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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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