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4

It is a bit of a moot point whether you include separability in the axioms for a Hilbert space. If you do then you get the notion of the Hilbert space (e.g., as represented by $l_2$) that you will find in the early literature on the subject. However, I think it is more usual these days not to require separability, so we tend to talk about Hilbert spaces. In ...


3

Sometimes is good to look at the simple examples first. Let $X=Y=\mathbb C^2$, and $$ f=\begin{bmatrix}1&1\\0&1\end{bmatrix}. $$ Take $X_0=\left\{\begin{bmatrix}x\\0\end{bmatrix}:\ x\in\mathbb C\right\}$, and $Y_0=f(X_0)=X_0$. Now you can check that $$ f'(Y_0)=\left\{\begin{bmatrix}x\\ x\end{bmatrix}:\ x\in\mathbb C\right\}\not\subset X_0. $$


2

If you know about compact selfadjoint operators, then you can study the operator $$ Lf = \frac{1}{i}\frac{d}{dt} $$ on the domain $\mathcal{D}(L)$ consisting of all absolutely continuous functions $f$ on $[0,2\pi]$ with $f(0)=f(2\pi)$ and $f' \in L^{2}[0,2\pi]$. $L$ is not compact, but its resolvent operator $(\lambda I-L)^{-1}$ ...


2

As Daniel Fischer noted, condition (P) is equivalent to the norm being strictly convex. Indeed, if the norm is not strictly convex, then the unit sphere contains a line segment, and removing the midpoint of this line segment from the closed unit ball creates a nonconvex set. Conversely, if the norm is strictly convex, then any $X$ as you described is ...


2

HINT: Use the equality: $$||\frac{x_n - x_m}{2}||^2 + ||\frac{x_n + x_m}{2}||^2= 1/2\,(\,||x_n||^2 + ||x_m||^2)$$ $M$ is convex so $\frac{x_n + x_m}{2}$ also have norm approaching $d$.


2

It is often difficult to prove that an operator is self-adjoint, which seems reasonable since you know a lot about an operator when it is self-adjoint. For instance it is unitarily equivalent to a multiplication operator by a real function. There are many possible approaches, one which is sometimes useful is the following Show that $D$ is symmetric, i.e. ...


2

This is not true: $$f\in L^1(\kappa)\iff f\in L^1(\lambda).$$ However, from $\kappa(E)\le c\lambda(E)$, you can deduce that $$ f\in L^1(\lambda)\quad\Longrightarrow\quad f\in L^1(\kappa). $$


2

There's an easy negative answer regardless to the way you want to think about it as a first-order structure: If $\kappa$ is an infinite cardinal such that $\operatorname{cf}(\kappa)=\aleph_0$ then there is no Banach space of cardinality $\kappa$. This is due to the Baire Category Theorem, and the fact that any subspace of smaller dimension is nowhere ...


2

I think that you are confusing basis with complete orthonormal set. The Fourier modes $\{e^{inx}\}_{n\in\mathbb{Z}}$ are a complete orthonormal set in $L^2[-\pi,\pi]$. This means that for any $f\in L^2$, its Fourier series converges in the $L^2$-norm to $f$. But it is not a basis of $L^2[-\pi,\pi]$. $L^4$ is not a Hilbert space, and it does not make sense ...


2

If you start with a unit vector $v \in \mathcal{H}$, then you can define a subspace $\mathcal{H}_{v}$ as the closure in $\mathcal{H}$ of all polynomials in $N$, $N^{\star}$ acting on $v$. $\mathcal{H}_{v}$ is invariant under $N$, $N^{\star}$. This defines a unitary map $$ \mathcal{F}_{v} : \mathcal{H}_{v}\rightarrow L^{2}_{\mu_{v}} $$ where ...


2

I take it by "complete" you mean that the $f_n$ have dense linear span or, equivalently, they have zero orthogonal complement. This we can see as follows. Suppose that $\langle g, f_n\rangle =0$ for all $n$. Then $$ |\langle g, e_n\rangle | =|\langle g, e_n-f_n\rangle | \le \|g\|\, \|e_n-f_n\| . $$ Take squares and sum over $n$. This yields $\|g\|^2 \le S ...


