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0

By definition, of the dual norm, $$ a^Tb\leq \|a\|\|b\|_*. $$ Then, using the arithmetic-geometric inequality (or $2xy\leq x^2+y^2$ if you want), $$ 2a^Tb\leq 2\|a\|\|b\|_*\leq \|a\|^2+\|b\|_*^2. $$


0

We will first show that, given $x\in X$, there exists $z\in N$ s.t. $\|x-z\|_X=d(x-z,N)$ and $p(x)=p(x-z)$. Let $\{z_n\}\subset N$ s.t. $\|x-z_n\|_X\to d(x,N)$. Then $\infty>\|x\|_X+\sup_{n}\|x-z_n\|_X>\sup_{m}\|z_m\|_X$. Since $N$ is finite dimensional, there exists a subsequence $z_{n_k}$ and $z\in N$ s.t. $z_{n_k}\to z$. Therefore, ...


1

Is $e_0$ supposed to have norm 1? Notice that if we suppose that, we obtain $1=|g_i(e_0)|\leq \Vert g_i\Vert_E\Vert e_0\Vert=\Vert g_i\Vert$, hence $\Vert g_i\Vert_E =1$.


3

Daniel Fischer's comment says yes, since $\ell^1=c_0^*$. One can give an explicit projection from $(\ell^\infty)^*$ onto $\ell^1$ by saying $$P\Lambda=(\Lambda e_1,\Lambda e_2,,\dots).$$


2

The set $S$ is convex and symmetric, i.e., $-x \in S$ for all $x\in S$. If $x$ is an interior point of $S$ and $B(x,\varepsilon)\subseteq S$ (where $B(x,\varepsilon)$ is the ball around $x$ with radius $\varepsilon$) you get $B(0,\varepsilon/2)\subseteq S$ since for $\|y\| <\varepsilon/2$ you have $y= \frac 12 (-x) +\frac 12 (x+2y) \in \frac 12 S + \frac ...


1

Look at this: http://planetmath.org/banachspacesofinfinitedimensiondonothaveacountablehamelbasis there is refference also. Example is any Banach space of infinite dimension .


0

In what follows below, I will denote the norm defined in (1) by $\left\|\cdot\right\|_{\alpha,r;q}^{(1)}$ and the norm defined in (2) by $\left\|\cdot\right\|_{\alpha,r;q}^{(2)}$. So the algebraic identity that I was missing is the following lemma, which allows us to write the $r^{th}$ difference $[T(t)-I]^{r}$ as a linear combination of $[T(2t)-I]^{r}$ and ...


0

Let $\theta \in (0,1)$ and consider $$\theta \int_0^1 \gamma_1(t)\,dt + (1 - \theta) \int_0^1 \gamma_2(t)\,dt = \int_0^1\theta\gamma_1(t) + (1 - \theta)\gamma_2(t)\,dt.$$ Now, to get the desired result you need to be able to prove that $\gamma(t) = \theta\gamma_1(t) + (1 - \theta)\gamma_2(t) \in \Gamma_0$. Clearly $\gamma$ has the correct fixed ends and is ...


0

Assuming $Q$ is a subspace, then indeed we can: Fix $t\in(0,1)$, and let $a_1=\int_0^1\gamma_1(s)ds$ and $a_2=\int_0^1\gamma_2(s)$ be elements of $l(\Gamma_0)$. Define the following homotopies: $H_1(s,y)=((t-1)y+1)\gamma_1(s)$, $H_2(s,y)=(-ty+1)\gamma_2(s)$, and $H_3(s,y)=(1-y)t\gamma_1(s)+y(1-t)\gamma_2(s)$. This shows that ...


1

To answer your second question: Assume that for some $x_n$ in your sequence it holds that $0 < \vert\vert x_n \vert\vert < 1$ (without loss of generality $0 < \vert\vert x_n \vert\vert$, since otherwise $x_n=0$ and $f(x_n)=0$, which cannot be the case for $n$ large enough). Now define a new sequence $y_n = \frac{x_n}{\vert\vert x_n \vert\vert}$ and ...


