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7

Assuming "polynomial" means "polynomial in $z$", i.e. of the form $\sum_{k=0}^n a_k z^k$, then no, they are not dense, and the function $\bar{z}$ is not in their uniform closure. The high-level reason is that polynomials are holomorphic on the unit disk, holomorphic functions are closed under uniform convergence, and $\bar{z}$ is not holomorphic. For a ...


6

1. We first talk a bit about the underlying method of the proof. We want look for $(I + \epsilon T)^{-1}$ as an infinite series, similar to the way we would expand $1/(1+x)$ into a power series for $|x| < 1$ over the real or complex field. We then show that our infinite series of operators converges. Along the way, we need to use the fact that $\|TS\| ...


5

This is an inner product space and accordingly it satisfies the parallelogram law. Hence you cannot show that it is not a Hilbert space be showing it fails to satisfy the parallelogram law. The reason it is not a Hilbert space is that it is not complete: there are some Cauchy sequences in it that fail to converge. For example, consider the sequence whose ...


4

Let $V$ be an $n$-dimensional vector space over the field $\mathbb{F}$. Given a linear map $T : V \to V$, there is an induced linear map $\bigwedge^nT : \bigwedge^n V \to \bigwedge^n V$ given by $\left(\bigwedge^nT\right)(v_1\wedge\dots\wedge v_n) = (Tv_1)\wedge\dots\wedge(Tv_n)$. As $\bigwedge^nV$ is one-dimensional, $\bigwedge^nT = ...


4

In terms of difficulty, it really depends on the level of instruction. Complex and functional analysis at an introductory graduate level are roughly comparable, and it really comes down to the student's own preferences and the quality of the instructor. But complex is often taught at an undergraduate level as well, while functional is typically not. As to ...


4

Let us assume that $V$ is complete (otherwise, this is probably false). Note that finite dimensional normed vector spaces are always complete. Also, let us assume that $T$ is not an arbitrary linear transformation, but a bounded linear transformation, which means that $$ \Vert T \Vert := \sup_{\Vert x \Vert \leq 1} \Vert Tx \Vert < \infty. \qquad ...


3

If $T$ is invertible, there is some $x$ such that $bx = Tx = I$, where $I$ is the identity matrix (why?). Hence?


3

It is always defined as $||f||=sup\{\frac{||f(x)||}{||x||},x\in X$, x is non-zero$\}$, and this can be transformed to the form in your question.


3

Let $q$ be a polynomial such that $\|f'-q\|_{\infty} \leq \varepsilon/2$ and set $$p(x) := f(0) + \int_0^x q(y) \, dy.$$ Obviously, $p$ is a polynomial and $p' =q$. Moreover, by the fundamental theorem of calculus $$\begin{align*} |f(x)-p(x)| &= \left| \left(f(0)+ \int_0^x f'(y) \, dy \right)- \left( f(0)+ \int_0^x q(y) \, dy \right) \right| \\ ...


2

Note that by linearity $$\eqalign{ \frac{h(\theta+\delta)-h(\theta)}{\delta}&=\frac{1}{\delta}\left(\xi(R(\theta+\delta))- \xi(R(\theta))\right)\cr &=\frac{\xi(R(\theta+\delta)-R(\theta))}{\delta}\cr &=\xi\left(\frac{R(\theta+\delta)-R(\theta)}{\delta}\right)\cr} $$ Taking the limit as $\delta\to0$, we get $$h'(\theta)=\xi(R'(\theta))\tag{$1'$}$$ ...


2

You don't have to. For linear functionals you can show that the above definition is equivalent to $$ \|f\|=\sup_{x\not=0}{\|f(x)\|_Y\over \|x\|_X}, $$ (where subscripts denote the norm from the appropriate space) as follows (using the absolute homogeneity property of norms and basic linearity for $f$): \begin{align} {\|f(x)\|_Y\over ...


2

The setup For completeness I want to state the parts from the book that are needed: We are given a (in general nonlinear) operator $$A: W^{1,p}(\Omega) \rightarrow W^{1,p}(\Omega)^*,$$ which is coercive (see Lemma 2.35 in the given literature). At that, coercivity means the existence of a mapping $\xi: \mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that $\xi$ ...


2

Let $e_n$ be the element of $\ell_\infty$ whose $m$'th coordinate is $1$ if $m=n$ and $0$ otherwise. The closed linear span of $\{e_n\mid n\in \Bbb N\}$ in $\ell_\infty$ is the space $c_0$ of sequences that tend to $0$. For your operator, we have $T(je_j)=e_j$; so the range of $T$ contains each $e_j$. Since the range of a linear operator is a linear space, ...


2

Suppose $T$ has inverse $T^{-1}$. Then $T$ is an injective map between finite dimensional vector spaces. The rank-nullity theorem then implies the map is surjective, so PhoemueX's argument may be applied.


2

There is no Schwartz space, just $C^\infty(\mathbb T^n)$. Indeed, $\mathcal S(\mathbb R^n)$ is an intermediate class between all smooth functions and compactly supported smooth functions. But on $\mathbb T^n$, everything is compactly supported. Since the Fourier transform of a function on $\mathbb T^n$ is a function on $\mathbb Z^n$, it does not make ...


