New answers tagged normed-spaces
1
You can do that by first showing that norms are equivalent in a finite dimensional space (see http://math.stackexchange.com/a/345661/10385) and then you have $M\in \Bbb R, \| \cdot\|_\infty \le M\|\cdot\|$ so if for a sequence of vectors $\left(v_n\right)_{n\in \Bbb N}$, $v_n\underset{n\to +\infty}{\longrightarrow}0$ for the norm $\|\cdot\|$, that is ...
0
Fix some $v\in \ell^1$ and note that $$\tag{1}\frac{\|x+tv\|-\|x\|}{t}=\frac{\sum_{i=1}^\infty(|x_i+tv_i|-|x_i|)}{t}$$
It follows from $(1)$ that $$\tag{2}\lim_{t\to 0}\frac{\|x+tv\|-\|x\|}{t}=\sum_{i=1}^\infty \lim_{t\to 0}\frac{(|x_i+tv_i|-|x_i|)}{t}$$
Now the question is: In what points the function $|\cdot|:\mathbb{R}\to\mathbb{R}$ is differentiable? ...
0
The statement is in general false taking the constant to be equal to $1$. The best possible constant in general is $2$.
It easy to prove that
\begin{equation}
\left\| \frac{x}{\| x \|} - \frac{y}{\| y \|} \right\|= \left\| \frac{(x-y) \| y \| + y(\| y \| - \| x \|)}{\|x \| \| y \|} \right\| \leqslant \frac{2 \|x-y\| \| y \|}{\|x \| \| y \|} \leqslant 2 \| ...
5
Let $f(x) = \frac 1{1+x^2}$, define $f_n\colon \mathbb R \to \mathbb R$ by $f_n(x) = f(x)$ for $x \in [-n,n]$, $f_n(x) = 0$ for $x \not\in [-n-1, n+1]$, affine-linear in between. Show that $(f_n)$ is Cauchy, but not convergent to some compact support function (as its limit in $\ell^\infty(\mathbb R)$ is $f$).
0
Since $$\max_{1\leq i\leq d}|x_i|^2\leq\sum_{i=1}^d |x_i|^2\leq d\ \max_{1\leq i\leq d}|x_i|^2$$ one has
$$\|{\bf x}\|_\infty\leq\|{\bf x}\|_2\leq\sqrt{d}\>\|{\bf x}\|_\infty\qquad(x\in{\mathbb R}^d)\ .$$
Therefore the identity map
$${\rm id}:\quad\bigl({\mathbb R}^d,\ \|\cdot\|_\infty\bigr)\ \to\ \bigl({\mathbb R}^d, \ \|\cdot\|_2\bigr)$$
is Lipschitz in ...
-1
I think
$$\displaystyle\sum_{j=0}^{\infty} |a_n| + \displaystyle\sum_{j=0}^{\infty} |b_n| = \displaystyle\sum_{j=0}^{\infty} |a_n+b_n|$$
is not correct for all situations.
Consider the following
$$\left\{a_n\right\}_{n=0}^{\infty}=\left\{-\frac{1}{2^{n}}\right\}_{n=0}^{\infty} $$
...
2
There is already a very nice answer, but I thought I could give an example of a function which is discontinuous as a function of two variables, but continuous in each variable. Consider the function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$,
$$
f(x,y) = \begin{cases}
\frac{xy}{x^2 + y^2} & (x,y) \neq (0,0) \\
0 & (x,y) = (0,0) \\
...
3
HINT: There are at least two reasonable approaches.
Start with a Cauchy sequence $\langle x^n:n\in\Bbb N\rangle$, where $x^n=\langle x_1^n,\dots,x_k^n\rangle$ for each $n\in\Bbb N$. Now look at the real-valued sequences $\langle x_i^n:n\in\Bbb N\rangle$ for $i=1,\dots,k$. Show that each is Cauchy in the usual metric on $\Bbb R$, let $x_i$ be the limit ...
