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16

$S^3$ is compact, while $\mathbb{R}^3$ is not. Since any continuous function $f:S^3\rightarrow \mathbb{R}^3$ maps compact subsets of $S^3$ to compact subsets of $\mathbb{R}^3$, it can't be surjective (or else $f(S^3)=\mathbb{R}^3$ is also compact).


6

I thought I would just offer an alternate solution using Minkowski's integral inequality applied to: $$ \| Tf \|_2 = \left\{ \int_0^{\infty} \left( \int_0^x x^{-1} f(t)\,dt \right)^{2}\,dx \right\}^{1/2} $$ But first we do a variable substitution $t \rightarrow xt$ so the inner integral is integrating over a fixed space: $$ \| Tf \|_2 = \left\{ ...


6

For the boundedness, observe that \begin{align*} \| Tf \|_2^2 &= \int_{0}^{\infty} |Tf(x)|^2 \, dx \\ &\leq \int_{0}^{\infty} \frac{1}{x^2} \int_{0}^{x} \int_{0}^{x} |f(u)||f(v)| \, dudvdx \\ &= \int_{0}^{\infty}\int_{0}^{\infty} |f(u)||f(v)| \left( \int_{u \vee v}^{\infty} \frac{dx}{x^2} \right) \, dudv \\ &= ...


6

$p$ is not needed for the proof, but rather it's an important part of the formulation of the theorem. Not only can the linear functional be extended, but the extension is still dominated by $p$. Just an extension is not hard to produce, but the theorem actually tells you that you still have control over the extension knowing that the original functional is ...


5

The dual space of $L^\infty$ is the space of "finitely addditive" signed measures and not the space of finite signed measures. The usual Radon-Nikodym theorem does not apply.


4

Every metric space is Hausdorff and $T_4$: Hausdorffness is easy to check. For $T_4$, we use the reciprocal of Urysohn's Lemma: take $F,K \subset X$ be disjoint closed sets. Then: $$f: X \to \Bbb R \\ f(x) = \frac{{\rm d}(x,F)}{{\rm d}(x,F)+{\rm d}(x,K)}$$ is well-defined, continuous, and $f^{-1}([0,1/2[) \supset F$, $f^{-1}(]1/2,1]) \supset K$ are disjoint ...


4

Hint: Both of these spaces are separable. Note that $\ell^2$ is isomorphic to its dual space, but the dual space of $\ell^1$ fails to be separable. Note then that a surjection from $\ell^2$ to $\ell^1$ would induce an injection from $(\ell^1)'$ to $(\ell^2)'$.


4

Well I feel a little silly for asking since I just came up with a really nice counter-example after thinking about $2\times 2$ matrices and what the involutions are in that setting. Let $(e_i)$ be an orthonormal basis for $\ell^2$. Define $D(T) = \operatorname{span}\{e_i: i\ge 0\}\subseteq \ell^2$ (which is clearly dense) and define $T$ via its matrix ...


4

No, this is not true. If $D$ is a Dedekind finite set with a Dedekind finite power set, then $\ell_1(D)$ is a Banach space which has a Hamel basis which is also a Schauder basis, and every linear operator from $\ell_1(D)$ to a normed space is continuous. But if $D$ is Dedekind finite, then $|D|^{\aleph_0}>|D|$. So it suffices to assume that an infinite ...


4

Not unless you're specifically talking about the closed span of $\varepsilon$. For example in $\ell^2$ the span of the standard (Schauder) basis is the set of all sequences with finite support, which is not closed.


4

Note that $$ \| Tx\|^2 = \sum_{j=1}^{\infty} \frac{|x_j|^2}{j^2} \leq \sum_{j=1}^{\infty} |x_j|^2 = \|x\|^2 $$ Hence $\|Tx\|^2 \leq 1 \cdot \|x\|^2$, which implies that $\|Tx\| \leq 1 \cdot \|x\|$ and therefore $\|T\|\leq 1$ (1). On the other side, since $e_1=(1,0,\cdots)$ is such that $\|e_1\|=1$, then $$ \| T \| = \sup_{\|x\|=1}\{ \|Tx\|\} \geq ...


