New answers tagged

2

Let: $B: C[a,b] \times C[a,b] \to \Bbb R$ be the (check that it is) bilinear form given by: $$B(x,y) = \int_a^b x(t)y(t) dt$$ We have that: $$|B(x,y)| \le \int_a^b |x(t)||y(t)| dt \le (b-a)\|x\|_{\infty} \|y\|_{\infty} $$ Hence, $B$ is continuous and in particular $\|B\| \le b-a$. Now, $f(x) = B(x,x)$ is the composition of two continuous functions, ...


0

Without the requirement $\|B\|=1$ the exact value of the infimum is known: $$ \inf \{\|A-B\| \colon \operatorname{rank} B = p\} = \sigma_{p+1} $$ where $\sigma_1\ge \dots \ge \sigma_n$ are the singular values of $A$, in nonincreasing order. This is the min-max principle for singular values. If you require $\|B\|=1$, the singular value $\sigma_{p+1}$ still ...


1

Yes, it does. Indeed, let $g_n(x) : = \inf_{y\in B_{1/n}(x)}f(y)$. Then it is easy to see that $g_n(x)\le g_{n+1}(x)$ and hence $$g(x) = \sup_{n \in \mathbb{N}} g_n(x) = \lim_{n \to \infty} g_n(x) = h(x).$$


2

If $X = \{p\}$ then $X$ is connected. If $X \neq \{p\}$ then $X$ is not connected. This follows from the fact that in a metric space singletons are closed, together with the fact that in a connected space the only sets that are both open and closed are the empty set and the set itself.


0

None of the cross terms survive in the following: \begin{align} \frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)\overline{g(t)}dt & = \frac{1}{2\pi}\int_{-\pi}^{\pi}\sum_{n=\infty}^{\infty}A_n e^{int}\sum_{n=-\infty}^{\infty}\overline{a_n}e^{-int}dt \\ & =\sum_{n=-\infty}^{\infty}A_n\overline{a_n}\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{int}e^{-int}dt \\ ...


2

The Spectral Theorem for $A$ is given in terms of a Borel Spectral measure $E$ $$ Ax = \int_{-\infty}^{\infty}\lambda dE(\lambda)x, $$ and $x \in \mathcal{D}(A)$ iff $$ \int_{-\infty}^{\infty}\lambda^2 d\|E(\lambda)x\|^2 < \infty. $$ The operator $e^{iA^2}$ is defined through the functional calculus as $$ e^{iA^2}x = ...


0

How about a bounded quasinilpotent operator $T \in \mathscr{L}(L^2[0,1])$ defined by $$ (Tf)(x) = \int_{0}^{x}f(t)dt. $$ This operator has $\sigma(T)=\{0\}$. The resolvent cannot have a pole at $0$ because $T$ is not nilpotent of any order. So the resolvent has an essential singularity. You can solve for the resolvent by solving ...


0

A way to interpret the delta function in this context is through an integral. Start with the example of the Fourier transform composed with its inverse. $$ f = \frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(t)e^{-ist}dt\right)e^{isx}ds. $$ This may be correctly written as $$ ...


1

Sure, you can do it this way. If $y \neq 0$ is killed by all of the $f_i$, then if $x$ is in a basic weak neighborhood corresponding to the $f_i$, then $x+ty$ is also in this neighborhood for all $t$, since $f_i(x) = f_i(x+ty)$. In particular, the weak neighborhood will be unbounded. Ultimately, I think the point is that the kernel of a linear functional ...


2

This is in fact never true; suppose you have a matrix $A$ of rank $k$. Then, we can take linearly independent $v_1, \dots, v_k \in \text{Ran}(A)$, so there exist $u_1, \dots, u_k$ such that $Au_i = v_i$ for each $i$. Suppose for a contradiction the statement is true for some $p < k$, so there is a sequence $(B_n)$ of $n \times n$ norm 1 matrices of rank ...


1

Orthornormality refers to the basis $e_i$. When a basis is orthonormal it means the inner product between any two elements of the basis $e_i,e_j$ is $\langle e_i, e_j \rangle = \delta_{ij}$ (see kronecker delta). More generally, two vectors $u,v$ are orthogonal if $\langle u, v \rangle = 0$. The normality part comes from elements of the basis having norm ...


