# Tag Info

9

By the Cauchy Schwarz Inequality, for any integrable function $f(x)$: $\displaystyle\left(\int_a^b f(x) \cdot f(x)^2\,dx\right)^2 \le \left(\int_a^b f(x)^2\,dx\right) \left(\int_a^b (f(x)^2)^2\,dx\right)$ $\displaystyle\left(\int_a^b f(x)^3\,dx\right)^2 \le \left(\int_a^b f(x)^2\,dx\right) \left(\int_a^b f(x)^4\,dx\right)$ But by the given conditions, we ...

6

Not only is $f$ constant, that constant is either $0$ or $1$. \begin{align} \int_a^b\left[f(x)^2-f(x)\right]^2\,\mathrm{d}x &=\int_a^b\left[f(x)^4-2f(x)^3+f(x)^2\right]\,\mathrm{d}x\\ &=0 \end{align} Thus, $(f(x)-1)f(x)=0$ for almost all $x\in[a,b]$. Since $f$ is continuous, we have either $f(x)=0$ for $x\in[a,b]$ or $f(x)=1$ for $x\in[a,b]$.

6

These are some reasons that I can see at the moment (In the case I recall something else it shall be added to this list): 1) In fact any signal in reality is a function in $L^2(I)$ where $I$ is a time interval, since its energy or power is finite, i.e. $$\int_I |x(t)|^2 {\rm d}t < \infty$$ 2) The Fourier series intrinsically means that any periodic ...

6

It's a matter of convention. Indeed, when you say $f(x) = 1/x$, you've not really specified $f$ (it might, for example, only be defined on the domain $x > 4.7$), but by convention, we treat the domain as "as much of the reals as possible" and infer that it's therefore all of $\mathbb R$. (Slightly amended) When you say that $f \circ g (x) = x$, you've ...

5

Hint. Let $x = a-b$, and use $\langle y,y\rangle = 0 \iff y = 0$.

5

The matrix $$A= \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$ is a counter-example: $$\|A\|_2^2 = \|A^TA\|_2^2 = \|A^T\|_2\|A\|_2=1,$$ but $$\|A^2\|_2=0$$

5

The statement is equivalent to the Continuum hypothesis. Indeed, taking a quotient cannot increase the density character, and the density character of $l_\infty$ is $c$. This gives one implication. If $\aleph_1=c$, then pick a dense subset of cardinality $\aleph_1$ in the unit ball of $l_\infty$. Map the standard basis vectors of $l_1(\aleph_1)$ ...

5

No. Nonlinear transformations and weak convergence go together like drinking and driving. For example, let $r_k$ be the $k$th Rademacher function on $[0,1]$, that is $r_k = \operatorname{sign}\sin ( 2^k \pi x)$. Then $2^p r_k \rightharpoonup 2^{p-1}\mathbf {1}$ in $L^1$, where $\mathbf{1}$ is the constant function equal to $1$. On the other hand, ...

5

A concrete example, inspired by comments: define the metric $$d(z,w) = \begin{cases}|z-w|, \quad &\text{ if } |z|,|w| \le 1 \text{ or } z=w \\ |z-w| + 1 & \text{ otherwise} \end{cases}$$ The triangle inequality is easy to check. The multiplication $z\mapsto 2z$ is not continuous because, e.g., $1-1/n \to 1$ but $2-2/n\not\to 2$. Similar for ...