4

It should be obvious enough that $\|Tx\|\leq\|x\|$, which means that $\|T\|\leq1$. On the other hand, since $\|Te_1\|=\|e_1\|=1$, the norm is actually attained. More detail: One typically shows that $\|T\|=C$ by showing the following two things: For every $x$, $\|Tx\|\leq C\|x\|$ There is an $x$ such that $\|Tx\|=C\|x\|$. Statement 1 says that the norm ...


4

By Cauchy Schwarz, $$|f(x)| =\left| \sum \frac{x_n}{n^2}\right|\le \sqrt{\sum \frac{1}{n^4}} \|x\|_2 = \sqrt{\frac{\pi^4}{90}}\|x\|_2$$ (See here for the calculation). Note that equality holds when $x_n = \frac{1}{n^2}$. (Indeed, if you think of $f\in \ell^2(\mathbb N)^*$, then $f(x) = \langle x, y\rangle$, where $y_n= \frac{1}{n^2}$. Thus $\|f\| = ...


4

For all $x \in [0,1]$ we have $$ |f(x)-g(x)|\leq |f(x)-h(x)|+|h(x)-g(x)|\leq \| f-h\|+\|h-g\|$$ and the last bound doesn't depend on $x$ so $\sup _{x\in [0,1]}|f(x)-g(x)|\leq \| f-h\|+\|h-g\|$.


3

Doing only 3). For each $n \in \mathbb{N}$ we define the operators $T_n: \ell^2 \to \ell^2$ as follows, for each $x=\{ x_j\}_j \in \ell^2$ put $$ T_n(x) = \left( \sum_{k=1}^\infty a_{1k} x_k, \cdots, \sum_{k=1}^\infty a_{nk} x_k, 0 ,\cdots \right) $$ By Hölder, for any $j\in \mathbb{N}$ $$ \left|\sum_{k=1}^\infty a_{jk} x_k\right| \leq \left( ...


3

If $P$ and $Q$ are bounded, then \begin{align} n\|Q^{n-1}\| = \|nQ^{n-1}\| & = \|PQ^{n}-Q^{n}P\| \\ & \le \|P\|\|Q^{n}\|+\|Q^{n}\|\|P\| \\ & = 2\|P\|\|Q^{n}\| \\ & \le 2\|P\|\|Q\|\|Q^{n-1}\|. \end{align} If $\|Q^{n-1}\| \ne 0$, then $$ n \le 2\|P\|\|Q\|. $$ This is impossible for large ...


3

Every space which has a Schauder basis is separable. We have that $\ell^\infty$ is not separable, hence it can't have a Schauder basis. You can see a proof of this here, for example. Using the axiom of choice, we have that every vector space has a Hamel basis - in particular, $\ell^\infty$ has a Hamel basis, but for infinite dimensional vector spaces we ...


3

The weak convergence tells you that, for every $y\in H$, $\langle x_n-x,y \rangle \rightarrow 0$. But, you can't chose $y=x_n-x$, because it depends on $n$. To prove the strong convergence with $\|x_n\| \rightarrow \|x\|$, simply see that : $\|x_n-x\|^2 = \langle x_n-x,x_n-x \rangle = \|x_n\|^2+\|x\|^2 - 2\langle x_n,x \rangle$ $\|x_n\|^2 \rightarrow ...


3

Consider any unitary operator $U:L^2[0,2]\to L^2[0,1]$. Then the set of $f_k:=Ue_k$ is an orthonormal base in $L^2[0,1]$. An example of such operator is $$(Uf)(s)=\sqrt 2 f(2s),\ f\in L^2[0,2],\ s\in[0,1].$$


3

I suspect there is a confusion about what "pull over" means. There are two natural ways to get a metric on $(0, 1)$ from the usual metric $d$ on $\mathbb{R}$. The first is restriction: since $(0, 1)\subseteq\mathbb{R}$ we can take the metric $d_{(0, 1)}$, which is just $d$ itself, limited to points in $(0, 1)$. This is the "usual metric" on $(0, 1)$, and ...