2

The indices are different because the sums are independent from one another. If one intends to multiply the sums, then this distinction is a critical one. Suppose that we were to multiply the summations $\sum_{n=1}^2a_n$ and $\sum_{n=1}^2b_n$ and naively failed to make this distinction. Then, we would have incorrectly $$\begin{align} ...


1

That's because it is notationally bad and depending on how you read it, it may even give different results. For example, consider simple vectors in $\mathbb{R}^{3}$ given by $v=a_{1}e_{1}+a_{2}e_{2}+a_{3}e_{3}=\sum_{n=1}^{3}a_{n}e_{n}$. Then if we use same index, \begin{equation} \begin{aligned} ...


0

No. The determinant is a continuous map on $\mathbb{R}^{n\times n}$. If $A$ is a matrix with nonzero determinant (full rank), then any matrix $B$ sufficiently close to $A$ will also have nonzero determinant (and thus will also be full rank), by continuity.


0

You have already basically proved what you want when you wrote $|T_n(x)| \le \left(\sum_{k=1}^{n}|a_k|^2\right)^{1/2}\|x\|$. All you have to note is that $\{ a_k \} \in \ell^2$, which gives $|T_n(x)| \le \|a\|_2\|x\|_2$ for all $n$ and fixed $x$. That's enough to apply the uniform boundedness principle. The uniform boundedness principle is the simplest way ...


3

An alternate approach is induction on $n=\dim(W)$. The base case $n=0$ is clear, so the hard part is the induction step. For this, it's enough to prove the following result: if $M$ is a closed subspace of $V$ and $x\in V,x\not\in M$, then $M+\mathbb{C}x$ is also a closed subspace. Indeed, by the Hahn-Banach theorem there is a continuous linear functional ...


0

This is not true. Let $A=C[0,1]$. Consider the increasing sequence of closed sets closed sets $F_0=\varnothing$, $F_i=[0,2^{-i}]$, and set $I_i=\left\{f\in A:f|_{F_i}=0\right\}$. Then we are in the setting you described. We can identify $A/I_i$ with $C(F_i)$, via the map $A/I_i\to C(F_i)$, $f+I_i\mapsto f|_{F_i}$. With this $A/I_{i+1}\to A/I_i$ is ...


1

HINT (I don't know if it works): Take $\phi\in \mathcal S'$ (a continuous linear functional) with the property that $$\tag{1}\langle \phi, G\rangle = 0\qquad \forall G(x)=Ae^{\frac{|x-x_0|^2}{2}}, $$ where $x_0\in\mathbb{R}^n, A\in\mathbb{R}$ are arbitrary. Claim (to be proved): $\phi=0$. Once this is proved, the exercise is done by standard duality ...


4

I’d attack it much more directly. HINT: Suppose that $a=\langle a_n:n\in\Bbb Z^+\rangle\notin\ell_\infty$. Then $a$ has a subsequence $\langle a_{n_k}:k\in\Bbb Z^+\rangle$ such that $|a_{n_k}|\ge k$ for each $k\in\Bbb Z^+$. For each $k\in\Bbb Z^+$ let $$x_{n_k}=\frac1{ka_{n_k}}\;,$$ and let all the other terms of $x$ be $0$. Show that $x\in\ell_1$, ...


4

Notice that for each vector $x$, one has $$\|T^* T^2(x)\|^2 = \langle T^* T^2(x),T^* T^2(x) \rangle = \langle TT^*T^2(x), T^2(x)\rangle = \langle T^*T^3(x),T^2(x) \rangle = \langle T^3(x),T^3(x) \rangle = \|T^3(x)\|^2.$$ Thus $$\|T^3\| = \sup_{\|x\|=1} \|T^3(x)\| = \sup_{\|x\|=1}\|T^*T^2(x)\| = \|T^*T^2\|.$$


0

In polar coordinates, since the spherical surface form is $R\sin \theta d\theta d\phi $, in order for the integration to be one, the dirac delta should be: $1/(R\sin \theta_0) \delta(\theta-\theta_0,\phi -\phi_0)$


1

Diagonalization will help here. (When in doubt and working with normal matrices, try utilizing diagonalization!) Write $T = UDU^*$, then $T^* = UD^* U^*$, giving that $T^*T^2 = UD^*D^2U^*$. However $T^3 = UD^3 U^*$. Since unitary conjugation does not change the operator norm, this boils down to considering $D^*D^2$ and $D^3$. Here $Dg(x) = f(x)g(x)$ for ...


