8

No, there are counterexamples. For instance, there exists a space $X$ with a basis, $X^*$ separable, and yet $X^*$ fails approximation property. See Lindenstrauss-Tzafriri's book Theorem 1.e.7 for a discussion of this (it is an existence proof).


6

M. Zippin showed that for a Banach space $X$ with a basis, if every basis of $X$ is boundedly complete or if every basis of $X$ is shrinking, then $X$ is reflexive. The result of Zippin answers you question in the negative (or, rather, the answer is, "not necessarily".) However to complete the picture let us recall the earlier result of R.C. James (alluded ...


4

One can transfer the continuous $L^p$ theory to this discrete setting without difficulty. Let's normalise $\sum_k |a_k|^p = \sum_n |b_n|^q = 1$. Consider the two quantities $$ X_1 := \sum_{k \neq n} \frac{a_k \overline{b_n}}{\lambda_k - \lambda_n}$$ $$ X_2 := \sum_{k, n} p.v. \int_{{\bf R}^2} \varphi(s) \varphi(t) \frac{a_k \overline{b_n}}{(\lambda_k+s) -...


4

Expand your resonance in spherical harmonics: $\psi = \sum_{\ell=0}^\infty \sum_m \psi_{\ell m}(r) Y_{\ell m}(\theta,\phi)$. Then each coefficient satisfies the radial Schrödinger equation $$ -\frac{1}{r^2} \frac{\partial}{\partial r} r^2 \frac{\partial \psi_{\ell m}}{\partial r} + \frac{\ell(\ell+1)}{r^2}\psi_{\ell m} + V(r) \psi_{\ell m} = 0 . $$ Since $...


4

Great question! What you need is Sandy Grabiner's approximation lemma: Lemma: Let $E$ and $F$ be Banach spaces, let $T \in B(E,F)$, and let $M > 0$ and $0 < r < 1$. Suppose that for each $y \in [F]_1$ there exists $x_0 \in [E]_M$ with $\|y - Tx_0\| \leq r$. Then for each $y \in F$ there exists $x \in E$ with $\|x\| \leq \frac{M\|y\|}{1-r}$ and $Tx =...


3

Most antiderivative of $g$ are not in $L^1(\mathbb{R})$, in your case only one antiderivative $G_0$ will be in $L^1(\mathbb{R})$, the one actually equal to $f$. All the other antiderivatives $G$ are equal to $G_0 + c$, with $c \neq 0$, which is not in $L^1(\mathbb{R})$. To proof that there exist one antiderivative $G_0$ in $L^1(\mathbb{R})$. You start by ...


2

You can find an overview of methods to obtain conservation laws from a wave equation in On the structure of conservation laws of (3+1)-dimensional wave equation. Noether's method requires that the PDE follows from a variational principle for a Lagrangian (as pointed out by Willie Wong). A direct algorithmic method to obtain conservation laws from a PDE ...


1

I realise I was too optimistic in my comment: no such ineqaulity holds in general. Indeed, consider $f(x) = \prod_j f_j(x_j)$. Then $$ K_{(\alpha_1,\ldots,\alpha_d)} f(x) = \prod_j K_{\alpha_j} f_j(x_j) $$ and so $$ \|f\|_p = \prod_j \|f_j\|_p , \qquad \|K_{(\alpha_1,\ldots,\alpha_d)} f\|_q = \prod_j \|K_{\alpha_j} f_j\|_q $$ So by the usual Hardy–Littlewood—...


1

Yes the theorem is true in the general case. The semigroup is assumed to be contractive for the sake of simplicity. Of course the assumption is not restrictive since there is always an equivalent norm which makes the semigroup contractive (this is the same idea in the proof of Hille-Yosida theorem to pass from the uniformly bounded case to the contractive ...


1

In the same idea a paper "SHARP NORM INEQUALITIES FOR THE TRUNCATED HILBERT TRANSFORM" by Enrico Laeng .


1

The answer is: Weyl, H. 1919, Annalen der Physik, 365, 481 doi: 10.1002/andp.19193652104 Though it's difficult to find in there if you don't understand German


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