3

Think to the space of odd, square-integrable functions over $(-\pi,\pi)$, equipped with the inner product: $$ \langle f,g\rangle = \int_{-\pi}^{\pi}f(x)\,g(x)\,dx. $$ The sequence given by: $$ f_n(x) = \frac{1}{\sqrt{\pi}}\sin(nx),\quad n\geq 1 $$ gives an orthonormal base, but if we consider any discontinuous function in our space, it cannot be written as a ...


3

Necas' book "Direct Methods in the Theory of Elliptic Problems" is a wonderful guide for the topics that you mentioned above, although the proofs are very abstract and most steps are omitted. Evans' book is more understandable and also includes the topics above, but not in the most general settings of the theorems. If you just start studying the Sobolev ...


3

All finite dimensional vector spaces $R^n$ are Hilbert space, and the existence of an inner product gives us the possibility to speak of orthogonal vectors, and many other nice properties. So our usual $3$ dimensional space is an Hilbert space, but I suppose that your question is about infinite dimensional Hilbert spaces. These ''space of functions'' are ...


3

Absolutely cloddish inequality: If $a,b \ge 0,$ then $(a+b)^2 \le 4a^2 + 4 b^2.$ Proof: If $a\le b,$ then the left side is $\le (2b)^2 = 4b^2,$ same idea of course if $a\ge b.$ So $$\sum (x_n+y_n)^2 \le \sum (|x_n|+|y_n|)^2 \le \sum (4|x_n|^2 + 4|y_n|^2)$$ and that does it.


3

Hint. $(f_n)$ is not equicontinuous at $0$ as for all $n \ge 1$ $$\vert f_n(1/n) - f_n(0) \vert =1/2$$ Jean-Pierre - http://www.mathcounterexamples.net/


3

The short answer is yes, you can think of the integrals as iterated double integrals as in elementary calculus. Whether there is a normalization constant like $\dfrac{1}{\text{Area}(\Omega)}$ will depend on context (and that is not the only kind of "weighted" area measure, so more context is needed if you want to ask about it). Area measure means ...


3

We have $$f(X)\subseteq\{0\}\cup f(\mathrm{supp}(f)),$$ which is compact in $\mathbb{R}$ (since $\mathrm{supp}(f)$ is compact and $f$ is continuous), hence bounded.


3

As suggested in the comment, it is not true in general that $L^1(X)\cap L^\infty(X)$ is dense in $L^\infty(X)$. Indeed, we can use "the standard way" to deal with your question. First of all, for each $g \in L^1(X)\cap L^\infty(X)$, consider $$\text{sgn}(f) g =\begin{cases} g & \text{ if } f>0 \\-g & \text{ if } f<0\end{cases}$$ Then ...


3

Note that $$\langle (A-mI)Ax,x\rangle=\langle AAx,x\rangle-m\langle Ax,x\rangle,$$ so $$\langle AAx,x\rangle-m\langle Ax,x\rangle \leq \|A\|\langle Ax,x\rangle-\|A\|m.$$ Since $\langle AAx,x\rangle = \|Ax\|^2$, because $A$ is symmetric, we have $$ \|Ax\|^2-m\langle Ax,x\rangle \leq \|A\|\langle Ax,x\rangle-\|A\|m$$ then $$ \|Ax\|^2 \leq ...


3

For separable $H$, fix an orthonormal basis $\{e_j\}\subset H$ and let $$ a_n=\frac1{\sqrt{n+1}}\,\sum_{k=n}^{2n}e_k, $$ For any $x\in H$ with $\|x\|\leq1$ we have \begin{align} |\langle x,a_n\rangle|&= \frac1{\sqrt{n+1}}\,\left|\sum_{k=n}^{2n}\langle x,e_k\rangle\right|\leq\frac1{\sqrt{n+1}}\,\left(\sum_{k=n}^{2n}|\langle ...



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