2

Yes, it is correct. The operators $T_N$ are of finite rank (hence compact) and converge to $T$. Thus, $T$ is compact as well.


3

This is studied in potential theory: the function $u$ is the Newtonian potential of $f$, $$u(x)=\int_{\mathbb{R}^n} K(x-y)f(y)\,dy$$ where $K(x)=c_n|x|^{2-n}$ for $n\ne 2$ and $K(x)=c_2\log|x|$ for $n=2$. In dimensions $n\ge 3$ the kernel $K$ decays at infinity, so $u(x)\to 0$ as $|x|\to\infty$ in this case, provided $f$ is reasonable (integrable and ...


1

Apart from a few typos, your derivation looks correct. There should be $e^{-\alpha^2 t}$ in the final equation and I get $x + \tan(x) = 0$ (which is $\cot(x) = -\frac{1}{x}$), as the defining equation for the $\alpha_n$'s. This follows from $$X(x) = \sin(\alpha x) \implies 0 = X(1) + X'(1) = \sin(\alpha) + \alpha \cos(\alpha) \implies \tan(\alpha) + \alpha ...


1

Let $C=\max\{\|L\|_{\infty}^2,1\}$. Assuming that $\alpha<\frac12$, we may compute \begin{align} \int_0^1\int_0^1|k(x,y)|^2\ \mathsf dy\ \mathsf dx &= \int_0^1\int_0^1 L(x,y)^2|x-y|^{-2\alpha}\ \mathsf dy\ \mathsf dx\\\\ &\leqslant C \int_0^1\int_0^1 |x-y|^{-2\alpha}\ \mathsf dy\ \mathsf dx\\ &= 2C\int_0^1\int_0^x (x-y)^{-2\alpha}\ \mathsf ...


0

always with the same problem, i want to calculate the solution with separate variable methode. That what i try: we put $$u(x,t)= X(x) T(t) \neq 0$$ then we have $$\dfrac{T'(t)}{T(t)}= \dfrac{X"(x)}{X(x)}$$ So we obtain twe equations $$ T'(t) + \lambda T(t)= 0 $$ and $$ X''(x)+ \lambda X(x)=0, X(0)=0, X(1)+ X'(1)=0 $$ this second equation admits eigenvalues ...


0

Your question seems to deal with convergence of sequences of functions. In your case, you have the sequence $(f_n)_{n\geq 0}$ where $f_n(x)=|x|^n+(-1)^n$ and you seem to be interested in the function this sequence converges to and in what sense (uniform, pointwise). Hence, the literature is convergence of function sequences.


1

Take any nontrivial, nonnegative, symmetric function $g \in C_c^\infty (\Bbb{R}^d)$. If we let $h := \mathcal{F}^{-1}(g)$, then $h$ is real-valued (why?) and $$ h(0) = \int g(x) \, dx > 0, $$ since $g \geq 0$ and $g \not \equiv 0$. By continuity of $h$ and by rescaling (i.e., replace $g$ by $Cg$ for some large $C>1$), we get $h \geq 1$ on $B_{2\delta} ...


0

You have the following typos: the display after "yields" should read $$ \|f_n(x)-f(x)\|<\epsilon $$ because you've taken the limit for $m\to\infty$ with $n$ fixed and you've used the continuity of the norm in $X$. You're missing a $\|$ bracket before $f(x)\|$ elsewhere. Now using the inequality in 1. and the triangle inequality (in the form ...


