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8

If such a function exists, then you may add any function whose integral between $0$ and $1$ is zero. Now you should find one particular solution. For this, a constant (with respect to $x$) function suffices.


6

There is no such operator. Note that $(C ([0,1]),\|\cdot\|_\infty ) $ is a Banach space, so that it suffices to show that $T $ has closed graph. Thus, assume $f_n \to f $ and $T f_n \to g $ (both with respect to $\|\cdot\|_\infty $). Then $f_n \to f $ and $T f_n \to g $ bith with respect to $\|\cdot\|_2$ (why?) and hence $T f_n \to Tf $ also with respect ...


4

You could not find a proof because it is not true. In fact $BL(X)$ is separable in the topology of uniform convergence if and only if $X$ is totally bounded. If $X$ is totally bounded, every Lipschitz function can be extended to its completion $\tilde{X}$, which is compact. Thus we obtain an isometric embedding $BL(X) \to C(\tilde{X})$. If $X$ is not ...


4

Triangle inequality $$|\, ||f_n ||_2 -|| f ||_2\, |\leq ||f_n-f||_2$$


4

since $\langle T(x+y),x+y\rangle =0$ that implies : \begin{eqnarray} \langle T(x+y),x+y\rangle &=&\langle Tx+Ty,x+y\rangle \\ &=&\langle Tx,x+y\rangle+\langle Ty,x+y\rangle\\ &=& \langle Tx,x\rangle+\langle Tx,y\rangle+\langle Ty,x\rangle+\langle Ty,y\rangle\\ \end{eqnarray} Then $$ \langle T x,y\rangle +\langle Ty,x\rangle=0 \qquad ...


3

Since $\|f - f_n \|_p \to 0$, we can extract a subsequence $f_{n_j}$ so that $\| f - f_{n_j} \|_p \le \frac{1}{2^j}$. Put $$g = \lvert f \rvert + \sum^\infty_{j=1} \lvert f - f_{n_j} \rvert.$$ Then $g \in L^p$ and by the triangle inequality we have $$\lvert f_{n_j} \rvert \le g \,\,\, (\text{almost everywhere}).$$ From this subsequence, you can extract a ...


3

If your example $\mathcal{B}$ were a real Banach algebra instead of a complex Banach algebra, then you would be right that there are four connected components, since $\mathcal{B}$ can be identified with $\mathbb{R}^2$ and the invertible elements split into four quadrants. But over $\mathbb{C}$, you have the complement in $\mathbb{C}^2$ of two (complex) one-...


3

A counterexample for $d=2$: let $\Omega$ be the disk $\{x:\|x\|<\exp(-\exp(\pi))\}$, and $$ f(x) = \sin \log \log \frac{1}{\|x\|} $$ This function is in $W_0^{1,2}(\Omega)\cap L^\infty(\Omega)$ (relevant calculations here) but has a discontinuity at $0$, and moreover cannot be made continuous by redefining it on a set of measure zero. On the other ...


2

Here is a simpler proof: $$ \sum_{n=1}^\infty| \langle x, e_n \rangle \langle y, e_n \rangle | \le \sum_{n=1}^\infty| \langle x, e_n \rangle|\cdot | \langle y, e_n \rangle | \\ \le \left(\sum_{n=1}^\infty| \langle x, e_n \rangle|^2\right)^{1/2}\left(\sum_{n=1}^\infty | \langle y, e_n \rangle |^2\right)^{1/2} \le \|x\| \cdot \|y\|. $$ First inequality is ...


2

According to Daniel Fischer: The left hand side cries for an application of the Cauchy-Schwarz inequality. And according to siminore: $$ x=\sum_n x_n e_n = \sum_n \langle x,e_n \rangle e_n \quad ; \quad y=\sum_n y_n e_n = \sum_n \langle y,e_n \rangle e_n $$ But we give it a twist: $$ x'=\sum_n |x_n| e_n = \sum_n \left|\langle x,e_n \rangle\right| e_n \quad ; ...


2

Hah! This is actually a specific example of something in my research! (My work attacks a more general set of integral equations, in some sense.) Let's go for something nontrivial (unlike previous answers/comments). If you consider what I like to call a diagonal kernel, i.e. $g(x,t) = f(xt)$ for some $f$ and assume $g$ is real analytic, then this is very ...


2

An easy solution can be obtained by making $g(x,t)$ degenerate: $$ g(x,t)={1\over2}\exp(-bx^4)\exp(at^4-|t|) $$ which is susceptible to the generalization $$ g(x,t)={1\over C}\exp(-bx^4)\exp(at^4)f(t) $$ where $f$ is integrable over $\mathbb R$ and $$ \int_{-\infty}^{\infty}f(t)dt=C\neq0 $$


2

As @AlexanderFrei pointed out, it should read $$\forall x: \lim_{n \to \infty} T_n(x) = T(x) \iff \forall K \subseteq X \, \text{compact}: \lim_{n \to \infty} \sup_{x \in K} \|T_n(x)-T(x)\| = 0.$$ The implication "$\Rightarrow$" is trivial, just choose $K= \{x\}$ for fixed $x \in X$. It remains to prove "$\Leftarrow$". Suppose that $T_n(x) \to T(x)$ for ...


2

Since $T - \lambda I$ is also normal, we have $$ \| T - \lambda I \| = \text{spr} (T - \lambda I) = 0, $$ showing that $T = \lambda I$. (I recently asked basically the same question (Self-adjoint operator with single point spectrum), but your formulation is more general so I thought it might be worth sharing the answer here.)


2

Any $*$-homomorphism between C$^*$-algebras is contractive. This is standard (i.e., it appears in every book on the subject) and is due to three things: The C $^*$-identity $\|a\|^2=\|a^*a\|$, which reduces the problem to norms of positives; The equality $\|a\|=\text {spr}\, (a) $ for $a $ positive; The fact that a $*$-homomorphism reduces the spectral ...


