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2

Suppose $T_n \rightarrow T$ and $U_n\rightarrow U$ are the convergent sequences of bounded linear maps. For any $x$, \begin{align} \|UTx - U_nT_nx\| &\leq \|(U - U_n)Tx\| + \|U_n(Tx - T_nx)\| \\ &\leq \|U-U_n\|\|Tx\| + \|U_n\|\|T-T_n\|\|x\|. \end{align} Take the limit as $n\rightarrow \infty$, the RHS tends to $0$.

3

$f(x+h)-f(x)-{\rm Id}\,h=0$ and so the derivative is the identity.

0

The orthogonal complement of $Y$ consists of all $g$ such that $$0 = (f,g) = \int_{-\pi}^{0}f(t)\overline{g(t)}+\int_{0}^{\pi}f(t)\overline{g(t)}dt \\ = \int_{0}^{\pi}f(t-\pi)\overline{g(t-\pi)}+f(t)\overline{g(t)}dt \\ = \int_{0}^{\pi}f(t)\overline{\{g(t-\pi)+g(t)\}}dt,\;\;\; f \in Y.$$ It follows that $$Y^{\perp} = \{ ... 1 Consider the norm |x| on \mathbb R. Clearly it is not differentiable at 0. An aside comment: Not exactly what you ask, but no norm on \mathbb R^n (and so on any normed vector space) is differentiable everywhere, although you ask at zero. The same happens in fact to any distance on \mathbb R^n. See the relevant paper of Rosenholtz at ... 2 The answer will not be (a). Consider$$ x = (1,-1,0,0,0,\dots) $$as a counterexample. To get an actual upper bound, note that$$ \|(Tx)\| = \sqrt{\sum_{i=1}^\infty |x_{i+1}-x_i|^2} \leq \sqrt{\sum_{i=1}^\infty |x_{i+1}|^2 + \sum_{i=1}^\infty|x_i|^2} \leq 2\|x\| $$1 Yes. Since K is a compact metric space it is separable; now if C is a countable dense subset of K the closed span of C is the same as the closed span of K. 0 Using your g(t) = t - \frac{\pi|{2}, for\, t > 0, and g(t) = t +\frac{\pi}{2}\,for\, t< 0, We have$$g(t) - t = -\frac{\pi}{2}\, for\, t > 0 $$and$$g(t) - t = \frac{\pi}{2} \, for\, t < 0$$So (g(t) - t) is odd function. so$$\int_{_pi}^{\pi} s(t)(g(t) - t)dt = 0 $$, for any s \in Y For any f \in Y, we have \int_{-\pi}^{\pi}f (g -h) ... 1 Simple example of two non-equivalent norms on infinietly-dimensional space: Consider space of all contnuously differentiable functions X = C^1 [0,1]. Then equipping it with the norm:$$ \|f \|_{C^1} = \sup \limits_{x \in [0,1]} |f| + \sup _{x \in [0,1]} |f'| $$gives us a complete space (Banach space), but if we consider norm:$$ \|f \|_\infty = \sup ...

1

Partial answer, for $p>2.$ Let $P_n(x)=\sum_{j=1}^n j x^j .$ We have $$\|P_n\|^p=\sum_{j=1}^n j^p<\sum_{j=1} ^n\int_j^{j+1}y^p dy=\int_1^{n+1}y^p dy=$$ $$=\frac{(n+1)^{1+p}-1}{1+p}<\frac {(n+1)^{1+p} }{1+p}.$$ $$\text {So }\; \|P_n\|<\frac {(1+n)^{(1+1/p)}}{(1+p)^{1/p}}\;\text { But }\; P_n(1)=(n^2-n)/2.$$ Let $Q_n=P_n/n^{1+2/p}.\;$ Then ...

