# Tag Info

6

The proof has nothing to do with the Schwartz space per se; nor with $i$ or $t$, or that $P$ and $Q$ are symmetric. If $P,Q$ are operators on a Hilbert space $H$ with domain $D$ and such that $PD\subset D$, $QD\subset D$, and $QP-PQ=\mathbb I$, then at least one of $P$ and $Q$ is unbounded. This applies to the case in the question because if we have ...

5

It's not quite as simple as that. The union $$N = \bigcup_{j,k,m \in \Bbb{N}} N_{j,k,m}$$ is a countable union, so properties of measures hold. If we take the most straightforward implementation of your replacement scheme, we need $$N = \bigcup_{\substack{j,k\in \Bbb{N} \\ 0 < \varepsilon < \varepsilon_0}} N_{j,k,\varepsilon}$$ which is no ...

5

If the $n$ real-valued continuous functions $f_1, \ldots, f_n$ separate points of $K$, then $(f_1, \ldots, f_n)$ is a homeomorphism from $K$ to a compact subset of $\mathbb R^n$. But not every compact metric space is homeomorphic to a compact subset of $\mathbb R^n$. For example, let $K$ be the Hilbert cube. For each $k$, $K$ has a subset $S_k$ ...

4

The first thing I thought of was a "soft proof": Suppose $f \to f'$ maps $H^\infty$ to $H^\infty.$ By the closed graph theorem, this linear map is continuous. Thus there exists $C$ such that $\|f'\|_\infty \le C\|f\|_\infty$ for all $f\in H^\infty.$ The functions $z^n$ show this fails. But it's easier to just note $f(z) = \sum_{n=1}^\infty z^n/n^2$ is a ...

4

The accepted answer is incorrect. To show that $S$ is closed, you must show that for any sequence of points $(x_n)$ in $S$ which converges to a limit $x\in \mathbb{R}$, the limit $x$ is also in $S$. You can prove this for your $S$ by contradiction: suppose $x_n\to x$ but $x\not\in S$. Now use the definition of limit (with $\epsilon=-x$) to show that ...

4

The only reason the question seems silly is that you include the answer! A locally convex TVS is one that has a basis at the origin consisting of balanced absorbing convex sets. The reason for the emphasis on "convex" is that that's what distinguishes locally convex TVSs from other TVSs: every TVS has a local base consisting of balanced absorbing sets. ...

4

All you need to do is a little algebra to substitute $f_1 + f_2$ into the new definition. To apply Minkowski inequality, let's denote the traditional $L_2$ norm of $f$: $\left(\int_0^1 |f|^2 dx\right)^{1/2}$ by $\|f\|_0$. We have: \begin{align} & \|f_1 + f_2\|^2 \\ = & \int_0^1 |f_1 + f_2|^2 dx + \int_0^1 |f_1' + f_2'|^2 dx \\ = & \|f_1 + ...

4

Suppose you have a non-trivial solution of $$T^*Tf = \int_{t}^{1}\int_{0}^{s}f(y)dy ds = \lambda f(t)$$ Then $\lambda \ne 0$ because the above would give $f=0$ after differentiating a couple of times. For $\lambda \ne 0$, any solution of the above must satisfy $$\lambda f'' = -f \\ f(1)=0,\;\; f'(0)=0.$$ Any ...

3

Suppose $y \not \in S; (y \in \mathbb R)$. Then $y < 0$. Can you show that $y$ is not a limit point of $S$? Let $\{x_n\} \subset S$ be any sequence. Then all $x_i \ge 0 >y$ so $|y - x_i| = x_i - y = x_i + |y| \ge |y|$. So for any $0 < \epsilon < |y|$ then $|y-x_i| \ge \epsilon$ for all $x_i$. So $x_n \not \rightarrow y$. So $y$ can not ...

3

The requirement $\epsilon \in (0, 1/2)$ can be replaced by an weaker one of $\epsilon \in [0, 1)$. Arguing by contradiction, suppose $\{v_i\}$ are linearly dependent. Hence, there exist $a_1, a_2, \ldots, a_n$, not all of them $0$, such that$$\sum_{i = 1}^n a_iv_i = 0.$$Without loss of generality, suppose$$|a_1| = \max\{|a_1|, |a_2|, \ldots, |a_n|\}.$$We ...

3

Then the cardinality is also $\mathfrak{c}$. To see this, note that every element may be written as $\sum_{i\in I}a_i e_i$ where $a_i$ are complex numbers such that at most countably many of them are non-zero and $e_i$ are elements of a fixed orthonormal basis. Then $|H|\leqslant \mathfrak{c}\cdot \mathfrak{c} = \mathfrak{c}$. On the other hand, it is clear ...

3

If "better" means better suited for certain purposes then yes. Fourier Series, for example, allow approximation which are extremely well suited for application in physics in general, also for numerical approximation (fast Fourier transform) or for analysing partial differential equations (since they map differential operators to multiplicatin operators). It ...

3

The sequence converges in the strong topology of a Hilbert space, but not in the uniform topology. You can start with continuous functions on $[0,1]$ and find a sequence of real continuous functions that converges up to $\chi_{(1/2,1]}$, while all of the functions remain uniformly bounded by $1$.