2

For the forwards direction, since $A$ is connected any two points $a, b \in A$ can be connected with a continuous curve $C$ in $A$ with end points, $a, b$. For all $x \in C$ there exists an open ball $$B_{\epsilon_x}(x) = \{y \in E\text{ }|\text{ }\|x - y\| < \epsilon_x\} \subseteq A$$ for some $\epsilon_x > 0$ as $A$ is open. Clearly $$\bigcup_{x \in ...


2

This is not a solution but just a suggestion. Consider the complete lattice of all topologies on $X$, where meet is intersection of topologies and join is “take the topology generated by the union”. Observe that: Condition (1) selects the lattice interval $I$ between $\tau_w$ and $\tau_n$. Conditions (1) and (2) together select a (complete) sublattice ...


2

Basically if $f$ satisfy the mixed derivatives condition, i.e.: $$\frac{d^2f}{dx dy} = \frac{d^2f}{dy dx}$$ And similarly for $x, z$ and $y, z$, then yes in your situation (i.e. smooth function and "contractible" domain). This is due to the Poincare Lemma.


2

For your example: $\dfrac{\partial f}{\partial x}=5x \implies f=\frac{5}{2}x^2+g(y,z)$ $\dfrac{\partial f}{\partial y}=\dfrac{\partial g}{\partial y} = 2y \implies g=y^2+h(z)$ $\dfrac{\partial f}{\partial z}=\dfrac{\partial g}{\partial z}=\dfrac{\partial h}{\partial z} = 8 \implies h=8z+c$ Then by substituting together, $f(x,y,z)=\frac{5}{2}x^2+y^2+8z+c$ ...


2

It may be helpful to notice that if function $f$ is lipschitz and $f(0)=0$ then $f\circ u$ is sobolev with $\partial_i (f\circ u)(x)=f'(u)\partial_i u$. Hence, you could use $$f(x):= \begin{cases} x & |x|<k\\ k & x>k\\ -k & x<-k \end{cases} $$ as your truncation function. Of course $f$ satisfies all conditions I wrote above and of ...


2

The second statement is essentially proven few pages later, p. 487 in your link, since in the case of functions defined on the real line Gâteaux derivative coincides with Fréchet. The first one follows from the estimate (which I assume is the same you used): $$ ...


2

If you ignore the inner product, $L^2$ space is just $L^p$ space with $p=2$. The inner product gives $L^2$ additional structure: it is not only a Banach space but also a Hilbert space. The subset of continuous functions in $L^p$ (for $1 \leq p < \infty$) does not form a Banach space because it is not complete. However, it is a dense subset of $L^p$, ...


2

Let $D=(x_n)$ dense in $X$ and let $\sigma(X^*,D)$ the locally convex topology in $X^*$ with countable local basis at $0\in X^*$ given by: $$U_n=\{x^*\in X^*: \sup_{1\leq j\leq n}|x^*(x_j)|< 1/n \}$$ Then $(X^*,\sigma(X^*,D))$ is a Hausdorff TVS. In fact, if $x^*\neq 0$, as $x^*$ is continuous, there is $j_0$ such that $x^*(x_{j_0})\neq 0$, so we have ...


2

The space $C_0$ is a Banach space. This implies that absolutely convergent sequences are convergent, i.e. if $$\sum_n \Vert \frac{p_n}{3^n} \Vert_\infty$$ is finite, then $\sum_n \frac{p_n}{3^n} \in C_0$. But projections have norm at most one, which means $$\sum_n \Vert \frac{p_n}{3^n} \Vert \leq \sum_n 3^{-n} = \frac{1}{1-1/3} < \infty.$$ This ...


1

Write $F(x) = \int_a^b f(y)1_{[a,x]}(y)\, dy$ and use Minkowski's inequality to obtain the estimate $$\|F\|_2 \le \int_a^b |f(y)| (b - y)^{1/2}\, dy$$ Finally, proceed with Cauchy-Schwarz to deduce $$\|F\|_2 \le \frac{b - a}{\sqrt{2}}\|f\|_2$$


1

I don't immediately see if your approach leads to a solution (in particular I think you need "normed" for this to be true). This is the way I would do it: assume that $\|\phi\|\leq c$ for all $\phi\in K$. Let $D=(x_n)$ be dense in the unit ball of $X$. Now define $$ d(\phi,\psi)=\sum_{k=1}^\infty2^{-k}|\phi(x_k)-\psi(x_k)|. $$ It is straighforward to check ...


1

The graph $G(A)$ of $A$ is a closed linear subspace of $\mathcal H ^2$. Any closed linear subspace of that is the graph of a closed restriction of $A$. So you want to take any closed linear subspace $S$ of $G(A)$ such that $P(S)$ is dense in $\mathcal H$, where $P(u,v) = u$ is the "first coordinate" projection of $\mathcal H^2$ on $\mathcal H$.


1

These bump functions, or test functions, are extremely important in distribution theory. They can be constructed using partitions of unity. I didn't find any good references online in this context, but if you can get hold of Hörmanders ``The Analysis of Linear Partial Differential Operators I'', then it's an excellent reference.


1

The second is the correct one.


1

The second one is the correct one. By the way, the first one is not a norm. (Take for instance $f$ constant equal to $1$ on $[-\pi,\pi]$, you don't have $|| 2 f || = 2 ||f||$.)



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