New answers tagged

0

As others have pointed out, there seems to be an issue with the statement of the problem. However, the general idea seems to be that the bounded linear functionals on $C[0,1]$ separate points. This follows from the Hahn-Banach theorem. Also, the statement of the problem is ``show that there exists...'' which leads me to believe an explicit example is not ...


1

You can get away with scaling and translating whenever your problem is invariant under transformations of the form $x\mapsto \alpha x + c$ where $\alpha$ is a real number (or, more generally, an element of the underlying field) and $c$ is a vector. Given how many problems respect such transformations, and how averse some mathematicians are to constants other ...


0

Let $\varepsilon$ be smaller than $\min\{b-a,1\}$ and $f(x)=g(x):=\mathbf 1_{[a,a+\varepsilon]}(x)$. Then $$\lVert fg\rVert_p=\left(\int_{[a,b]}\mathbf 1_{[a,a+\varepsilon]}(x)\right)^{1/p}=\varepsilon^{1/p}$$ and $$\lVert f\rVert_p\lVert g\rVert_p=\varepsilon^{2/p}.$$


0

I have good news and bad news. The good news is that you were correct. The bad news is that it was when you said: I'm really confused with norms and semi norms in $H^1$ and $L^2$. I'll try to clarify it a bit. First of all... The basic difference between a norm $\Vert\cdot\Vert$ and a seminorm $\vert \cdot\vert$ is that a norm can only be zero if ...


0

One way is by contradiction: suppose that for all $n\in\mathbb N$, there exists $u_n\in H^1(\Omega)$ such that $$\|u_n\|_{L^2(\Omega)}> n\|\nabla u_n\|_{L^2(\Omega)}+n\|u_n\|_{L^2(\partial\Omega)}.$$ Then $\|u_n\|_{L^2(\Omega)}>0$ for all $n$, and if we set $v_n=u_n/\|u_n\|_{L^2(\Omega)}$, then $\|v_n\|_{L^2(\Omega)}=1$ for all $n$, and also ...


1

Take some $f \in C[0,1]$. Set $$g(x) := \max(f(x),0),$$ then clearly $g \in B$. To see that this is the closest element in B, look at $$ \|f-g\|_2 = \int_{0}^{1} |f(x) - g(x)|^2 dx = \int_{[f\geq 0]} |f(x)-g(x)|^2 + \int_{[f<0]} |f(x)-g(x)|^2$$ Obviously, on the set $[f\geq 0]$ we have $f=g$ such that the first integral vanishes. But on set where f is ...


6

The proof has nothing to do with the Schwartz space per se; nor with $i$ or $t$, or that $P$ and $Q$ are symmetric. If $P,Q$ are operators on a Hilbert space $H$ with domain $D$ and such that $PD\subset D$, $QD\subset D$, and $QP-PQ=\mathbb I$, then at least one of $P$ and $Q$ is unbounded. This applies to the case in the question because if we have ...


2

There are one or two things you have to check before the following proof. Since you don't assume $C$ to be closed, what happens if $x_0\in\overline{C}$. (What is a closed ball of radius $0$?) This is a special case which should be dealt with separately. Edit: (A little note on balls of radius $0$.) Let $x\in X$, and let $r>0$. Denote the open ball of ...


1

At first I thought one or the other inequality must be obvious, but I don't see it after a little thought. Big Hint: It's trivial from the Closed Graph Theorem.


2

You can push the norm $\|\cdot \|_Y$ back to $X$ using the bijection $T$. That is, defining a norm on $X$ by $\|x\|_{T}:=\|Tx\|_Y$. Then using the assumption that every norm on $X$ is equivalent to $\|\cdot\|_X$ to obtain $$c_1\|x\|_X\leq \|x\|_T=\|Tx\|_Y\leq c_2\|x\|_X$$ for some non-zero constants $c_1,c_2$, this implies that $T$ and $T^{-1}$ are ...


1

As $T$ is bijective and linear, $\|\cdot\|_Y$ induces a norm on $X$. By assumption, this norm is equivalent to $\|\cdot\|_X$, hence induces the given topology on $X$. So the topologies on $X$ and $Y$ are "the same" via $T$.


0

In Rudin's PMA (Theorem 6.27) $\gamma$ is assumed continuously differentiable. The proof given there works for general normed spaces with no changes: see below. However, it invokes the integral triangle inequality $$ \left\|\int_a^b \mathbf f(t)\,dt \right\| \le \int_a^b \left\|\mathbf f(t)\right\|\,dt \tag{1}$$ with $\mathbf f = \gamma'$. Since the ...


0

Any linear transformation is determined by its values on a basis. As $T$ is defined on $\left \{ e_n \right \}_{n\in \mathbb N}$, the natural thing to do is to define $T$ on all of $V$ by $Tv=T(\sum_{i=1}^{n}c_ie_i)\doteq \sum_{i=1}^{n}c_iT(e_i),$ for any vector $v\in V$.You can check that this definition makes $T$ linear on $V$. Now, in your case, ...