3

No. Say $K$ is the "Hilbert cube", that is, the set of all sequences $x=(x_1,\dots)$ with $$0\le x_n\le 1/n$$for all $n$, and the metric $$d(x,y)=\left(\sum(x_n-y_n)^2\right)^{1/2}.$$The question is equivalent to asking whether there exists $N$ and a continuous injective mapping from $K$ to $\Bbb R^N$. And the answer to that is no. Let $$K_N=\{x\in ... 3 It is not true that T_n \to I. They only converge pointwise, but not in the operator norm. Here is an easier proof: Let Y be an incomplete normed space and let X be the scalar field. Then B(X,Y) is isometric to Y, so it is incomplete. 3 No, the sequence (g_n) need not be Cauchy in the L^2 topology. Consider g_1 = \begin{cases} 1, && x \in [0,1] \\ 0, && \text{otherwise}\end{cases}. Then g_n = \begin{cases} n, && x \in [0, \frac 1 n] \\ 0, && \text{otherwise}\end{cases}. One way of showing that (g_n) is Cauchy is showing that \| g_{n+p} - g_n \| ... 2 It's enough to show that$$-f_{0}(z)+p(z+u)\ge -f_{0}(y)-p(-y-u)$$for every y,z\in X_0. But$$f_0(z)-f_0(y)=f_0(z+u)+f_0(-y-u) \le p(z+u)+p(-y-u).$$2 An example of a frame operator depends of course on the frame. For a frame (f_n)_{n \in \mathbb{N}} in a Hilbert spaces \mathcal{H}, the frame operator S : \mathcal{H} \to \mathcal{H} is, as you mentioned, just defined as$$ S f = \sum_{n \in \mathbb{N}} \langle f, f_n \rangle f_n. \quad \quad (*)$$To my knowledge, this is the only form in which ... 2 You are definitely on the right track: I would consider something like$$ f = \begin{cases} \frac{1}{x^a(\log x)^b} & \text{if}\ x \in (0,0.5) \\ 1 &\ \text{if}\ x \in [0.5,1) \end{cases} $$so that you don't need to worry about integrability at 1. Now you only need to play around with a and b. A choice that works is the following: 2 There are one or two things you have to check before the following proof. Since you don't assume C to be closed, what happens if x_0\in\overline{C}. (What is a closed ball of radius 0?) This is a special case which should be dealt with separately. Edit: (A little note on balls of radius 0.) Let x\in X, and let r>0. Denote the open ball of ... 2 If T:X\to Y is a linear operator with graph G, then note that G is a vector subspace of X\oplus Y. Using the fact that X is finite dimensional try to convince yourself that G is finite dimensional and hence closed. Now, use the Closed Graph Theorem. 2 The closest element in U to (1,-1) is (1,0) in the Euclidean metric, but in the metric ||\cdot ||_\infty, all points in the set \{(x,0) \mid 0 \leq x \leq 2\} are equally close to (1,-1). Recall that ||(x,y)-(x',y')||_\infty = \max\{|x-x'|,|y-y'|\}. Can you use this definition to show (a) all elements in \{(x,0)\mid 0 \leq x \leq 2\} are ... 2 I did not check all details, but you can try this: For the Lebesgue measure \lambda on [0,\infty) an application of Fubini's theorem gives,for p>1, \|u\|_{L^p}^p=p \int\limits_0^\infty x^{p-1} \lambda(\lbrace u>x\rbrace)dx. Then approximate f(x)= \lambda(\lbrace u>x\rbrace)- \lambda(\lbrace v>x\rbrace) by ploynomials to get ... 2 In THIS ANSWER and THIS ONE, I provide primers on the Dirac Delta. For any smooth test function \phi, we have for any \nu >0, there exists a \delta>0 such that |\phi(x)-\phi(0)|<\nu whenever |x|<\delta. Then, we can write$$\begin{align} \lim_{\epsilon \to 0}\int_{-\infty}^\infty ...

2

Now that I've actually read it carefully, let me try again. He throws away those $A_k$ in order to be able to say that $f$ is bounded on $E^c$. Your proof appears correct as well. My guess is that the main reason for using those $A_k$ was to shorten the proof, because this way he didn't have to use any epsilons at all. Your $f$ may not be bounded on your ...

2

This duality holds if and only if $X^*$ has the Radon-Nikodym property with respect to the Haar measure on $\mathbb{T}$ which is equivalent to having the Radon-Nikodym property with respect to the Lebesgue measure on the unit interval. This is Theorem 1 on p. 98 in J. Diestel and J.J. Uhl, Vector measures. Mathematical Surveys, Vol. 15, Amer. Math. ...

2

Since your drawing is of low quality, see the figures below. Note that in the transformation to the log form, you forget a solution because it is not $\ln(x)$ but $\ln|x|$ (the absolute value of $x$). Do not forget the branch $\ln(-x)$ for the range $x<0$.

1

Consider this counterexample. Let $u:\mathbb{R}^2\to\mathbb{R}$ s.t. $$u(x,y)=\begin{cases} \frac{1}{\sqrt{x^2+y}} & (x,y)\in [0,1]\times[0,1]\\ 0 & \text{otherwise} \end{cases}$$ We can show this function is in $L^2(\mathbb{R}^2)$ (according to wolfram alpha, $$\int \vert u(x,y)\vert^2 dxdy=\int_0^1 \int_0^1 \frac{1}{x^2+y}dxdy \approx 2.26394 ... 1 What you have shown is that for any v\in \mathcal{C}, there is a sequence (v_n) of elements of \mathcal{C} which converges to V. For \mathcal{C} to be closed, it should be that any convergent sequence of elements of \mathcal{C} converges in \mathcal C. I suggest another way: notice let f:\mathbb{R}^2\mapsto \mathbb{R}^2, (x,y) \mapsto (x,y) ... 1 We've seen that f' need not be bounded. Applying Cauchy's Estimates in the disk D(z,1-|z|) shows that$$|f'(z)|\le\frac{||f||_\infty}{1-|z|}.$$Here's a simple example showing that that's best possible, at least up to a constant factor: Let$$f(z)=(1-z)^i\quad(|z|<1).To be specific, if $|z|<1$ then $1-z=re^{it}$ with $r>0$ and ...

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