# 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

For every function $u$, the set $\{u\le k\}$ is well-defined. But the elements of $L^p$ are not functions; they are equivalence classes of functions. Since you consider a Bochner space, let's recall what this means: $u\sim v$ if for almost every $t$, the equality $u(\cdot,t)=v(\cdot, t)$ holds for a.e. $x$. Then $\{u\le k\}$ is really an equivalence class ...

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

You started your proof by saying "$g$ is obviously continuous", but this is not true in general. However, it does follow if you assume that $f$ does not attain the value $M$. So if you begin your proof by saying: Assume $f$ does not attain the value $M$. Let $g(x) = \frac{1}{M-f(x)}$. Now $g$ is continuous... You now have a valid argument. Since ...

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

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

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 $E\subseteq ... 4 Suppose that$s(z)$has period$T$and is analytic on the strip$|\mathrm{Im}(z)| < a$for some$a>\pi T$. Then I claim that The derivatives$s^k(z)$are dense in the space of all$T$-periodic functions BUT Said derivatives are not linearly independent. I can not currently figure out whether the condition that$a>\pi T$is necessary or an ... 4 For every$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 $$\|fg\| = \sup_{x \in ... 4 For U \subset V, we have a natural (continuous) injection$$\iota^U_V \colon \mathscr{D}(U) \hookrightarrow \mathscr{D}(V).$$Its transpose,$$\rho^V_U \colon \mathscr{D}'(V) \to \mathscr{D}'(U)$$is called the restriction of the distributions on V to distributions on U. For regular distributions, that corresponds to the restriction of the locally ... 4 Take X=c_{00}---the space of all sequences which are almost everywhere 0 and as x_n---the sequence having \frac{1}{2^n} on n-th place and 0 elsewhere. 4 No, this cannot be done. If \phi\in C_c^\infty(R), then its Fourier transform extends to be an entire analytic function (cf. Paley--Wiener theorem). Should it be constant for (real) x in some neighborhood of x_0, it would be identically constant by the uniqueness theorem for analytic functions, and hence zero, because the Fourier transform of a smooth ... 3 I don't see any use of any Hardy-Littlewood inequality here. You have a measure space (Q,\delta^\alpha \,dx\,dt), which has finite total measure. I will denote the measure by \mu for simplicity. The assumption (2.7) says that u is in the weak L^{\hat q}( d\mu) space. Then it's just a matter of interpolation to get that u\in L^q( d\mu) for every ... 3 The norm on the direct sum A \oplus B of two C*-algebras is the maximum norm$$\lVert (a,b) \rVert = \max\{\lVert a\rVert, \lVert b\rVert\},$$not the \ell_1-norm that you write: an example that helps me remember this is to consider A=C(K), B = C(L) and to note that A \oplus B should be isomorphic to C(K \sqcup L). Since Murphy identifies ... 3 Yes, it is possible. For example, Serge Lang exhibits the basics of a theory of differential forms on Banach manifolds in Chapter V of his Differential and Riemannian Manifolds. (Non-)Separability is not an issue. According to Lang, a p-form on a Banach space E is simply a continuous alternating p-linear map on E. This yields a notion of p-forms ... 3 By Jensen's inequality \int |f|^2 \log |f|=\int |f|^2 \cdot \frac{1}{p-2}\log |f|^{p-2} = \frac{1}{p-2}\cdot\int |f|^2\log |f|^{p-2} \leq \frac{1}{p-2}\log (\int |f|^{p-2}\cdot |f|^2) = \frac{1}{p-2}\cdot \frac{p}{2}\log (\int|f|^p)^\frac{2}{p}=\frac{1}{p-2}\cdot \frac{p}{2} \log ||f||_p^2 because \frac{1}{p-2}=\frac{n-2}{4}, ... 3 A bounded linear operator A on a separable Hilbert space X with orthonormal basis \{ e_{j} \}_{j=1}^{\infty} is a Hilbert-Schmidt operator if$$ \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 You can show that$c = 1+|\lambda|$and that this cannot be improved. The last statement first: if$u = 1$for$\lambda \geq 0$($u = -1$for$\lambda <0$) and$v = 0$the the RHS is$c$and the LHS is$1+\lambda$, so$c\geq 1+\lambda$. To show the converse, recall that$|a +b|\leq |a| + |b|$and that$|\int f \mathrm dt |\leq \int |f|\mathrm d\tau$. 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 This counterexample is in terms of continuous but non-differentiable functions because it's easy to describe. At the points of non-differentiability one can smooth out the function to the degree required, e.g. by convolution with mollifiers etc. Take$f\equiv 1$.$J\$ is a function whose graph is an infinite sequence of hats (by "hat" I mean like the graph ...

3

here is my short list of Visual / intuitive books about Topology : Intuitive Concepts in Elementary Topology. - Arnold From Geometry To Topology - H. Graham Flegg Classical Topology and Combinatorial Group Theory - John Stillwell Three-Dimensional Geometry and Topology - Bill Thurston & Silvio Levy The shape of space. - Jeff Weeks or ...

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