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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

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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