2

(Huge) hint: Is it complete? Why or why not? (Even bigger hint below.) Follow-up:


2

No for $p=1$, yes for $1<p<\infty$. If $\phi\in \Delta C^\infty_c$ then $\int\phi=0$; this shows that $\Delta C^\infty_c$ is not dense in $L^1$. One might think at first that this shows the same thing for other $p$, but it doesn't, because the integral is not a bounded linear functional. Suppose from now on that $1<p<\infty$. Suppose that $K\...


2

There are a few subtle points. Typically, the domain for $L$ would consist of all absolutely continuous $f \in L^2$ such that $f' \in L^2$. Absolute continuity is required in order to integrate by parts, for example. This operator $L$ is closed, and that's useful to know in cases like this. And $\mathcal{D}(L)$ is dense in $L^2$. The adjoint $L^*$ consists ...


2

I don't know about "geometrical view", but the condition essentially follows from assuming that multiplication is continuous. Note the "essentially"; we'll see below what I mean by that. Say we have an algebra with a norm, and multiplication is (jointly) continuous. Continuity at $(0,0)$ shows that there exists $\delta>0$ so that $$||xy||\le1\quad(||x||,|...


1

You cannot apply Banach-Alaoglu for $L^1(\Omega)$, since $L^1(\Omega)$ is not the dual space of a normed space. Rather you have to embed $L^1(\Omega)$ into larger spaces $L^\infty(\Omega)^*$ or $C(\bar\Omega)^*$ to obtain a weak-star convergent subsequence. To see that for $p=1$ the assertion is not true, consider the sequence $f_n(x)=n \chi_{0,1/n}(x)$ on $...


1

Continuing where you left off, you have two things to show: Does the new sequence $f$ belong to $X$? For this, note that $(f_n)$ is Cauchy and hence norm-bounded. Thus, $\exists R>0$ such that $$ \sum_{k=0}^{\infty} (k+1)|f_n(k)|^2 \leq R \quad\forall n\in \mathbb{N} $$ For each $m\in \mathbb{N}$, this implies $$ \sum_{k=0}^m (k+1)|f_n(k)|^2 \leq R $$ ...


1

In general, consider the polynomial $(x-r_1)(x-r_2)\cdots(x-r_n)$. Expanding, we have $$x^n - (r_1+r_2+\ldots+r_n)x^{n-1} + \ldots$$ So the sum of the roots appears as the negative of the $x^{n-1}$ term. Note that this applies to monic polynomials (i.e., the coefficient of $x^n$ is $1$); otherwise, first factor out the coefficient of the $x^n$ term.


1

A quadratic is of the form $x^2 - (\alpha + \beta)x + \alpha \beta$ where $\alpha$ and $\beta$ are its roots. This can be seen by expanding $(x-\alpha)(x-\beta)$. So the sum of your roots is given by the negative of the coefficient of $x$, i.e: $3m-1$.


1

(You don't say how you got the second equality; since it is not trivial, I'm not sure how you did it and so it is done below) Since $A^*A$ is positive and compact, it is orthogonally diagonalizable (spectral theorem): $A^*A=U^*D^2U$ for some unitary $U$ and $D$ diagonal with diagonal $s_1(A),s_2(A),\ldots$ Assume $s_1(A)\geq s_2(A)\geq \cdots$ Since $U$ ...


1

I recently wrote up a solution to this very problem, which I've copied below. Note that in this context, $\mathfrak A$ is the space of bounded operators on $B$. I hope you find this helpful. $ \newcommand{\f}{\mathfrak} \DeclareMathOperator{\rad}{r} \newcommand{\eps}{\varepsilon} \DeclareMathOperator{\spec}{spec} \newcommand{\lp}{\left(} \newcommand{\rp}{\...


1

Note that the two equalities $A=AA^*A$ and $A^*=A^*AA^*$ are the same, since you can obtain one from the other by taking adjoints. Assume first that $A=AA^*A$. By multiplying by $A^*$ on the left, we get $$ A^*A=(A^*A)^2.$$ It follows that the eigenvalues of $A^*A$ all satisfy the equation $\lambda=\lambda^2$, so only $0$ and $1$ are possible. Conversely,...


1

No, it's not a coincidence. The definition of "$\lim (T-T_k)=0$" is "$\lim||T-T_k||=0$". No, you can't do the same for any operator. You can't show that $||T-T_k||\to0$. (Let $Tx=x$. What is $||T-T_k||$?)


1

using spectral theorem since $T$ is normal it exist a spectral measure $E$ such that $$ Tx=\int_{\sigma(T)}tdE(x)=\int_{\{\lambda\}}tdE(x)=\lambda E(\{\lambda\})(x)=\lambda E(\sigma(T))(x)=\lambda I (x)=\lambda x $$


1

Define $x(t) = w(t) e^{i \zeta t}$. We want to show that $x \in C$, i.e, $\lim_{|t| \rightarrow \infty} x(t) = 0$ according to (2), or, equivalently, $|x(t)| \rightarrow 0$ as $|t| \rightarrow \infty$. Note that $e^{i \zeta t} = e^{i (Re \zeta + i Im \zeta) t} = e^{-(Im \zeta) t} e^{i (Re \zeta) t}$, and note also that the text requires $|Im \zeta| < c$, ...


1

Well, $$||x||^{m^2}=(||x||^m)^m\le(K||x^m||)^m=K^m||x^m||^m\le K^mK||(x^m)^m||=K^{m+1}||x^{m^2}||.$$Now your argument for $n\le m$ also works for $n\le m^2$...



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