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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 ...


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

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

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 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

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, ...


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

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

This follows from Ptolemy's inequality, which states that for any quadrilateral $ABCD$, we have $$\overline{AB}\cdot \overline{CD}+\overline{BC}\cdot \overline{DA} \ge \overline{AC}\cdot \overline{BD}.$$ Identifying $A,B,C,D$ with $0,z,x,y$, this gives $$\left\|z-0\right\|\left\|y-x\right\|+\left\|x-z\right\|\left\|0-y\right\|\ge ...


3

I'm not in this area but I can say you that the main issue here is the application of this kind of Mathemathics to Quantum Mechanics. Indeed, even if Hilbert didn't started studying the argument with this in mind it was soon finded that this branch of mathemathics was really suitable to modelize Quantum phenomenas. Indeed what happened is that soon after ...


2

I'm just going to apologize in advance. If there is any lesson to be learned here, it's that clearly stating your goals and cast of characters, as well as good notation both help immensely. SCENE I Enter Vector Space $P$, the space of polynomials with real coefficients and variable $t$, with his servant the Mapmaker. Vector Space $P$: Mapmaker! I ...


2

Consider first the functions $$h_1(x)=\frac{|\pi-x|}{2},\,\,h_2(x)=-2\ln\left|\sin\frac{x}{2}\right|.$$ It is shown here that the Fourier series of $h_1$ on $(-\pi,\pi)$ is $\sum_{j=1}^{\infty}\frac{\sin jx}{j}$, and the Fourier series for $h_2$ on $(-\pi,\pi)$ is $\sum_{j=1}^{\infty}\frac{\cos jx}{j}$. Therefore, the Fourier series of the function $\sin ...


2

No, the last assumption does not follows from the first two one. To see this consider operators $T_n f = f\left(x^n \right).$


2

It does not have a particular name. It is the general form of a linear partial differential operator of order $n$.


2

I recently saw it asserted that the second dual of $C[0,1]$ was $\mathcal L^\infty[0,1]$, the space of bounded Borel functions. I was surprised at this, tried to prove it, couldn't quite make the details work. Finally saw a simple proof it was not so. (See the Amusing Note below regarding what $\mathcal L^\infty[0,1]$ actually is.) Started to post something ...


2

Since $K$ is self-adjoint, we have $\langle Ku,v\rangle=\langle u,Kv\rangle$. If $Ku=\lambda u$ and $u=v\ne0$, then $$ \lambda \langle u,u\rangle=\langle \lambda u,u\rangle=\langle u,\lambda u\rangle=\overline\lambda\langle u,u\rangle, $$ where the second identity comes from the fact that $K$ is self-adjoint. Since $\langle u,u\rangle=\|u\|^2\ne0$, we ...


2

From weierstrass approximation theorem, for any $\epsilon > 0$, there exists a polynomial $P_{\epsilon}(x)$ such that $$\int_0^1 |f(x) - P_{\epsilon}(x)| dx \le \epsilon$$ Now, from the condition of this question, $$\int_0^1f(x)P_{\epsilon}(x) dx = 0 $$ So, (using fact f is continuous on closed interval) $$\int_0^1f^2(x)dx = \int_0^1f(x)(f(x) - ...


2

Separation of variables works on regions that are rectangular in some particular coordinate system. The underlying operator must be separable in that coordinate system. For the Laplacian, that generally means an orthogonal coordinate system such as spherical coordinates, cylindrical coordinates, elliptic coordinates, etc.. A rectangular region in spherical ...


2

An alternative answer (based more on OP's knowledge): In a metric space $(X, \Vert \cdot \Vert_X)$ and $A \subset X$ we have the following assertion: $$ x \in \bar{A} \iff \exists (x_n)_{n \in \mathbb N} \subset A: x_n\to x \in X $$ This means in words: an element $x$ of a subspace $A \subset X$ is in the closure of $A$ (the smalles closed set containing ...


2

For the second part, take a bounded subset $B$ of $C[a,b]$. We can find a constant $R$ such that $|v(x)|\leqslant R$ for each $x\in [a,b]$ and each $v\in B$. To prove that $M(B)$ is equi-continuous and bounded, use the fact that $g$ is uniformly continuous (and bounded) on $[a,b]\times [-R,R]$.


2

Of course this is quite simple, as Jonas showed. Here's a fun way to look at it. Let $K=\{x_n'\}\cup\{x'\}$. Then $K$ is a compact metric space. Regard $x_n$ and $x$ as functions from $K$ to $\Bbb C$. Uniform Boundedness shows that $||x_n||$ is bounded, and this shows that our family of functions from $K$ to $\Bbb C$ is equicontinuous. And $x_n\to x$ ...


2

Let $$ X=\{f:\mathbb{R} \to \mathbb{R} \} $$ Then $X$ is a function space, as mentioned in the comments. Note that $M \subset X$. Also note that $M$ is a function space in its own right. What would it mean to call a subset of $X$ open? Well, there's a field of math called topology that studies how one would go about defining this. But if you haven't ...


2

No it is not. The dual of $C[0,1]$ is the space $\mathfrak{M}([0,1])$ of complex (or signed) regular Borel measures on $[0,1]$. The dual of $\mathfrak{M}([0,1])$ is the set $\mathcal L^\infty([0,1])$ of bounded Borel functions on $[0,1]$. For example, $\chi_{\{0\}}$, the function which vanishes everywhere, except at $x=0$, where it is equal to $1$, ...


2

To construct counterexamples we need a better understanding of "diagonally constant." Let $g(x)$ be given, then $f(x,y)=g(x-y)$ is "diagonally constant" since $g((x+a)-(y+a))=g(x-y)$. Conversely if $g(x)=f(x,0)$ with $f$ diagonally constant, then we have $f(x,y)=f(x-y,0)=g(x-y)$. If $g(x)$ is $\ell$ periodic then $g(x-y)$ is $\ell$ periodic separately in ...


2

Let $\epsilon > 0$ and $x, z\in X$, and $d(x,z) < \delta$. Then, since $A$ is closed, you know that the infimum in the function definition is actually achieved at some $y_0$ (for $x$ and $y_1$ for $z$).($d(x,y)$ is bounded below and continuous for fixed $x$ so it achieves its $\inf$ on the closed set $A$). WLOG assume $d(x,y_0) \ge d(z,y_1)$. (It is ...


2

Your convergence statement is false, because Dirichlet's test assumes monotonicity. Indeed $x_n=(-1)^n \frac{1}{n}$ is in $c_0$ but $\sum_{n=1}^\infty (-1)^n x_n = +\infty$.



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