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7

The problem is that $L^p$, for $p \in (0,1)$ is not locally convex. Your professor showed that its dual is zero, hence it wouldn't make sense to talk about weak topology as it is defined using elements in the dual space. An easy way to convince yourself such spaces are not locally convex is to define the $L^p$ norm in $\mathbb R^2$ and draw the unit ...


5

One need not to know the whole dual space of $C[0,1]$ to see the lack of reflexivity. Simply note that $C[0,1]^*$, opposed to $C[0,1]$, is non-separable because for each $t\in [0,1]$ the map $$\langle f, \delta_t\rangle = f(t)\quad (f\in C[0,1])$$ is a norm-one linear functional on $C[0,1]$ and $\|\delta_t - \delta_s\| =2$ for distinct $s,t$. To see this, ...


5

or is there a simple sequence of functions which respect my problem condition? There is a simple such sequence. Since we only look at $x < 1$ in the condition $\int_0^x f_n \to \int_0^x f$, if we let the support of $f_n$ shrink towards $1$, we will have $\int_0^x f_n = 0$ for all large enough $n$. So let's look for example at $$g_n(x) = ...


5

You are right, the partial derivatives of distributions on higher-dimensional spaces are defined as you surmised, $$\frac{\partial T}{\partial x_i} \colon \varphi \mapsto -T\biggl[\frac{\partial \varphi}{\partial x_i}\biggr],$$ more generally $$D^{\alpha} T \colon \varphi \mapsto (-1)^{\lvert\alpha\rvert} T[D^{\alpha}\varphi]$$ for higher derivatives. ...


4

Not necessarily. You might have $(W, V)=0$ and $(PW, V)\ne 0$ for example. Consider in $\mathbb{R}^2$ the vectors $W=(-1, 1), V=(1, 1)$ and the projector $$P(x, y)=(x, 0).$$


4

Consider the space $H = L^2([0,1],\mu)$, where $\mu$ is the Lebesgue measure. Define $T$ to be the multiplication with the identity function, i.e. $$(Tf)(x) = x\cdot f(x).$$ Since the identity function is bounded, $T$ is bounded ($\lVert T\rVert \leqslant 1$), and since it is real-valued, $T$ is self-adjoint. Clearly $T$ has no eigenvalues, since $$(T - ...


4

Note: I completely rewrote my previous attempt - posted in another (now deleted) answer, since the previous version was incorrect. Thanks to Asaf Karagila for pointing out the problem with my previous proof. And also to Eric Wofsey for simplifying some steps in the proof. Let us hope that this time I have avoided mistakes. The formulation of the version of ...


4

A possible proof, using a different argument, is this: Assume $T$ is not bounded. Then for all $k\in\mathbb{N}$ there is $z_k\in X$ with $\Vert z_k\Vert_X=1$ and $\Vert Tz_k\Vert_Y>k^2$. Now, consider the sequence $(x_k)_k$ s.t. for all $k$, $ x_k=\frac{z_k}{k^2}$. Then: $\Vert Tx_k\Vert_Y>1$ for all $k$. The series $\sum_{k=1}^\infty x_k$ is ...


4

It is incorrect. Let $S$ the left shift operator. Then $TX=(x_2-x_1,x_3-x_2,x_4-x_3,...)=(x_2,x_3,...)-(x_1,x_2,..)=SX-IX=(S-I)X$. Thus $T=S-I$. Since $S$ and $I$ are bounded operators, and the space of bounded operators are a vector space, then $T$ is bounded. Now, if $X_n=(0,0,..,0,1,0,0,0)$, where the $1$ is in the $n$th-place, $n\ge 1$, then ...


4

Yes. Ignoring questions of measurability: Suppose $f:[0,1]\to X$ and $\int_0^1||f(t)||_X^2\,dt<\infty$. Say $X_n$ is an increasing sequence of finite-dimensional subspaces of $X$ so that $\bigcup X_n$ is dense in $X$. For each $n$ choose $f_n:[0,1]\to X_n$ such that, say, $$||f_n(t)-f(t)||_X\le 2 d(f(t),X_n)$$for all $t\in [0,1]$. Note that ...