2

Define $\phi:(B+I)/I\to B/B\cap I$ by $$\phi(b+j+I)=b+B\cap I,\ \ \ \ \ b\in B,\ j\in I.$$ Of course we need to check that this is well-defined. If $b_1+j_1=b_2+j_2$, then $$ b_1-b_2=j_2-j_1\in B\cap I, $$ so $b_1+B\cap I=b_2+B\cap I$. The map is obviously linear, multiplicative, $*$-preserving, and onto. As for injectivity, if $b_1+B\cap I=b_2+B\cap I$, ...


1

Take a positive, nonzero $u \in C^{\infty}_c(\mathbb{R}^d)$ and consider $\{u_{\epsilon}\}$, where $u_{\epsilon}(x) = u(x\epsilon^{-1})$. For simplicity, let $s = 1$ and notice that (by a change of variables) $$\|u_{\epsilon}\|_{L^{2^*}} = \Big(\int |u_{\epsilon}|^{2^*}\Big)^{\frac{1}{2^*}} = \epsilon^{\frac{d}{2^*}}\|u\|_{L^{2^*}}$$ $$\|\nabla u\|_{L^2} = ...


2

Given $f$ and $\epsilon$, choose a polynomial $p$ with $\Vert f-p\Vert_{\infty,X}<\epsilon$ (where $\Vert\cdot\Vert_{\infty,X}$ is the supremum norm oin $X$). Now see the corresponding polynomial function in $\mathcal{A}$, $p:\mathcal{A}\to\mathcal{A}$. (Remember: the functional calculus respects this notation, i.e., $p(a)$, in the functional calculus, is ...


1

If you have $y \in H^1(\Omega) \cap C(\bar\Omega)$ you can use a truncation argument to obtain $x \in H^1(\Omega) \cap C_c(\bar\Omega)$ such that $\|x-y\|_{H^1}$ is arbitrarily small.


2

Defining $\phi_\lambda(x)=\phi(\lambda x)$ for smooth $\phi$, the requirement is $$ u(\phi_{\lambda})=\lambda^{-m-d}u(\phi)\quad\forall\phi\in C_0^{\infty}. $$ If $u$ happens to be a continuous function (and hence $u(\phi)=\int u(x)\phi(x)$), this is equivalent to what you wrote.


0

Note: There is nothing about completeness of $\mathcal{H}$ needed to carry out of the following steps. Because $P_n$ is monotone, then $(P_nx,x)$ is monotone in $n$ for each fixed $x$, and is bounded above by $(x,x)$, which forces convergence of $\lim_n(P_n x,x)$ for all $x$. Then, using polarization, the following expression must also have a limit in $n$ ...


0

I haven't worked out the details - also I suspect we still haven't been given all the relevant definitions. But in case it helps, here's how the argument "must" go in outline, if it's by R-N: Somehow we reduce to the case $\mu(\Omega)<\infty$. Define a complex measure $\nu$ by $$\nu(E)=x^*(\chi_E).$$Detail: Something shows somehow that $\nu$ is in fact ...


0

As suggested by daw, take $u(x) = 1$. If you have some regularity of $\Omega$, then $$\int_{\mathbb R^n} u \nabla v \, \mathrm{d}x = \int_{\partial\Omega} \frac\partial{\partial n} v \, \mathrm{d}s$$ for $v \in C_c^\infty(\mathbb R^n)$. The right hand side cannot be written as $\int_{\mathbb{R}^n} w \, v \, \mathrm{d}x$ for some $w \in ...


2

You want to show: $\int_a^b \frac{1}{L} e^{-2 \pi inx/L}e^{2\pi imx/L}dx=0$ \begin{align} <a_n|a_m> & = \int_a^b \frac{1}{L} e^{-2 \pi inx/L}e^{2\pi imx/L}dx \\ & = \frac{1}{L} \int_a^b e^{2 \pi i x(m-n)/L} dx\\ & = \frac{1}{L} \frac{L}{2 \pi i(m-n)} \left( e^{2 \pi i (m-n) b/L}-e^{2 \pi i(m-n)a/L}\right) \\ & = \frac{1}{2 \pi ...