0
Without loss of generality, we can assume $a=0$. Let $x \in \text{int} \, \overline{B}(0,r)$ and $\delta>0$ such that $B(x,\delta) \subseteq \overline{B}(0,r)$. For $y :=t \cdot x$, $t>1$ we have $$y \in B(x,\delta) \Leftrightarrow \|\underbrace{y}_{t \cdot x}-x\|<\delta \Leftrightarrow \|x\| \cdot (t-1)< \delta$$ i.e. we can choose $t>1$ such ...
0
Let $x$ be an interior element of the closed ball. Then there is some $\delta > 0$ so that $B(x, \delta)$ lies in the closed ball. Every element of $B(x, \delta)$ lies in the open ball. You can use the triangle inequality to see this.
7
The norm $\|\cdot\|$ is a function from $V$ to $\Bbb R$. There are many topologies that can be placed on $V$ that make this function continuous. Let $\mathscr{T}$ be the set of all such topologies. Then it turns out that there is a topology $\tau_0\in\mathscr{T}$ such that $\tau_0\subseteq\tau$ for all $\tau\in\mathscr{T}$. That is, every topology on $V$ ...
2
think about $f(x) = \frac 1x$.
On any closed interval $[a,b]$ with $a>0$ the function is bounded and I can approximate it very well with a polynomial.
But on the open interval $(0,1)$ it's unbounded. The reason I can get away with this is because it's not defined at $0$. Any continuous function on a closed interval is bounded. That's not true on ...
7
Weierstrass theorem tells that on a compact interval, given a continuous function, there is a sequence of polynomials converging uniformly to this function. But the polynomials are not necessarily of the form $\sum_{j=0}^{N_n}c_jx^j$, because the coefficients depend on $n$. They are indeed of the form $\sum_{k=0}^{N_n}c_{n,k}x^k$.
To get a concrete example, ...
2
Let $f : \left\{ \begin{array}{ccc} \ell^{\infty} & \to & \mathbb{R} \\ (x_n) & \mapsto & \limsup\limits_{n \to + \infty} x_n \end{array} \right.$ and $g : \left\{ \begin{array}{ccc} \ell^{\infty} & \to & \mathbb{R} \\ (x_n) & \mapsto & \liminf\limits_{n \to + \infty} x_n \end{array} \right.$.
Show that $f$ and $g$ are ...
1
Hint 1: Convergence in $\ell_\infty$ and as a consequence in $c$ is a uniform convergence on $\mathbb{N}$
Hint 2: If $\{a^{(n)}:n\in\mathbb{N}\}\subset c$ is a uniformly convergent sequence, then
$$
\lim\limits_{i\to\infty}\lim\limits_{n\to\infty}a_i^{(n)}=
\lim\limits_{n\to\infty}\lim\limits_{i\to\infty}a_i^{(n)}
$$
1
Hint: a subset $A$ of a metric space $X$ is closed if and only if for any sequence $a_n\in A$ converging to an $x\in X$ you have $x\in A$.
1
Norm $1$ and $3$ are equivalent, so $C[0,1]$ is also complete in the norm $3$.
$||f||_{\infty}^{0,1}=||f||_{\infty}+|f(0)|+|f(1)|\le 3||f||_{\infty}$
1
At the moment no one knows the proof of this fact without sladgehammer like Hahn-Banach or something equivalent. On the other hand, to prove it is not only known but even impossible to find a direct proof is a difficult task. Such a proof that proof does not exist requires digging into foundations of mathematics.
Finally Hahn-Banach is so much important ...
0
It's hard to know what you're trying to achieve. When possible, avoid coordinates until the end. What do you know here and what do you not know? What do you mean by "solve"? Your first equation would be my preference, dotting $Y-X\beta$ with itself, but you have to get the formula correct.
3
$L^2(\Bbb R)$ is the space of square-integrable real functions:
$$L^2(\Bbb R) = \left\{ f: \Bbb R \longrightarrow \Bbb R \mid \int_{-\infty}^\infty |f(x)|^2 \, dx < \infty \right\}.$$
Note that the above is not quite right; we say two functions $f, g \in L^2(\Bbb R)$ are equivalent if they take the same values outside of a set of measure zero.