2

For a given $p\gt1$, consider $$a_{n,k}=\left\{\begin{array}{cl} \dfrac1{n^{1/p}}&\text{if }1\le k\le n\\ 0&\text{if }k\gt n \end{array}\right.$$ Then \begin{align} \left(\sum_{k=1}^\infty a_{n,k}^p\right)^{1/p} &=\left(\sum_{k=1}^n\frac1n\right)^{1/p}\\[6pt] &=1 \end{align} while \begin{align} \sum_{k=1}^\infty a_{n,k} ... 3 Hint Define a sequence of polynomials by P_n(x)= \frac{1}{\log(\log(n))}\sum_{k=0}^n \frac 1{k+1}x^k $$1 Consider the subspace V_\infty of \mathscr l^2(\mathbb N) where only a finite amount of terms in a series is non-zero. This is an infinite dimensional normed vector space. Define also the subspace V_n where only the first n terms of a series are non-zero. V_n \cong \mathbb R ^n with the standard norm. As such there is a sequence of compacta ... 2 No, the last assumption does not follows from the first two one. To see this consider operators T_n f = f\left(x^n \right). 0 The key issue is the continuity of point evaluation. For every x\in X we have the evaluation functional \phi_x : V\to\mathbb{C} defined by \phi_x(f)=f(x). If these functionals are continuous in \tau, they extends to the completion \overline{V} and therefore allow us to interpret the elements of \overline{V} as functions on X. On the other ... 1 Your notation is a little idiosyncratic. More precisely, \|x\| = \inf \{ r \ge 0 | x \in r B \}. Suppose x = 0, then x \in rB for all r >0, hence \|x\| = 0. If \|x\| = 0, then there are r_k \ge 0 such that r_k \to 0 such that x \in r_k B. Since \cap_k r_k B = \{0\} (from 5.), we see x = 0. Suppose \lambda = 0, then 0 = \| ... 3 Note that f is bounded if and only if it is continuous. So let f be unbounded. This means for any n \in \mathbb N we have an x_n \in X so that f(x_n) ≥ n \|x_n\|. By rescaling set \|x_n\|=1. Now let z be in X. From the construction of the x_n it follows that z_n:= z - \frac{f(z)}{f(x_n)}x_n is a sequence that converges to z. But ... 1 I'm going to revise the notation to make things easier to follow. Say X^* is the space of bounded linear functionals on X, and similarly for Y^*. I'm going to write x and y for elements of X and Y and I'm going to write x^* and y^* for elements of X^* and Y^*. Define the adjoint T^*:Y^*\to X^* as usual:$$T^*y^*=y^*T.$$We can use ... 0 We use the Closed Graph Theorem. Assume x_n \to x and Tx_n \to y. We need to prove that Tx = y. By Hahn Banach this is true if f(y) = f(Tx) for all f \in Y^*. But f(Tx) = \lim_n f(Tx_n) = f(y). So we are done. (In case X,Y are Banach.) 1 First, f_n Cauchy in ||\cdot||_{\infty} norm implies that for every x\in \mathbb{R} f_n(x) is Cauchy, so by completeness of \mathbb{R} f_n\to f pointwise. Next as f_n is Cauchy given \varepsilon>0 there is N such that m,n\ge N implies ||f_m-f_n||_{\infty}<\varepsilon/3. Taking limit as m\to\infty we get$$ ...

1

The sequence $f_n$ is uniformly convergent because of the norm you have chosen: $$\Vert f_n-f\Vert=\sup_{x}|f_n(x)-f(x)|.$$ Thus, for any $y$, $|f_n(y)-f(y)|\leq \sup_{x}|f_n(x)-f(x)|=\Vert f_n-f\Vert.$ So, if you insist that $\Vert f_n-f\Vert<\varepsilon/3,$ then $|f_n(y)-f(y)|<\varepsilon/3$ for every $y$.

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$\newcommand{nrm}[1]{\left\lVert{#1}\right\rVert}$ Long story short: yes, and it happens quite often. For instance, let $I=(0,1)$. Consider the map \begin{align}\psi:W^{1,p}(I)&\hookrightarrow L^p(I)\\u&\mapsto u\end{align} Since $\nrm{u}_{W^{1,p}}=\nrm{u'}_p+\nrm{u}_p$, it holds $\nrm\psi\le1$. But $\psi\left(W^{1,p}(I)\right)=F$ contains ...

5

Edit: In fact there's a very simple theorem here that gives the whole truth: Given a bounded linear bijection $T:X\to Y$, where $X$ is complete, $Y$ is complete if and only if $T^{-1}$ is bounded. (If $Y$ is complete the open mapping theorem shows that $T^{-1}$ is bounded. On the other hand if $T^{-1}$ is bounded it's trivial to show that $Y$ is complete: A ...

1

What you need to show is that if $X$ and $Y$ are normed linear spaces with $X$ complete and $Y$ not complete then they are not equivalent. Let $(y_n)_n$ be a Cauchy sequence in $Y$ with no limit point. Suppose there exists bounded linear bijections $B:Y\to X$ and $A:X\to Y$ with $A=B^{-1}.$ Then $(B(y_n))_n$ is a Cauchy sequence in $X,$ so it has a limit ...

