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8

Here is an outline for proving your statement. $\ell^p$ is a space of (some) functions $\mathbb N\to\mathbb R$, and so is $L^p$ (with the choices indicated by OP). (I chose to work with the reals, but this choice is irrelevant.) The counting measure gives nonzero measure to all nonempty sets, so each equivalence class in $L^p$ only consists of one ...


7

We have the following characterization of adjoint operators: Suppose that $X$ and $Y$ are normed spaces. If $T:X\rightarrow Y$ is a bounded linear operator, then $T^*$ is weak*-weak* continuous. Conversely, if $S$ is a weak*-weak* continuous linear operator from $Y^*$ to $X^*$, then there is a bounded linear operator $T:X\rightarrow Y$ such that $T^*=S$. ...


6

Hints: 1)There is no norm on on $C^{\infty}[0,1]$ which makes the derivative continuous. This is because, $||\frac{d}{dx}(e^{nx})|| \leq c||e^{nx}||$ for all values of $n$ isn't possible. 2) Show that in the above metric the derivative map is continuous.


4

In fact, any infinite-dimensional Banach space $X$ has a compact subset that does not lie in any finite-dimensional linear subspace. Namely, define a sequence $x_n$ inductively such that $x_{n+1}$ is not in the linear span of $\{x_1, \ldots, x_n\}$ but $\|x_{n+1}\| < 1/n$. Then $\{0\} \cup \{x_1, x_2, \ldots\}$ is your set.


3

Let $x$ be a non-zero element of $X$. Consider the closed linear $Y$ span of $\{T^n(x)\colon n\in \mathbb{N}\}$. Then $Y$ is a proper subspace of $X$ which is left invariant by $T$.


3

In the following, I will assume that $\mu$ is $\sigma$-finite for simplicity. One can probably drop that assumption with a bit of extra work. Also, I do not claim that my proof is one of the simpler ones. Let us define $$ \left\Vert k\right\Vert _{L^{p,q}}:=\left(\int\left(\int\left|k\left(x,y\right)\right|^{p}d\mu\left(x\right)\right)^{q/p}\, ...


3

I assume you mean $C[0,1]$ with sup-norm. Hint: Try to show that for each $n$ and arbitrary $\varepsilon>0$ there exists $g\notin E_n$ such that $\|g\|_\infty<\varepsilon$. Try to show that $f\in E_n$ and $g\notin E_{2n}$ implies $f+g\notin E_n$. Using these two facts you should be able that if $f\in E_n$ then in any ball $B(f,r)$ there is a ...


3

$$\int_0^1 \underbrace{|k(s,t)-k(s',t)|}_{\leq \sup_{r \in [0,1]} |k(s,r)-k(s',r)|} \underbrace{|f(t)|}_{\leq \sup_{r \in [0,1]} |f(r)|} \, dt \leq \sup_{r \in [0,1]} |k(s,r)-k(s',r)| \cdot \|f\|_{\infty} \cdot \int_0^1 \, dt.$$


2

Yes, because if $x$ is invertible and $\|y - x \| \le \|x^{-1}\|^{-1}$, $$ y = x (1 - x^{-1} (x-y)) $$ and $1 - x^{-1} (x - y)$ is invertible with $$ (1 - x^{-1}(x-y))^{-1} = \sum_{j=0}^\infty (x^{-1} (x-y))^j$$ Moreover $$ \|y^{-1} - x^{-1}\| \le \sum_{j=1}^\infty \|x^{-1}\|^{j+1} \|x - y\|^j = \dfrac{\|x^{-1}\|^2 \|x - y\|}{1- \|x-y\| \|x^{-1}\|}$$


2

Consider the space $X = C_b[0,\infty)$ of bounded continuous functions on $[0,\infty)$ endowed with the uniform norm $$\|f\|_{\infty} := \sup_{x \geq 0} |f(x)|.$$ Define $$T_t f(x) := f(x+t),\qquad x \geq 0, t \geq 0.$$ Obviously, $(T_t)_{t \geq 0}$ has the semigroup property (i.e. $T_t T_s = T_{t+s}$) and $\|T_t\| \leq 1$. This means that $(T_t)_{t \geq ...


2

A common way to prove that some operator $T$ is not compact is to exhibit an infinite-dimensional subspace $M$ on which $T$ has a lower bound: that is, there exists $c>0$ such that $$\|Tx\|\ge c\|x\|,\quad \forall\ x\in M \tag{1}$$ If (1) holds, then the image of unit ball under $T$ contains a ball of radius $c$ in the infinite-dimensional subspace $TM$, ...


2

It can be done fairly easily in any infinite dimensional normed space $X$. For a proof, see Lemma 1.4.22 in Robert E. Megginson's An Introduction to Banach Space Theory.


2

Let $g(x)=\lvert x\rvert$ in $[-1,1]$, and extend $g$ to be periodic in $\mathbb R$, with period $2$. Set $$ g_{k,\ell}(x)=\frac{1}{k} g(\ell x). $$ It is not hard to see that $$ g_{k,\ell}\in E_n \quad\text{iff}\quad n\ge \frac{\ell}{k}. $$ Fix now $n\in\mathbb N$. We shall show that $E_n$, which is a closed subset of $C[0,1]$, has empty interior. Let ...


2

Take a sequence $(f_n)_{n\in\Bbb N}\to f$ uniformly with $(f_n')_{n\in\Bbb N}\not\to f'$ uniformly. Try with $f_n'(x)= x^n$, $f_n(x)=\cdots$


2

The two definitions are not equivalent. Let $(P)$ be the property of being infinite-dimensional. If $F$ and $E/F$ have this property, then $E$ also has this property (it is sufficient that one of $F$ and $E/F$ has that property for $E$ to have it too). But if $E$ and $F$, resp. $E$ and $E/F$ are infinite-dimensional, then $E/F$ resp. $F$ need not be ...


