varepsilon/2$ you have $y= \frac 12 (-x) +\frac 12 (x+2y) \in \frac 12 S + \frac ...


1

The first statement is true both ways. Specifically, suppose $(X, ||\cdot||)$ is a normed linear space. Then the norm $||\cdot ||$ is induced by an inner product iff the parallelogram law holds in $(X,||\cdot||)$. For the second statement, this is not true. Call the condition $d(x,y)=d(x+a,y+a)$ translation invariance, and the condition $d(x,y)=d(ax,ay)$ ...


1

A local contraction is just a function that is Lipschitz around a point with Lipschitz constant less than one. Differentiability is not required.


1

To complement the other answers: in many regards distributions can be thought of as "functions of a real variable", yes. (That's why I like Gelfand-etal's name "generalized functions" better. If it were just some dual space, it would not have the same usefulness...) But, yes, some cautions are necessary (as are necessary in other more conventional parts of ...


1

The common representation that you mention is a non rigorous description that originates in the work of Physicists, before the precise mathematical theory of distributions had been developed. This theory does not treat the delta distribution as a "function", and the fact that it is nonetheless called by that name (that is, the so-called "Dirac Delta ...


1

Well the answer was contained in the question linked by rschwieb (Prove the approximate identity from the unitization), I'll write it up here. If a C* Algebra $\mathscr{U}$ does not contain a unit then approximate identities will not be approximate identities in the C* Algebra $\mathscr{\tilde U}$ with a unit adjoined. Every approximate identity in a C* ...


1


1

One classic example that can help you here is the set of functions of the form $$ f_n(x) = \sqrt{x^2 + 1/n} $$ for $n \in \Bbb N$. Note that the functions $f_n(x)$ form a Cauchy sequence, but their derivatives do not.


1

If $M$ is a vector space, then we refer $A$ as an endomorphism. There are differences when a function takes an element from one space to an element in the same space. Take for example: $f:C[0,1]_{\|\cdot\|_1}\to C[0,1]_{\|\cdot\|_2}$, where $\|\cdot\|_1 \to$ integral norm $\|\cdot\|_2 \to$ maximum norm Note that $C[0,1]$ is complete under maximum ...


1

As was pointed out to me in the comments, the easiest way to go about proving the claim is to consider the family of functionals $\Phi_y: \mathscr{C}([0,1])\rightarrow\mathbb{K}$, $g \mapsto g(y)$ where [0,1] is the index set. The pointwise definition of both addition of functions and multiplication of a function by a scalar immediately gives linearity, and ...


1

Is $e_0$ supposed to have norm 1? Notice that if we suppose that, we obtain $1=|g_i(e_0)|\leq \Vert g_i\Vert_E\Vert e_0\Vert=\Vert g_i\Vert$, hence $\Vert g_i\Vert_E =1$.


1

Let $y_0$ be some nonzero element of $V$. Take a sequence of elements of $S$, call it $x_i$. Let $e_i(x)=y_0$ if $x=x_i$ and $0$ otherwise. Prove that the $e_i$ for $i=1\dots,n$ are linearly independent and yet do not span $F(S,V)$.


1

The Spectral Mapping Theorem allows you to more easily compute the spectrum of some operators. If you know that you can write an operator $A$ as $A=f(a)$ for $f\in hol(a)$ and $a\in\mathcal{A}$, where you already know the spectrum of $a$, you can compute the spectrum of $A$, since $\sigma(A)=f(\sigma(a))$.


1

$$ \begin{align} &d_\infty[T(x_1,x_2)-T(y_1,y_2)]\\ &=d_\infty\left[\left(\frac{x_1+2x_2}5-1,\frac{x_1-2x_2}7+1\right)-\left(\frac{y_1+2y_2}5-1,\frac{y_1-2y_2}7+1\right)\right]\\ &=d_\infty\left[\left(\frac{(x_1-y_1)+2(x_2-y_2)}5,\frac{(x_1-y_1)-2(x_2-y_2)}7\right)\right]\\ ...



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