1

There’s a theorem which will come to the rescue: Theorem. Every normed space over a complete field has a dual space which is complete under the operator norm. $\mathbb{R}$ is, of course, complete.

2

$f:(0,1]\rightarrow R, f(x)=1/x$.

2

Lemma (Exercise 3.29, Brezis). Let $E$ be a normed vector space with a uniformly convex norm and fix $p > 1$. If $x$, $y \in \overline{N(0, M)} =: B$ are at least $\epsilon > 0$ apart, then there is some $\delta$ such that$$\left\|{{x + y}\over2}\right\|^p \le {{\|x\|^p + \|y\|^p}\over2} - \delta.$$ Suppose not for the sake of contradiction. Then ...

0

Suppose $(x_n)_{n\in N}$ is a Cauchy sequence. Let $(a_n)_{n\in N}$ be any monotonically decreasing sequence of positive numbers. Let $g(1)$ be the least (or any) $n$ such that $\forall n'>n\;(|x_n-x_{n'}|<a_1).$ Recursively, let $g(j+1)$ be the least (or any) $n>g(j)$ such that $\forall n'>n\;(|x_n-x_{n'}|<a_{n+1}).$ The subsequence ...

0

Hint: $L(X \times Y;Z)$ is the vector space of linear maps $T: X \times Y \rightarrow Z$. For each you need to find a set of linear maps that form a basis of the respective spaces. So for $L(X \times Y;Z)$ you need to find a collection of linear maps $T: X \times Y \rightarrow Z$ that form a basis of the vector space $L(X \times Y;Z)$. The number of such ...

1

The definition of Cauchy sequence begins by saying "for every $\varepsilon>0$". Whatever is true of EVERY positive number is true of $1/2^kn$.

1

Given a basis for $A$, the components in that basis give a vector $x(y)$ of $\mathbb R^n$ for each $y\in A$, where $n$ is the dimension of $A$. Note that a sequence $y_n$ in $A$ is a Cauchy sequence if and only if $x_n=x(y_n)$ is a Cauchy sequence in $\mathbb R^n$. From now on you can thus proceed as if $A$ was $\mathbb R^n$. This implies that $A$ is ...

1

Hint: equivalence preserves completeness, and the dual space is always complete...

4

Note: I completely rewrote my previous attempt - posted in another (now deleted) answer, since the previous version was incorrect. Thanks to Asaf Karagila for pointing out the problem with my previous proof. And also to Eric Wofsey for simplifying some steps in the proof. Let us hope that this time I have avoided mistakes. The formulation of the version of ...

2

Only a partial answer: If you take the norm $\|x\|'=\|x\|+\alpha |x|$ as gerw said in the comments, it is easy to see that since $\|.\|\sim |.|$, i.e $\exists \underline c,\overline c>0: \underline c\|x\|\leq |x|\leq \overline c \|x\|,\forall x\in E\,\,$ you will have $$\|x\|\leq \|x\|+\alpha|x|\leq (1+\alpha \overline c)\|x\|,\forall x\in E$$. It is ...

1

Because $\sum_{n=1}^{\infty} |a_n|^p < \infty,$ $|a_n|^p \to 0.$ Thus there exists $N$ such that $|a_n|^p< 1$ for $n\ge N.$ For such $n$ we can say that, since $q/p > 1,$ $$|a_n|^q = (|a_n|^p)^{q/p}\le |a_n|^p.$$ Thus $\sum_{n=1}^{\infty} |a_n|^q$ converges by the comparison test.

0

The result is not true. Counter example: take X=Y=R(real line).the topology and norm are taken as usual.Now take E=[0,1] and take f(x)=x for all x except at 1/4 it is 2. and A={1/2}. It will not be a continuous,though condition will be satisfied.