...
2
Let $f:X\rightarrow R$ be an equivalence class of functions that are equal to each other almost everywhere. Then $L^2(X,\mu)$ is the space of all these equivalence classes $f$ such that $\int_X |f|^2 d\mu<\infty$.
On the other hand, let $a = (a_1,a_2,a_3,\dots)$, $a:N\rightarrow R^N$, be an infinite sequence of real numbers. Then $\ell^1(N)$ is the set ...
1
There's the books by Triebel: Theory of Function Spaces, II, III.
3
The Minkowski integral inequality states that for $ r \geq 1 $ and other appropriate conditions we have
$$ \Bigl(\int_X \Bigl|\int_Y f(x,y)dy\Bigr|^r dx\Bigr)^{\frac{1}{r}} \leq \int_Y\Bigl(\int_X |f(x,y)|^rdx\Bigr)^{\frac{1}{r}} dy $$
So you substitute $ f = |g|^q,\ r = p/q $ obtain the inequality and then take $1/q $ power on both sides. This will end up ...
3
Normally, one proves that two normed linear spaces are isometric in a constructive™ way, by exhibiting an isometry between them. One common exception: since it's widely known that any two separable Hilbert spaces are isometrically isomorphic to each other, an explicit isometry is unnecessary in this case.
One elementary way to show that two spaces ...
1
These two spaces are not isometric. Unit sphere of the first space contains straight segment while unit sphere of second space doesn't. You can draw them to see this. But property of unit sphere to contain segments preserved under linear isometries, so there is no isometry between these two normed spaces. You can also check similar question asked earlier.
1
Your argument correctly shows that $\|x\|_1\le \sqrt{n}\|x\|_2$, and thus disproves the opposite inequality stated in the title.
Bonus content: in the opposite direction, $\|x\|_2\le \|x_1\|$ holds. Indeed, squaring both sides we get $\sum_i x_i^2$ on the left and $\sum_i x_i^2 +\sum_i\sum_j |x_i|\,|x_j|$ on the right.
3
Consider $$f_n(x) = \begin{cases} 0, & \text{if $x \leq 0 \leq \frac{1}{2}$} \\
n\left(x - \frac{1}{2}\right), & \text{if $\frac{1}{2} \leq x \leq \frac{1}{2} + \frac{1}{n}$}\\
1, & \text{if $\frac{1}{2} + \frac{1}{n} \leq x \leq 1$}. \end{cases}$$
Then $f_n(x)$ converges in the $L^1$ norm to the function that is $0$ on $[0,0.5]$ and $1$ on ...
4
Calling $e^{(n)}$ the sequence whose all terms are $0$, except the term $n$ with is $1$, and defining
$$x^{(n)}:=\sum_{j=1}^n2^{-j}e^{(j)}.$$
It's a Cauchy sequence in $c_{00}$ for each $\lVert\cdot \rVert_p$ norm, but it doesn't converge to an element of this space.
A deeper reason for which $c_{00}$ is not complete for any norm is that it has a ...
2
Just think of the ordinary scalar product in $2$ or $3$ dimension. This is where these concepts (inner product, norm) originally started out..
If we have an inner product on a space, then geometrically it's more or less nothing else but the ability to talk about angles and lengths (length=norm). By the Cauchy-Schwarz inequality, we have
$$\langle ...
2
If you have an inner product space $\left(E, \varphi\right)$, it has a natural structure as a normed vector space: $\left(E,x\mapsto \sqrt{\varphi(x,x)}\right)$ but the other way around isn't true. There are norms that do not come from inner products.
And example with $E=\Bbb R^2$
If you take ...
2
Every inner product $<-,->$ gives rise to a norm defined by $\|x\|^2=<x,x>$. But a norm in a normed space does not necessarily arise from an inner product. There are normed spaces whose norm can't be induced in the way above from any inner product. So, the class of normed spaces is truly larger than the class of all (normed spaces induced by the ...