2

It usually just means "suppose $T$ is linear". In other words $$T(\sum_{\alpha} x_{\alpha}e_\alpha) = \sum_{\alpha}x_\alpha T(e_\alpha).$$ The key point is that defining how a linear map behaves on a basis defines how it behaves on the entire space. Now, suppose that $T$ is bounded, so that $\|Tx\|\leq C\|x\|$ for some positive constant $C$. Note that ...


2

The first matrix norm is called the Frobenius norm, it's natural as in that it's the default Euclidean norm if the matrix were interpreted as a vector in $\mathbb{R}^{m \times n}$. The second matrix norm for $A \in \mathbb{R}^{m\times n}$ is the operator norm given by the linear operator $x \mapsto Ax$, it is naturally induced by the norms you choose for ...


1

Functional analysis has a lot to do with spaces of functions, as its name suggests (more so historically, but it still does). Using single bars for the norm of a function is ambiguous because $|f|$ also means the function $|f|(t) = |f(t)|$. Double bars $\|f\|$ eliminate the ambiguity. By the way, in some function spaces it is important to observe that ...


-3

This question is not true: $\mathbb{R}$ is homeomorphic to $(0,1)$. $\mathbb{R}$ is a complete metric space whit Euclidean norm and $(0,1)$ is a normed vector space which isn't complete!


2

In any metric space a compact sub space is closed and bounded. (The reciprocal is not true in general). Take $0\neq x\in F$ and $n\in\Bbb{N}$. One has $$\|n\cdot x\|=n\cdot\|x\|$$ And the sequence $y_n=n\cdot x\in F$ is not bounded ; therefore $F$ is not bounded henceforth not compact.


0

A convex function must be subdifferentiable on the relative interior of its domain (where it is finite). So the only points where one can have an issue is the boundary of its domain. The function constructed by Kevin Holt is one where all of the domain is boundary--hence Theorem 23.4 in Rockafellar's Convex Analysis cannot be used. As another illustration ...


0

An example in front: Let $T:X\hookrightarrow Y$ be the inclusion map, for $X$ a proper subspace of $Y$. Answer: An inverse exists and is bounded when restricting attention to the subspace $\operatorname{Im}T\subset Y$. Assume $Tx=0$, then the norm estimate implies $x=0$, hence $\ker T = \{0\}$. Thus $T$ defines a linear bijection from $T$ onto ...


0

Let $\|\cdot \|_1$ and $\|\cdot \|_2$ be equivalent norms on a normed vector space $S.$ That is, the norms generate the same topology. Then the norms are uniformly equivalent : There exist positive $A, B$ such that for all $v\in S$ we have $A\|v\|_1\leq \|v\|_2\leq B\|v\|_1$. Otherwise, with $(i,j)=(1,2)$ or $(i,j)=(2,1) ,$ there exists $T=\{v_n\}_{n\in ...


3

A slight correction, you'll need to define $$(Fg)(t) = \sin(t) +\int^t_{0} K(s,g(s)) ds;$$ that is, the upper bound needs to be $t$, not $T$. Take $g,h \in E$. We see \begin{align*} \lvert (Fg)(t)-(Fh)(t) \rvert &= \left \lvert \int^t_{0} [K(s,g(s)) - K(s,h(s))] ds\right \rvert \\ &\le \int^t_0 \lvert K(s,g(s)) - K(s,h(s)) \rvert ds \\ &\le ...


1

If you you expand $(T-\lambda I)^n$ you get $R+(-\lambda)^nI$ with $R$ compact, and now you can apply the same argument as before.


2

They are equal. Let $C$ be the closed convex hull of $A$, defined as the intersection of all closed convex sets containing $A$. (Closed half-spaces are enough here.) Let $B$ be the convex hull of $A$, defined as the intersection of all convex sets containing $A$. (Half-spaces, either open or closed, are not enough here.) Clearly $B\subset C$. Since $C$ is ...


0

This is because $\lvert x_i\rvert=\sqrt{x_i^2}\le\sqrt{\sum_{i=1}^nx_i^2}=\lVert x\rVert_2.$


0

Let $f\in X$. For the sake of simplicity I'll write $\int f$ for $\int_0^1 f(t)dt$. Let's use the scalar product on $X$ : $f\cdot g=\int fg$ $||Tf||^2=\int f^2 g2=f^2\cdot g^2$ so by the Cauchy Schwarz inequality, $||Tf||^4\le\int f^4\int g^4$. We are searching for $||T||=\sup||Tf||,||f||=1$ so let $f'=\frac{f}{||f||}=\frac{f}{\int f^2}$. Then $\forall ...