0

It is the same to show that $f(A)^-1$ is open in Y. If f is continuous in norm than of every open ball containg f(x) exists an open ball in X (with the distance inducted by the norm ) containg x and such that f(B)

1

Let $\overline{A}$ denote the closure of $A$ in $(X,\Vert\cdot\Vert)$. We have to prove that $x\in \overline{A}$ if and only if there exists a sequence $\{x_{n}\}_{n}\subset A$ converging to $x$. Suppose that $x\in\overline{A}$. By definition, it means that any open subset containing $x$ also contains a point of $A$. Let $B_{n}=\{y\in X\mid \Vert ... 1 As a base of neighborhoods of$x$is given by the open balls$B(x,r),$you have that $$x\in \overline{A}\iff\forall r>0,B(x,r)\cap A\neq\varnothing.$$ Choosing a such element for all$r=\frac{1}{n},n\in\mathbb{N}^*,$you will get your sequence. For the other part, John Ma has given you the idea. 1 Let$(F, ||\cdot||)$be a finite dimensional normed space over$\mathbb{F}$(where$\mathbb{F} = \mathbb{R}$or$\mathbb{F} = \mathbb{C}$). Choose some linear isomorphism$T \colon \mathbb{F}^n \rightarrow F$and define a norm$||\cdot||_1$on$\mathbb{F}^n$by$||v||_1 := ||Tv||$. Then$T \colon (\mathbb{F}^n, ||\cdot||_1) \rightarrow (F,||\cdot||)$... 2 I think there's no general inequality which holds: The sequences $$a_0 = 2, a_k = 0, k > 0$$ and $$b_0 = \frac{1}{2}, b_k = 0,k > 0$$ should show that (take i.e.$p = 1, q = 2$). To prove$l^p \subseteq l^q$, you have to take a sequence$(a_k) \in l^p$and show that it is in$l^q$. First, observe that we can assume without loss of generality that ... 0 The$x$is an element that you can take norm of, and$0$is an neutral element under addition (ie such that$0+f = f$). In your example it would translate to:$||f||_1\ge 0||f||_1 = 0$iff$f(t)=0$for all$t\in[a,b]||\alpha f||_1 = |\alpha|\cdot||f||_1||f + g||_1 \le ||f||_1+||g||_1$So for example for the first we note that since$|f(t)|\ge0$... 1 The first condition says$||x||\geq0$for all$x\in X$. In your case$X=C[a,b]$, such that elements of$X$are continuous functions$f:[a,b]\to\mathbb{R}$. In other words, you have to prove$||f||\geq0$for all$f\in X$. Since$|f(t)|\geq0$for any$t\in[a,b]$, you have$\sup_{t\in[a,b]}|f(t)|\geq0$, proving$||f||\geq0$. I think you'll be able to prove ... 1 I'd say the main argument would be that this formulation is not really advantageous. The thing is that for$\dim X < \infty$,$B[0,1]$and$S$are compact and$\left\Vert Tx \right\Vert$is continuous. Hence, there exists some$x_0 \in S$such that $$\left\Vert T \right\Vert = \left\Vert Tx_0 \right\Vert$$ making the supremum a maximum. The supremum over ... 3 Hint: Since$x$is an accumulation point of$A$for each$n$in there exist an$x_n\in A-\{x\}$such that$||x-x_n||<\frac{1}{n}$. 1 Let$U_n=\{y\in E: \|y-x\|<1/n\}$, for$n\ge1$integer. Then$U_n$is a neighborhood of$x$and, by definition of accumulation point, there is$x_n\in U_n\cap A$,$x_n\ne x$. Prove that the sequence$(x_n)$converges to$x$. 0 You said it yourself, if a series converges absolutely then the series converges. So you cannot have a divergent series that converges absolutely. 5 Sure. Take a (Hamel) basis$\{v_i\}_{i\in I}$of your vector space$V$(its existence is guaranteed by Zorn's lemma) and define an inner product by setting $$\langle v_i,v_j\rangle = \delta_{ij}$$ and extending linearly over$V$. Now take the induced norm$\|w\| = \langle w,w\rangle$. However, this structure is not interesting at all, as it basically only ... 4 Yes (as long as you have an absolute on the field, and you do not use non-common axioms schemes for your set-theory). Just recall that every vector space has a basis, fix one, and define the norm, e.g., as the$\infty$-norm of the coordinates in this basis. 2 Just use the same method as in the usual proof of Hahn-Banach to extend your functional to each point of a countable dense subset one at a time by induction. You then get that the functional is defined on a dense subspace of$X$, and so then you can extend it to all of$X$by just taking limits (and can check that this is well-defined because the functional ... 0 Define a series of functions$s_n$to be $$s_n(x)= \begin{cases} \mathrm{sign}(g(x)) &\mbox{if } |g(x)|\geq n^{-1} \\ \text{linear} & \mbox{O.W.} \end{cases}$$ Meaning it takes the sign of$g(x)$whenever$g(x)$is far enough from$0$. To make sure it is continous we define it be linear (connecting the different signs). Obviously if$g \equiv 0\$ ...

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