4

Almost always one studies closed operators. In fact, you rarely can do much of anything with an operator that is not closable. Closed operators that are everywhere defined on a Banach space are continuous by the closed graph theorem. So the best you can hope for is that the operator is densely-defined; that leaves you having to specify the precise domain, ...


4

The integral diverges. To see this, we can write $$\int_0^n \left(1-\frac{3x}n\right)^ne^{x/2}\,dx=\int_0^{n/3} \left(1-\frac{3x}n\right)^ne^{x/2}\,dx+\int_{n/3}^n \left(1-\frac{3x}n\right)^ne^{x/2}\,dx \tag 1$$ We will present two parts. In Part $1$, we will show that the first integral on the right-hand side of $(1)$ converges. In Part $2$, we will ...


4

The sixth volume is "Representation Theory and Automorphic Functions" by Gelfand, Graev, and Pyatetskii-Shapiro. It does not say in the title of the English translation that this is the sixth volume of the series by Gelfand and coauthors, but this is indicated inside the book and the original Russian version is clearer on this point. If you can read Russian, ...


3

You can just take the coefficients of the polynomials in $E$ to have $E$-valued polynomials. Trying to mimic the Stone-Weierstraß proof for $E$-valued functions may however be tricky. But from uniform continuity of continuous functions on $[0,1]$ you immediately get that you can uniformly approximate all functions in $C([0,1],E)$ by piecewise affine ...


3

Seems like the answer is yes. Take $f\in C([0,1],E)$. First extend $f$ to a function in $C(\Bbb R, E)$, say by making $f$ constant on $[1,\infty)$ and constant on $(-\infty,0]$. Now say $\phi_n$ is a smooth (real-valued) approximate identity; then the convolution $f*\phi_n$ should be differentiable and it should be that $f*\phi_n\to f$ uniformly on $[0,1]$. ...


3

The corrected version is $$ u(x)=-\frac{e^{-x}}{2}\int_{-\infty}^{x}f(t)e^{t}dt-\frac{e^{x}}{2}\int_{x}^{\infty}f(t)e^{-t}dt. $$ If $f$ is compactly supported in $\mathbb{R}$, then $u(x) = Ce^{x}$ for large negative $x$, where $C=-\int_{-\infty}^{\infty}f(t)e^{-t}dt$; and $u(x)=De^{-x}$ for large positive $x$, where ...


3

Let $T=\frac{1}{i}\frac{d}{dt}$ be defined on the domain $\mathcal{D}(T)$ consisting of all absolutely continuous functions $f \in L^2[0,1]$ for which $f(0)=0=f(1)$. More precisely, $f \in \mathcal{D}(T)\subset L^2[0,1]$ is an equivalence class of functions equal a.e. with one element $\tilde{f}$ of the equivalence class that is absolutely continuous on ...


3

It isn't true. The standard counterexample is to look at $L^2((0,1))$ with Lebesgue measure and take $f_n(x) = \sqrt{2} \sin(n \pi x)$. The functions $f_n$ are orthonormal in $L^2$, so by Bessel's inequality they converge weakly to 0. But pointwise, the sequence $\{f_n(x)\}$ diverges for every $x \in (0,1)$.


3

Define $\Lambda:E^*\to\Bbb C$ by $\Lambda f = \lim f(x_n)$. Then $\Lambda$ is certainly linear. You say you've shown $||x_n||$ is bounded; that implies that $\Lambda$ is bounded. So $\Lambda $ is a bounded linear functional on $E^*$, which is to say $\Lambda\in E^{**}$. But $E$ is reflexive, which says that $E^{**}=E$. Except of course that's not a literal ...


3

Let $f_n(x)=\left(1-\frac{3x}{n}\right)^ne^{x/2}$. Then $$\lim_{n\to\infty }f_n(x)\lambda_{[0,n]}(x)=\lim_{n\to\infty }f_n(x)\cdot \underbrace{\lim_{n\to\infty }\lambda_{[0,n]}(x)}_{=\lambda_{[0,\infty [}(x)}=\lambda_{[0,\infty [}(x)\lim_{n\to\infty }f_n(x).$$ Therefore $$\int \lim_{n\to\infty }f_n(x)\lambda_{[0,n]}(x)\mathrm d x=\int \lambda_{[0,\infty ...