0

Your argument looks correct, although it's a bit long and drawn out. I might argue this way: From your definition we have $f_k$ rapidly Cauchy iff $$\sum_{k=1}^{\infty} |f_{k+1} - f_k|^{1/2} < \infty.$$ (Never heard of "rapidly Cauchy" before). Now $$f_{k+1} - f_k = \frac{(-1)^{k+1}}{k+1} - \frac{(-1)^{k} }{k} = (-1)^k\left ( -\frac{1}{k+1} ...


1

Pick $a_k = \frac{1}{2^k}$. Then $$ b_n = \frac{1}{n}\sum_{k=1}^n a_k = \frac{1}{n}\sum_{k=1}^n \frac{1}{2^k} = \frac{1}{n}\bigg(1 - \frac{1}{2^{n+1}}\bigg) > \frac{1}{2n}, $$ so $\sum b_n$ diverges


5

It fails for the sequence $1,0,0,\dots $


0

If $(X,d)$ is a metric space, each closed ball $\overline{B}(x,r):=\lbrace y \in X : d(x,y) \leq r \rbrace$ is a closed, i.e. $\overline{\overline{B}(x,r)}=\overline{B}(x,r)$. Recall the distance function $\mathrm{dist}(x,Y):=\inf_{y \in Y} d(x,y)$. His property is $d(x,Y)=0$ iff $x \in \overline{Y}$, since $x \in \overline{Y}$ iff $B(x,r) \cap Y \neq ...


0

The closed ball is always closed in $X$ independent of the dimension. Let $x_0\in W$, $r>0$ and $W$ as above. Let $(x_n)_{n \in \mathbb{N}} \subseteq W$ which converges in $X$ to $l\in W$. To verify the closedness of $W$ we need to check that $l\in W$. As $x_n \rightarrow l$ we can choose for every $n\in \mathbb{N}_{\geq1}$ a $m_n\in \mathbb{N}$ such that ...


3

You can construct the completion of a normed vector space $X$ by using a construction of the completion of a general metric space and then defining a normed vector space structure on that completion. But there's a better way, or at least a way that a lot of people might regard as simpler and more elegant: If $X$ is a normed vector space then the dual $X^*$ ...


2

I assume you want this $\forall K>0$, not all $t$. Since $f(t)^Tf(t)\geq0$, you can take $\beta=(\int_0^{\infty}f(t)^Tf(t)\,dt)^{1/2}.$


1

Just note that $\|f\|_K^2=\int_0^K f(t)^Tf(t) \, dt\leq \int_0^\infty f(t)^Tf(t) \, dt$ for all $K$, since $f(t)^Tf(t)$ is always nonnegative.


1

Note that $$ \left(\frac{x}{e^x}\right)'=\frac{e^x(1-x)}{e^{2x}} $$ vanishes at $x=1$ (which is a maximum of the function). Hence, $$ p(f)=\frac1e\sup_{\|v\|=1}f(v)=\frac1e\|f\|^*. $$


4

Notice that for each vector $x$, one has $$\|T^* T^2(x)\|^2 = \langle T^* T^2(x),T^* T^2(x) \rangle = \langle TT^*T^2(x), T^2(x)\rangle = \langle T^*T^3(x),T^2(x) \rangle = \langle T^3(x),T^3(x) \rangle = \|T^3(x)\|^2.$$ Thus $$\|T^3\| = \sup_{\|x\|=1} \|T^3(x)\| = \sup_{\|x\|=1}\|T^*T^2(x)\| = \|T^*T^2\|.$$


1

Diagonalization will help here. (When in doubt and working with normal matrices, try utilizing diagonalization!) Write $T = UDU^*$, then $T^* = UD^* U^*$, giving that $T^*T^2 = UD^*D^2U^*$. However $T^3 = UD^3 U^*$. Since unitary conjugation does not change the operator norm, this boils down to considering $D^*D^2$ and $D^3$. Here $Dg(x) = f(x)g(x)$ for ...


1

Yes. First $f$ is surjective. Indeed, let $y \in Y$. $\frac{y}{||y||} \in B_Y = f(B_X)$ hence there is $x \in B_X$ s.t $f(x)=\frac{y}{||y||}$ and so $f(||y||x)=y$ by linearity. We conclude that $f$ is surjective and hence bijective (we already know it is injective by assumption) Because of the facts that $B_Y = f(B_X)$ and that $f$ is bijective, we have ...


0

This is also exercise 1) page $111$ points a)-b) in Functional Analysis by Rudin. By a discussion in another network the proof it's correct. For second part, it is follows essentially from ii) of convex separation theorem, taking $\lbrace w_0 \rbrace$ as a compact set, and $\overline{B}_E$ as closed set, equivalently considering corollary in A).


1

The notion of equivalent norms that I have is the following: Two norms $||\cdot||_1,||\cdot||_2$ over a vector space $V$ are equivalent if there are constants $C_1,C_2>0$ such that for every $x\in V$, $$C_1||x||_1\leq ||x||_2\leq C_2||x||_1.$$ Is it the same definition that you have? If so, under this definition, we can show the norms $||\cdot ...



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