Hot answers tagged

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 ...


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 ...


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.


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 ...


4

A possible proof, using a different argument, is this: Assume $T$ is not bounded. Then for all $k\in\mathbb{N}$ there is $z_k\in X$ with $\Vert z_k\Vert_X=1$ and $\Vert Tz_k\Vert_Y>k^2$. Now, consider the sequence $(x_k)_k$ s.t. for all $k$, $ x_k=\frac{z_k}{k^2}$. Then: $\Vert Tx_k\Vert_Y>1$ for all $k$. The series $\sum_{k=1}^\infty x_k$ is ...


3

Hint: Since $x$ is an accumulation point of $A$ for each $n$ in there exist an $x_n\in A-\{x\}$ such that $||x-x_n||<\frac{1}{n}$.


3

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


3

Hint Define a sequence of polynomials by $$ P_n(x)= \frac{1}{\log(\log(n))}\sum_{k=0}^n \frac 1{k+1}x^k $$


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 ...


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} ...


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 $$ \|(Tx)\| = \sqrt{\sum_{i=1}^\infty |x_{i+1}-x_i|^2} \leq \sqrt{\sum_{i=1}^\infty |x_{i+1}|^2 + \sum_{i=1}^\infty|x_i|^2} \leq 2\|x\| $$


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 ...


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 ...


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 $0$.


2

No, the last assumption does not follows from the first two one. To see this consider operators $T_n f = f\left(x^n \right).$


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 ...


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 ...


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 ...


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

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 $x$.


1

If it were Frechet, then there would be a (bounded) linear operator $A$ so that $$ \|v\|=A(v)+o(\|v\|). $$ Now insert $-v$ and compare.


1

Rewrite the inequality that you want to prove as $$\|u\|_{p}^{p/2}\|(-\Delta)^{\sigma/4}|u|^{\frac{p+m-1}{2}}\|_{2}\geq C\|u\|_{\frac{N(2p+m-1)}{2N-\sigma}}^{\frac{2p+m-1}{2}}$$ Set $v:=|u|^{\frac{p+m-1}{2}}$. The LHS above becomes $$\|v\|_{q}^{\alpha}\|(-\Delta)^{\sigma/4}v\|_{2}$$ with $q=\frac{2p}{p+m-1}$ and $\alpha=\frac{p}{p+m-1}$. Apply the NGN ...


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

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 $K$.


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 = \| ...


1

Hint Consider a Cauchy sequence $\{x_n\}$. See what happens with $$\left\{\frac1{\|x_n\|}x_n\right\}$$ (Note that you also have to consider sequences with zeros).


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 $T \colon (\mathbb{F}^n, ||\cdot||_1) \rightarrow (F,||\cdot||)$ ...


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 ...


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.



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