0
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1
The question is correctly stated (but boundedness has to be assumed). If $\phi\in Y^*$, then, by definition, $T\phi\in X^*$ and hence $[T\phi](x)$ makes sense. Also $Tx\in Y$ and hence, $\phi(Tx)$ makes sense. The way $T^*\phi = \phi\circ T$ is also ok.
1
There is nothing wrong with the problem as it is phrased. The only thing is that it forgets to say $T$ is bounded. But that should be clear since it mentions $\|T\|$.
1
It is a general fact that the space $\mathcal{K}(X)$ of non-empty compact subsets of a separable metric space $(X,d)$ is separable with respect to the Hausdorff metric. In fact, if $D$ is a countable dense subset of $X$ then the set $\mathcal{F}(D)$ of non-empty finite subsets of $D$ is countable and dense in $\mathcal{K}(X)$.
To see this, let $K$ be any ...
1
Even if the question is loosely formulated, I would say that the answer is no. I'm thinking at the following examples: in the vector space $\mathbb{R}$, consider the usual absolute value $\lvert\cdot\rvert$, which indeed is a norm and induces the ordinary topology of the real line. We can define a metric on $\mathbb{R}$ in terms of $\lvert\cdot\rvert$ as ...
0
Go ahead. Use triangle inequality, and interchange the summation. Then you find $||T||_{op}\le\max_{1\leq j\leq n}\sum_{i=1}^m |T_{ij}|$. Let the max be attained at $j=j_0$. Now choose $x$ to be $e_{j_0}$.
1
It might be helpful to get a geometric understanding of this transformation. This makes 1,2,3 obvious.
For 4, use the observation above to translate this in $\mathbb{C}$.
You will see that for $\alpha\not\in 2\pi\mathbb{Z}$, $U$ is a strict contraction ($M$ Lipschitz for some $0<M<1$). Of course, 4 is false when $\alpha\in 2\pi\mathbb{Z}$.
If ...
2
Show that, for $v\in B[(0,0),1]$, $||U(v)||< k||v||$ for some $k\in (0,1)\subset\mathbb{R}$.
This will mean that $U$ is a contration; therefore its image does not abandon $B[(0,0),1]$.
If you don't know what a contraction is, try showing that
$$
||U^n(v)||< k^n||v||
$$
and since $k<1$, when $n\rightarrow\infty$, $||U^n(v)||\rightarrow 0$. Now, ...
1
To prove $\bar C_{00}=C_{0}$
Let $x\in C_0$. Let us choose $\epsilon >0. $Then by definition there exist $k \in \mathbb{N}$ such that $\forall n\geq k$ $|x_n| <\epsilon$. Now consider $y=\{y_n\}$ by $y_n=x_n$ for $n<k$ and otherwise zero. Then $\|y-x\|_\infty<\epsilon$.
1
For $f\in L^2((0,1))$, we have, by Cauchy-Schwarz:
$$
\int_0^1|f(x)|dx\leq \sqrt{\int_0^1|f(x)|^2dx}.
$$
In particular, $L^2((0,1))$ is contained in $L^1((0,1))$.
Now if the $L^2$ norm was dominated by the $L^1$ norm, on $L^2(0,1)$, we would have in particular, for $f_n(x)=\frac{1}{\sqrt{x}}1_{(1/n,1)}$,
$$
\sqrt{\log ...
3
Take for instance the space $\mathcal C([0,1])$ of continuous functions on $[0,1]$. The $\mathbb L^2$ norm is dominated by the uniform norm but they are not equivalent, since convergence in $\mathbb L^2$ does not imply uniform convergence.
1
Let $(X,\lVert \cdot\rVert)$ be an infinite dimensional normed space and $f\colon X\to \Bbb R$ be a linear non continuous form (for the mentioned norm). Define $N(x):=\lVert x\rVert+|f(x)|$; then $N$ is a norm which is not equivalent to $\lVert\cdot\rVert$.
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