2

The answer to both of your questions is based on the linearity of $f$. For the first question, notice that, if there is even a single $x$ with $f(x)\neq0$, then by multiplying $x$ by a large positive real number $r$, you get $|f(rx)|=r|f(x)|$, which gets arbitrarily large if you take $r$ large enough. So the only way $f$ could be bounded everywhere (rather ...


1

I think that a direct proof is possible: Let $(u_k)$ be a Cauchy sequence in $\mathcal{B}$. Then, $(u_k)$ is Cauchy in $L^2(0,T; H^1_0)$ and $(\partial_tu_k)$ is Cauchy in $L^2(0,T; H^{-1})$. Since these are Banach spaces, we conclude that there exists $u\in L^2(0,T; H^1_0)$ and $w\in L^2(0,T; H^{-1})$ such that $$\left\{\begin{align}u_k\to ...


0

The assertion is wrong: Given the interval $[-1,1]$. Consider the Borel measure: $$\mu:\mathcal{B}([-1,1])\to\overline{\mathbb{R}}:\quad\mu:=\lambda+\delta$$ Regard the Borel measure: $$\rho:=1-1_{\{0\}}:\quad\mu_\rho(A):=\int_{[-1,1]}\rho\,\mathrm{d}\mu=\lambda$$ It has Borel support: $$\sigma_\rho=\sigma(\lambda)=[-1,1]$$ But it is not surjective: ...


0

The linearity of the limit means the following: for every $l$ from βN∖N, and for every two sequences ($a_n$),($b_n$) and their continuous completion over βN ($A$),($B$) B-lim (A+B)($l$) = B-lim A($l$) + B-lim B($l$)


1

Looking for biorthogonal functionals seems like a good place to start. We want $\Lambda_n\in c_0^*$ so that $\Lambda_n s_n=1$ and $\Lambda_n s_m=0$ for $m\ne n$. Now $\Lambda_n$ is given by some $\ell^1$ function, and it's not hard to see what the coordinates must be, turns out $$\Lambda_n x=x_n-x_{n+1}$$works. So if we're going to write $x=\sum a_ns_n$ the ...


0

In what follows, $M>0$ is such that $\left\|T(t)\right\|\leq M$ for all $0\leq t<\infty$ (such a constant exists by hypothesis that $\left\{T(t):t\geq 0\right\}$ is equibounded). We state without proof the following lemma. A proof can be found in Proposition 3.1.4, pg. 162 of Butzer and Huber. Lemma 1. For any $r\in\mathbb{N}$, $D(A^{r})$ is dense ...


4

Here is my favorite proof, which I think is simpler than both the one suggested by David C. Ullrich and the one I had given earlier, elaborating on David Mitra’s hint. It uses only the Hahn–Banach theorem, but not Riesz’s lemma. It is based on the hint presented in Exercise 5.25, Folland (1999, p. 160). If $X^*$ is separable, let $\{f_n\}_{n\in\mathbb N}$ ...


2

Third version - maybe this one is right. Lemma. If $K$ is a separable compact Hausdorff space then $C(K)$ is separable. (And it follows that $K$ is metrizable, but we won't be using that.) Proof. Wlog $K$ is infinite. Say $p_1,\dots$ are dense, $p_j\ne p_k$ for $j\ne k$. Choose $f_{j,k}:K\to\Bbb R$, such that $f_{j,k}(p_j)\ne f_{j,k}(p_k)$. The algebra ...


3

Not sure what $E$ is doing there, but here is how the proof usually goes : The map $$ \tau : f \mapsto (f(x))_{x\in X} $$ defines a continuous injection from the unit ball of $B'$ to $U$. Let $A$ denote the image of $\tau$, so it suffices to prove that $A$ is closed in $U$. Suppose $y \in U$ such that $\tau(f_{\alpha}) \to y$. Then, for every $x \in B$ ...


