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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 ...


5

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


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 $$\|T^3\| = \sup_{\|x\|=1} \|T^3(x)\| = \sup_{\|x\|=1}\|T^*T^2(x)\| = \|T^*T^2\|.$$


3

It is not true that $T_n \to I$. They only converge pointwise, but not in the operator norm. Here is an easier proof: Let $Y$ be an incomplete normed space and let $X$ be the scalar field. Then $B(X,Y)$ is isometric to $Y$, so it is incomplete.


3

Suppose each $f_n$ is bounded, i.e. for each $n$, we have $\|f_n\|=\sup_{\|x\|=1}|f_n(x)|<\infty$. Define $T_n:X\to \ell^1(\mathbb N)$ by $x\mapsto \sum_{i=1}^n f_i(x)e_i$, where $\{e_i\}$ is the canonical basis for $\ell^1(\mathbb N)$. If $\|x\|=1$ then $$\sup_n\|T_n x\|=\sum_{i=1}^\infty|f_i(x)|<\infty, $$ so by Banach-Steinhaus we have ...


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 ...


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 $\beta=(\int_0^{\infty}f(t)^Tf(t)\,dt)^{1/2}.$


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 ...


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 ...


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 ...


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 ...


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.


2

What you did for the first part is correct. For the converse, however, there is a problem. You have to prove that the sequence $\left(M_n\right)_{n\geqslant 1}$ is bounded. Your $K$ depends on $X$, and the task is to prove that $K$ is finite. You can solve the question using the uniform boundedness principle: since $Tx$ belongs to $\ell^\infty$ for each ...


1

Here is a solution, sorry that it is rather long, I'm sure a lot can be cut. Let $x_1,x_2 \in [y_1,y_2]$, then since $[x_1,x_2]$ is the convex span of $x_1$ and $x_2$ you have $[x_1,x_2] \subset [y_1,y_2]$. Similarly $y_1,y_2 \in [x_1,x_2]$ implies $[y_1,y_2] \subset [x_1,x_2]$. In the following $\alpha, \tilde \alpha, \beta, \gamma \in [0,1]$. Assume ...


1

We can. Suppose $T$ is not continuous, then $T$ is not bounded, so there is some sequence $(x_n)$ in $E$ such that $T(x_n) \rightarrow \infty$ as $ n \rightarrow \infty$. We can assume WLOG that all $T(x_n) \neq 0$. We surely have some $p \notin \ker{T}$ (otherwise $T \equiv 0$), and so we can define $y_n = p - \frac{T(p)}{T(x_n)}x_n$, where the fraction is ...


1

Suppose $T$ is not continuous. So $T$ is not bounded. i.e. $\exists$ a sequence $x_n$ such that $T(x_n) \to \infty$ as $n\to \infty$. Let $a\notin \ker T$. Then defining $$x_n' = a - \frac{T(a)}{T(x_n)}x_n ,$$ it is clear that $T(x_n') = 0$ and so $x_n'\in \ker T$. Also $x_n' \to a \notin \ker T.$ So $\ker T$ is not closed. Hence $\ker T$ closed implies ...


1

Denote by $y = (y^n)_{n \in \mathbb{N}}$ the sequence $y = (1, \frac{1}{2}, \frac{1}{3}, \dots)$ (we will use upper indices for the terms of an element in $\ell^{\infty}(\mathbb{N})$ in order to not confuse ourselves when considered sequences of elements in $\ell^{\infty}(\mathbb{N})$). Then we have $$ \left( y - \sum_{k=1}^n x_{\pi(k)} \right)^i = \left( y ...


1

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 ...


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 ...


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.


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 $T$.


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 ...


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.


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


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

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, $$ p(f)=\frac1e\sup_{\|v\|=1}f(v)=\frac1e\|f\|^*. $$



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