2

Example, $$ A = \frac{1}{i}\frac{d}{dx} $$ on the domain $\mathcal{D}(A)$ of absolutely continuous functions $f \in L^2[0,1]$ for which $f' \in L^2[0,1]$ and $f(0)=0$. Then $A^*$ is the same as $A$ except that the condition $f(0)=0$ is replaced by $f(1)=0$. Then $A^{\star\star}=A$ because $A$ is closed and densely-defined. However, ...


1

Setting $v=u_1-u_2$, define $I(t)=\frac{1}{2}\int_0^1v(x,t)^2dx$. Then $$\frac{dI}{dt}=\int_0^1v(x,t)v_t(x,t)dx=\int_0^1vv_{xx}dx$$ Integrating by parts, this becomes: $$\frac{dI}{dt}=\left[vv_x\right]_0^1-\int_0^1v_x^2dx$$ Now, note: $vv_x\vert_{x=0}=0,vv_x\vert_{x=1}=-v(1,t)^2$ by the conditions given, so: ...


1

If $\Omega$ is bounded, it suffices to take $u(x)\equiv 1$. Then $\hat u$ cannot be in $W^{1,p}(\mathbb R^n)$ for $p>n$ since it is discontinuous. (Sobolev embedding)


0

Consider the two cases of $V_n$ being monotone increasing and monotone decreasing in $n$. In the first case you have $P_n P_{n+1}=P_n$ and in the second you have $P_n P_{n+1}=P_{n+1}$. At any rate you have either $P_n(P_{n+1}- P_{n})=0$ or $P_{n+1}(P_{n+1}-P_{n})=0$. First consider $V_n$ is increasing. So $(P_{n+1}-P_n)(z)$ is in $V_n^\perp$. This means ...


1

Assume $V_n \subseteq V_{n+1}$ for all $n$. Then $P_{n+1}P_n=P_n$. Using adjoint, $P_nP_{n+1}=P_n$ must also hold because $P_k^*=P_k$ for all $k$. Therefore, for all $x$, $$ (P_nx,x) = (P_nx,P_nx)=\|P_nx\|^2=\|P_nP_{n+1}x\|^2 \le \|P_{n+1}x\|^2=(P_{n+1}x,x). $$


0

In every metric space $(X,d)$ a sequence $x_n$ converges to $x$ if and only if every subsequence $x_{n_k}$ has a further subsequence converging to $x$ (easy proof by contradiction). All you need to now is thus that convergence in measure is convergence in a metric space (e.g. $d(f,g)=\int \min\lbrace 1,|f(x)-g(x)|\rbrace \, d\mu(x)$ is a suitable metric on ...


2

A linear operator is continuous if and only if it is bounded. By one of the comments above, it is possible to show that the sequence $a_k$ must be bounded (for otherwise, $Tx$ for $x\in\ell^1$ would not itself be an element of $\ell^1$). Therefore, $|a_k|\leq C$ for some $C\geq 0$. Next, recall that a linear operator $T:\ell^1\rightarrow\ell^1$ is ...


2

By the monotone convergence theorem, you always will have $$\int f^p = \lim_{k \rightarrow \infty} \int f_k^p$$ Taking $p$th roots (and using continuity of the function $x \rightarrow x^{1 \over p}$), one therefore has $$||f||_p = \lim_{k \rightarrow \infty} ||f_k||_p \tag 1$$ Since the $f_k$ increase to $f$, the limit in $(1)$ is an increasing limit. So one ...


-1

Oh, I just read your updated version of the question. If fk->f in Lp, we know that f is in Lp. Considering fk is a non-negative increasing sequence of functions, to converge at f it must be that sup(k) ||fk||p=||f||p. f is in Lp, so ||f||p is finite.


-1

Well, fk is a sequence of increasing functions to f, so with sufficient work you should be able to show that sup(k) ||fk||p = ||f||p.


5

You need to show the three vectors are linearly independent. In this case I would use this trick; so that you don't need to worry about them being functions and the equality to hold for every value of $x$. If you consider $D: \mathcal{F} \rightarrow \mathcal{F}$, the derivative operator, is an endomorphism in $\mathcal F$ (i.e. a linear map from ...



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