2

Suppose that $X$ is not separable and fix $\theta\in(0,1)$. Let $\Omega$ be the set of countable ordinals. Fix $\alpha\in\Omega$ and using transfinite induction, suppose that $x_{\beta}\in X$ and $f_{\beta}\in X^*$ have already been defined for all $\beta\in\Omega$ such that $\beta<\alpha$ (if $\alpha=\min\Omega$, simply define $x_{\alpha}\equiv0$ and ...


1

Is it true that an open linear map necessarily is reversely bounded, i.e.: $c \|x\| ≤ \|Tx\|$ No. Let $T$ be the projection of $\mathbb R^2$ onto $\mathbb R$ defined by $T(x,y)=x$. Set $z=x+iy$. It is clear that $c \lvert z\rvert \leq \lvert x\rvert$ does not hold. It seems that we can define a branch $R$ of $T^{-1}$ such that $\lvert ...


3

Yes, it fails. Consider $f : [0, \infty) \to [0, \infty)$ given by $$f(x) = \sqrt{1+x^2}.$$ Since $$f'(\xi) = \frac{\xi}{\sqrt{1+\xi^2}} < 1 \text{ for each } \xi \in [0, \infty),$$ by the mean value theorem $$|f(x) - f(y)| = f'(\xi) |x-y| < |x-y| \text{ for every } x, y \in [0, \infty).$$ But $f$ clearly has no fixed points.


1

Hint: by the Sobolev embedding theorem, if $H_0^2(0,2)$ were complete under either the second or the third norm, then the usual norm would be equivalent to either the $L^\infty$ norm or to the $L^2$ norm. Can you find counterexamples to this?


1

Argyros and Haydon constructed a separable Banach space $X$ such that each bounded linear operator $T\colon X\to X$ is of the form $T=cI_X + K$ where $K$ is compact. Let $Y$ be an infinite-dimensional subspace of $X$ that has infinite codimension. Then $Y$ is not the range of any operator on $X$. Indeed, if $T\colon X\to X$ were an operator with ${\rm ...


1

Yes if and only if $H$ is separable. Yes if and only if $H$ is separable. If I remember correctly yes. Please check this survey paper.


1

No. Kakutani proved that a Banach space $X$ is isometric to a Hilbert space if and only if each two-dimensional subspace $Y$ is complemented by a norm-one projection. Now suppose that $X$ is not isometric to a Hilbert space and take a two-dimensional subspace $Y\subset X$ such that each projection $P\colon X\to X$ with ${\rm im}\,P = Y$ has norm $>1$. ...


1

Certainly the image of that map is closed. It's the space of all $L^p$ functions which are constant on each one of those intervals. (Or: The image is closed because that map is an isometry.) And the image is complemented: a natural choice for $S$ is the space of all $L^p$ functions $f$ such that the integral of $f$ over each one of those intervals is $0$.


3

It doesn't matter whether the last step is legitimate or not, because you're simply not proving what needs to be proved. As you point out, you need to show that $f'=g$. But there's no $g$ anywhere in your supposed proof of this! There's no $g$ visible. There's also no implicit hidden $g$. By definition $g=\lim_nf_n'$, and there's also no $f_n'$ in the ...


5

No. The baire category theorem guarantees that no normed space with dimension equal to the cardinality of the natural numbers can be a Banach space.


1

A late answer, but maybe still helpful. All you have to keep in mind are the natural identification of $\ell^1$ and $c_0^\ast$ resp. $\ell^\infty$ and $(\ell^1)^\ast$. I will write $\ast$ for the product in the three steps of the construction of the Arens product so that there is no confusion with pointwise multiplication. For $a,b\in c_0, \omega\in\ell^1$ ...


1

First off, I think your statement of the theorem is a bit off. It should go something like: Given $A:\mathbb{X}\rightarrow\mathbb{R}$ (not $\mathbb{L}\rightarrow\mathbb{R}$) sublinear and $B:\mathbb{L}\rightarrow\mathbb{R}$ linear such that $B(u)\le A(u)$ for all $u\in\mathbb{L}$, there exists a linear extension of $B$ (call it $C$) from ...