3

Hint Define a sequence of polynomials by $$ P_n(x)= \frac{1}{\log(\log(n))}\sum_{k=0}^n \frac 1{k+1}x^k $$


3

You need to put the absolute value sign within the double integral. It is well possible for the integral of $f-g$ to be zero without $f$ being equal to $g,$ even for the nicest imaginable continuous functions with compact support. You also need to account for functions that are almost everywhere equal - in fact the usual procedure is to define a ...


3

We will prove that $\exists x_0\in S(0,1)=\{x\in H:\|x\|=1\}:\|Tx\|=\|T\|$. You know that $\|Tx\|\leq \|T\|\,\forall x\in B(0,1)\Rightarrow \exists$ a sequence $\{x_n\}\subset B(0,1): \|Tx_n\|\to \sup\limits_{x\in B(0,1)} \|Tx\|=:\|T\|$. From this sequence you can chose a weakly convergent subsequence, still denoted by $\{x_n\}$, in $B(0,1)$ , say ...


3

Yes, this is standard. Let $T\in A$ selfadjoint, i.e. with $T=T^*$. Note first that $\phi(T)$ is real: since $T+\|T\|\,\text{id}$ is positive, we have that $$ \phi(T)+\|T||\in\mathbb R, $$ so $\phi(T)\in \mathbb R$. Now, as $-T+\|T\|\,\text{id}\geq0$, we get $-\phi(T)+\|T\|\geq0$, so $$\phi(T)\leq\|T\|.$$ Since $-T$ is also selfadjoint, we can also get ...


3

The unit ball of $C^0([0,1])$ has only two extreme points (the constant functions with value $\pm 1$) but the one of $\mathbb R \oplus C^0([0,1])$ with the norm $|t|+ \|f\|_\infty$ has (at least) four. Therefore, the spaces are not isometric.


3

If $T$ is not onto, this is not true. Consider the embedding $T \colon \def\R{\mathbf R}\R \cong \R \times \{0\} \to \mathbf R^2$, and the subset $A := \{x \in \mathbf R^2: x_1^2 + x_2^2 = 1, x_2 \ge 0\}$, the upper half of $S^1$. Then $A$ is contractible, but $T^{-1}[A] = \{(1,0), (-1,0)\}$ is not. On the other hand, if $T$ is onto, this is true: Choose a ...


3

Note that $f$ is bounded if and only if it is continuous. So let $f$ be unbounded. This means for any $n \in \mathbb N$ we have an $x_n \in X$ so that $f(x_n) ≥ n \|x_n\|$. By rescaling set $\|x_n\|=1$. Now let $z$ be in $X$. From the construction of the $x_n$ it follows that $z_n:= z - \frac{f(z)}{f(x_n)}x_n$ is a sequence that converges to $z$. But ...


3

Depending on the domain of your function, either no such function exists or it does exist but is not unique. If the domain is $\mathbb C$, then note that both $\lim_{n\to \infty}\frac{1}{\alpha n}=\lim_{n \to \infty}\frac{1}{\alpha n +1}=0$. So, since holomorphic functions are continuous, $f(0)=\lim_{n\to \infty} f(\frac{1}{\alpha n})=\lim_{n \to ...


3

The norm on the function space is given by the sup norm $\|f\|:=\sup\{|f(x)|x\in S\}$. Convergence of functions in sup-norm is called uniform convergence. It is a general statement that a uniform limit of continuous functions is again continuous. To see that let $x_n \to x$, let $f_m \in C_b(S)$ and $f_m \to f \in C(s)$. Consider $\epsilon>0$. Since ...


3

Assume $X$ is a Banach space and $A$ is a bounded linear operator on $X$. $\lambda$ is in the point spectrum iff $\mathcal{N}(A-\lambda I) \ne \{0\}$. $\lambda$ is in the continuous spectrum iff $\mathcal{N}(A-\lambda I)=\{0\}$ and $\overline{\mathcal{R}(A-\lambda I)}=X$. Everything else is the residual spectrum. You want $\lambda$ to be in the residual ...



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