2

Sure: the image of the unit ball in $\ell^2$ under the map that sends the standard basis element $e_n$ to $e_n/n$ (with $\ge 1$) is Hilbert-Schmidt, so a compact operator, etc. Indeed, any sequence going to $0$ replacing $1/n$ produces (pre-) compact image.


2

I agree that the book should have explained Corollary 2 a bit better, just as you explained. Nevertheless, we can prove Corollary 2 by contradiction (as far as I noticed, this book does not use Banach-Steinhauss/Uniform Boundedness Principle, which I will use): Suppose $M$ is an unbounded rcc (relatively countably compact). Let $(f_n)_{n=1}^\infty\subseteq ...


2

Simply use the Riesz lemma as $i(X)$ is norm-closed in $X^{**}$.


2

Let $J\colon E\hookrightarrow E^{\ast\ast}$ be the canonical embedding. For $x\in E$ and $\lambda \in E^\ast$, we have $$A^{\ast\ast}(Jx)(\lambda) = (Jx \circ A^\ast)(\lambda) = Jx(A^\ast\lambda) = (A^\ast\lambda)(x) = (\lambda\circ A)(x) = \lambda(Ax) = J(Ax)(\lambda),$$ so $A^{\ast\ast}\circ J = J\circ A$. Looking at the commutative diagram ...


1

We do not need to deal with Schauder bases; an elementary argument suffices. Let $C := \sup_i \|F_i\|$. Suppose to the contrary that $\|F_i A - A \| \not\to 0$. Passing to a subnet, we can assume there exists $\epsilon > 0$ such that $\|F_i A - A\| > \epsilon$ for all $i$. As such, by definition of the operator norm, for each $i$ there exists $x_i ...


1

Since $\phi$ is a linear isometry, $\phi(X)$ is a subspace imbedded in $Y$. Let $f\in X^{\ast}$. We want to show $\exists g\in Y^{\ast}$ such that $\phi^{\ast}(g)=f$. Given $f \in X^{\ast}$, define $h_{f}\in \phi(X)^{\ast}$ by $h_{f}(\phi(x))=f(x), x\in X$. $h_{f}$ is continuous since $\phi$ is an isometry. By Hahn-Banach extension theorem, $\exists g\in ...


1

The common classification of spectrum is the first one you gave. I'm not sure where $\sigma_{c}'(T)$ would have come from, or why it was defined that way, but it doesn't strike me as a standard or common definition, even though I can see a reason for the alternative definition. If $T$ is closed, then this alternative definition would probably be better ...


1

I am not sure what you mean by the equality $X^{***} = X$, if you mean that $X$ and $X^{***}$ are isometrically isomorphic then the answer is no, it does not imply reflexivity. There is a classical example $J$ of a Banach space constructed by R. C. James (the James space) with the property that $J$ is isometrically isomorphic to $J^{**}$, yet $J$ is not ...


1

To show $X = \bigcup_{k=1}^{\infty} kB_1$ we must check both $\subset$ and $\supset$ relations between these. $\subset$ : for every $x\in X$ there is $k$ such that $x\in kB_1$. E.g., $k$ could be any integer greater than $\|x\|$. $\supset$ : the Banach space $X$ is our Universe here; no elements from outside of it enter the proof. The unit ball ...


1

See if you can extend your operator to some $L^2$ space containing your functions. Maybe your operator is even self adjoint and still compact. Use the result for compact self adjoint operators on Hilbert spaces to find a basis of eigenvectors in that $L^2$ space. Show that in fact your eigenfunctions are already in your original space. Show that any ...


1

I think you are thinking about it backwards. Every Banach space is a vector space. Every vector space has a basis. If your eigenfunctions are: (1) linearly independent under the inner product (2) have the same dimension as the space. They must be an equivalent basis. However, neither (1) nor (2) are necessary conditions of eigenfunctions as far as I know. ...


1

Let $A : V \to \mathrm{C}[0,1]$ the operator $Au=u'$. In order to show that $A$ is bounded, it suffices, using Closed Graph Theorem, to show that, if $$ u_n\to u \,\,\&\,\, u_n'\to v, $$ with $\{u_n\}_{n\in\mathbb N}\subset V$, then $u\in \mathrm{C}^1[0,1]$ and $u'=v$. Since $V$ is closed subspace, then $u\in V$, hence $u$ is continuously ...


1

Compact operators on Banach spaces can in general be upper-triangularized---generalizing the Jordan canonical form. This means that the space is Schauder-spanned by the generalized eigenvectors. So what you would need to show is that, given any non-zero $\lambda \in \sigma(L)$, the finite dimensional subspace $\mbox{ker}(L - \lambda)$ is spanned by ...


1

It must mean "for the transpose of a compact linear operator $A\colon E\to E$". We can find an easy counterexample to the stronger claim: Let $E = c_0(\mathbb{N}) = \{x\in \ell^\infty(\mathbb{N}) : x_n \to 0\}$. We know that then $E^\ast$ is isometrically isomorphic to $\ell^1(\mathbb{N})$, so we identify the two spaces. If we take an operator $B$ on ...


1

If $E_\infty$ contains unbounded functions (without any continuity or similar requirements on the functions, that means $\Omega$ is infinite), then the topology of uniform convergence is not a vector space topology on $E_\infty$. One sees that then the balls around $0$, $B_\varepsilon(0)$, are not absorbing, hence the scalar multiplication is not continuous. ...



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