2

A function is said to be an extension of another one if, given the original function $f_1 : A \to X$, there is a function $f_2:B \to X$ such that $A\subset B$ and $f_1(x) = f_2(x) \,\,\,\forall x \in A$. Clearly, an extension doesn't have to be unique, it just has to be defined on a larger domain that contains the original one and agree with the first ...


1

The Daugavet equation is of consequence to the philosophy of linear approximation. Let $E$ be any finite dimensional subspace of $C([0,1])$. Then it follows that there cannot exist any non-zero operator $T:C[0,1]\to E$ such that $T(f)$ is a good approximation of $f$ for all $f\in C[0,1]$. Indeed, because of $\Vert I-T\Vert=1+\Vert T\Vert>1$ we find ...


2

Your claim is correct. The point is that in finite dimension all continuous operators are compact, while in infinite dimension you can have continuous operators which are not compact, as the following example shows: In $\ell_2(\mathbb{N})$ consider the operator $T(x)=(\sqrt{1 - \| x\|^2},x_1, x_2, \dots)$ defined for $\|x\| \leq 1$, where $x=(x_1, x_2, ...


0

Proving by contraposition doesn't seems very friendly... You can do it like that : Let $x_n$ be a Cauchy sequence that doesn't converge. First, notice that $x_n$ is bounded : Indeed, there exist $n$ such that $$\forall m>n, \ \|x_n - x_m \| < 1$$ So you have $$\forall m>n, \|x_m\| = \| x_n + (x_m - x_n) \| \leq \|x_n\| + \|x_m -x_n \| \leq ...


2

The conditions can not be weakend. We will show that each one is necessary Let $\left|\cdot\right|$ be any norm on $X$ and let $x\in X\setminus\left\{ 0\right\} $. By rescaling, we can achieve $\left|x\right|=1$. For any $\alpha,\beta>0$, we have that $\left\Vert \cdot\right\Vert _{1}:=\alpha\cdot\left|\cdot\right|$ and $\left\Vert \cdot\right\Vert ...


0

Great thanks to TrialAndError and Daniel Fischer!! For Hausdorff spaces: $$\overline{\tau(A)}\subseteq\overline{\tau({\overline{A}})}=\overline{\tau}(\overline{A})$$ Concluding compactness.


2

As pointed out by Clement C in the comments: Equivalent norms induce the same topology. Also the other direction of implication is true: When two norms induce the same topology then they are equivalent. Take two norms $\|\cdot\|_1$ and $\|\cdot\|_2$ on a vector space and you ask yourself: When are the topologies from those norms the same? This is the case ...


2

Some application of this result and more generally that all norms are equivalent on finite dimensional spaces: The compact subspaces are the closed bounded spaces. All linear maps are continuous. More generally all multilinear maps are continuous. All linear maps are bounded on the unit ball.


0

A random variable $X$ is symmetric if for each Borel set $B$ we have $\mathsf{P}(X\in B) = \mathsf{P}(-X\in B)$. Regarding your first question, we are looking for a measurable simple function $$f=-\mathbf{1}_{A_1} + 0\cdot \mathbf{1}_{A_2}+ \mathbf{1}_{A_3}$$ (once we have one, we have a sequence of independent copies) such that $\mu(A_1)=\mu(A_3)$ ...


0

Ok satisfactory enough: Given the Hilbert space $\ell^2(\mathbb{N})$. Extend to a Hamel basis: $$\mathcal{B}:=\{\delta_n:n\in\mathbb{N}\}\cup\{\beta,\beta',\ldots\}$$ Define an operator by: $$T(\delta_n):=n\quad(n\in\mathbb{N})$$ $$T(\beta):=0\quad(\beta,\beta',\ldots)$$ Then it is unbounded.



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