4

The quickest way is to note that $$\sup_{(x,y)\in\mathbb{R}^2\setminus \{(0,0)\}} \frac{(ax+by)^2}{x^2+y^2} = a^2 + b^2$$ is just the Cauchy-Schwarz inequality for the standard inner product on $\mathbb{R}^2$. Without using that, one can observe that $f(x,y) = \frac{(ax+by)^2}{x^2+y^2}$ is homogeneous of degree $0$, i.e. $f(tx,ty) = f(x,y)$ for all $t ... 4 Works with$C=4$. Can be improved a bit if someone wants to. Let$S=\sum_{n \geq 1} a_n$. Case 1: There exists$N$such that$a_N\ge S/4$. Then the right hand side of the above inequality is at least $$\Big(2^N (S/4)^2\Big)^{1/4}\Big(2^{-N} (S/4)^2\Big)^{1/4} = \frac{S}{4}$$ Case 2: There is no$N$as above. Then let$N$be the smallest integer such ... 4 I included a screenshot below. The concept of$C(\overline{\Omega})$is unambigious, as user161825 pointed out. If a continuous extension to the closure exists, it is unique and we may consider the function as already extended. Item 2) is a mess-up on the author's part. Just assume$\Omega$is a bounded open set, because it will be practically everywhere ... 4 Edit: Rewrote the answer for more clarity. Assume that$A$is symmetric and coercive: $$\langle Au,v\rangle = \langle Av,u\rangle\quad \forall u,v\in H^1_0(\Omega),$$ and $$\langle Au,u\rangle \ge \delta \|u\|^2 \quad \forall u\in H^1_0(\Omega)$$ with some$\delta>0$. We want to prove the following: Let$u^*\in H_0^1(\Omega)$. Then these two points ... 4 Let us try to estimate$|B|$from below. Every space$B_{n-1}^*$($n\geqslant 1)$is of the form$C(K_n)$for some compact Hausdorff space. For example,$K_1 = \beta\mathbb{N}$and$|K_1| = \beth_1$. In particular, by the Riesz–Markov–Kakutani representation theorem each space$B_n$is isometric to the space$M(K_n)$of Radon measures on$K_n$. Moreover, we ... 4 For every$x \in X $we have$|g(x)| \leq \sup_{z \in X} |g(z)|$by definition of the supremum, so for every$x \in X$we may observe that $$|f(x)g(x)| = |f(x)||g(x)|\leq |f(x)|\left(\sup_{z \in X}|g(z)|\right) =|f(x)|\|g\|,$$ Since this is true for every$x\in X$we may take the supremum on both sides of the equation to get $$\|fg\| = \sup_{x \in ... 4 This is known as Dini's theorem. Since f_n-f is also continuous, we may assume f_n\searrow 0. Given \varepsilon >0, consider the sets O_n=f_n^{-1}(-\infty,\varepsilon). Since f_n is continuous, each one is open. Prove that (\rm i) E=\bigcup O_n (\rm ii) O_n\subseteq O_{n+1} Since E is compact, you will find N such that E\subseteq ... 4 For U \subset V, we have a natural (continuous) injection$$\iota^U_V \colon \mathscr{D}(U) \hookrightarrow \mathscr{D}(V).$$Its transpose,$$\rho^V_U \colon \mathscr{D}'(V) \to \mathscr{D}'(U)$$is called the restriction of the distributions on V to distributions on U. For regular distributions, that corresponds to the restriction of the locally ... 4 C(K) will not generally be compact. However, the Arzela-Ascoli theorem, which characterizes some compact subsets, applies. Of course, C(K) is Hausdorff if its topology is taken relative to the uniform metric \|\cdot\|_\infty, which is often the case. In general, C(K) is taken to be a metric space of some kind, and all metric spaces are Hausdorff. ... 4 Take X=c_{00}---the space of all sequences which are almost everywhere 0 and as x_n---the sequence having \frac{1}{2^n} on n-th place and 0 elsewhere. 4 No, this cannot be done. If \phi\in C_c^\infty(R), then its Fourier transform extends to be an entire analytic function (cf. Paley--Wiener theorem). Should it be constant for (real) x in some neighborhood of x_0, it would be identically constant by the uniqueness theorem for analytic functions, and hence zero, because the Fourier transform of a smooth ... 3 The norm on the direct sum A \oplus B of two C*-algebras is the maximum norm$$\lVert (a,b) \rVert = \max\{\lVert a\rVert, \lVert b\rVert\},$$not the \ell_1-norm that you write: an example that helps me remember this is to consider A=C(K), B = C(L) and to note that A \oplus B should be isomorphic to C(K \sqcup L). Since Murphy identifies ... 3 Yes, it is possible. For example, Serge Lang exhibits the basics of a theory of differential forms on Banach manifolds in Chapter V of his Differential and Riemannian Manifolds. (Non-)Separability is not an issue. According to Lang, a p-form on a Banach space E is simply a continuous alternating p-linear map on E. This yields a notion of p-forms ... 3 By Jensen's inequality \int |f|^2 \log |f|=\int |f|^2 \cdot \frac{1}{p-2}\log |f|^{p-2} = \frac{1}{p-2}\cdot\int |f|^2\log |f|^{p-2} \leq \frac{1}{p-2}\log (\int |f|^{p-2}\cdot |f|^2) = \frac{1}{p-2}\cdot \frac{p}{2}\log (\int|f|^p)^\frac{2}{p}=\frac{1}{p-2}\cdot \frac{p}{2} \log ||f||_p^2 because \frac{1}{p-2}=\frac{n-2}{4}, ... 3 Yes, this is true. Let A = \{u \le 0\}, which is a measurable set of finite measure, so 1_A is a nonnegative function in L^2(\Omega). Therefore \int u_n 1_A \ge 0 and hence \int u 1_A = \lim \int u_n 1_A \ge 0. As u is nonpositive on A, it must be that u = 0 almost everywhere on A, which is to say u \ge 0 almost everywhere. The same ... 3 This follows from$$ \|f\| = \sup_{\|x\| = 1} |f(x)|. $$If you like, take a sequence x_k such that \|x_k\|=1 and |f(x_k)| \to \|f\|. Then set y_k = x_k / |f(x_k)|, so |f(y_k)| = 1, and \|y_k\| \to \frac{1}{\|f\|}. 3 A bounded linear operator A on a separable Hilbert space X with orthonormal basis \{ e_{j} \}_{j=1}^{\infty} is a Hilbert-Schmidt operator if$$ \sum_{j=1}^{\infty}\|Ae_{j}\|^{2} < \infty. $$This condition is true for one orthonormal basis iff it is true for every other orthonormal basis. If you're studying X=L^{2}[a,b], then an ... 3 "Locally" is ambiguous here f is locally Lipschitz in \Omega if and only if f \in W^{1,\infty}_{loc}(\Omega) The validity of this claim depends on interpretation of "locally Lipschitz". Does it mean every point of \Omega has a neighborhood in which f is Lipschitz, or there is L such that every point of \Omega has a neighborhood in ... 3 You don't want to show that Y is not closed, you want to show that Y is not compact. That is, you don't want a sequence (f_n) of members of Y that converges to a limit f such that f \notin Y - that would prove that Y is not closed, but Y is closed (why?). You want a sequence (f_n) that doesn't have any convergent subsequence and f_n(x) = ... 3 No chain of balls is needed in the case |x-y|\ge R. Just use the estimate$$ \frac{|u(x)-u(y)|}{|x-y|^\alpha}\le 2 R^{-\alpha} \sup_{\Omega'}|u|$$The appearance of$R^{-\alpha} $is not a problem, since it only depends on$\alpha$and on the distance of$\Omega'$to$\partial \Omega$. And$\sup_{\Omega'}|u|$is controlled by$\|u\|_{L^2(\Omega)}\$ by ...

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