2

I believe I have a counterexample to your statement about the equality $\|x\|=\|\bar{x}\|$. Let $S=\{a, b\}$ be a set with two entries. Define a complex-valued function $f$ on $S$ by $$f(a)=1+i, \quad f(b)=1$$ and consider the two-dimensional space $V=\operatorname{Span}\{f, \bar{f}\}$ over $\mathbb{C}$ equipped with the inner product $$(\phi, ...


2

I think your is just a matter of definition. A real vector space is an abelian group $(V, +)$ endowed with a scalar multiplication $\mathbb{R} \times V \longrightarrow V$, satisfying some axioms (associativity, distributivity, etc...). Given a vector space, its elements are called vectors. In particular, some function spaces (for example $L^2([0,1])$ and ...


2

A Hilbert space has been identified with its dual, if at places where one would expect an element from $H^*$ instead an element of $H$ is written. Let me explain this with some examples. (1) Consider a function $f:H\to \mathbb R$. If $f$ is Gateaux differentiable, then its first derivative $f'(x)$ is an element in $H^*$: it maps a direction (in $H$) to an ...


2

In finite dimensional complex spaces, you can find a branch of complex log function that doesn't intersect any of the eigenvalues. And that offers one explanation of why everything works so nicely. You can see it by using Cauchy's integral formula $$ \log(A) = \frac{1}{2\pi i}\oint_{C} \frac{1}{\lambda I-A}\log(\lambda)\,d\lambda. $$ You can prove that ...


2

I am not a fan of the term 'hermitian', because it is so often used haphazardly (sometimes it means symmetric, sometimes it means self-adjoint, sometimes it presupposes that the vector space is finite-dimensional). The term 'self-adjoint' seems to be used much more consistently, to mean a densely defined operator which satisfies $A=A^*$ (equivalently $A$ is ...


1

The commutation relation $[A,B]=iI$ (i.e., $AB-BA=iI$) is not satisfied by two linear operators on any finite-dimensional linear space. So Quantum Mechanics requires infinite-dimensional spaces. Even in infinite-dimensional spaces, the commutation relation cannot be satisfied by two bounded operators $A$ and $B$; at least one of them must be discontinuous. ...


1

Note that $$|\langle Ax, y\rangle |\le \|Ax\|\cdot \|y\|.$$ So, $$\frac{|\langle Ax, y\rangle |}{\|x\|\cdot \|y\|}\le \frac{\|Ax\|}{\|x\|}.$$ Thus $$||L|| = \sup_{x,y \in H, x,y \neq 0} \frac{|\langle Ax, y\rangle|}{||x||\cdot ||y||}\le \sup_{x\in H, x \neq 0}\frac{\|Ax\|}{\|x\|}. $$ On the other hand, for $x=y$ you get the desired equality.


1

I'm going to assume the facts that I proved in a previous answer to a question of yours: Spectral Measure Integration: Unbounded Functions? For any Borel function $f$, I defined an operator $T_{f}$ on $\mathcal{D}(T_{f})$ consisting of all $x \in \mathcal{H}$ for which $\int |f(\lambda)|^{2}\,d\|E(\lambda)x\|^{2}<\infty$. I showed that $T_{f}$ is a ...


1

In the finite case every operator is bounded and since the domain is the whole Hilbert space, every hermitian operator is self-adjoint. For the infinite case, the good definition of a hermitian operator is a densely defined operator $T$ such that $T\subset T^*$, i.e., it is symmetric and the adjoint $T^*$ is an extension, we have ...


1

Suppose that $(f,f_{n})=0$ for all $n$. Then $f=0$ because, if not, $$ \|f\|^{2}=\sum_{n}|(f,g_{n})|^{2}=\sum_{n}|(f,g_{n}-f_{n})|^{2} \le \sum_{n}\|f\|^{2}\|f_{n}-g_{n}\|^{2}< \|f\|^{2}. $$ Therefore the orthonormal set $\{ f_{n}\}$ must be complete because there is no non-zero vector $f$ which is orthogonal to every $f_{n}$.


1

This is impossible. For a proper convex and lower semicontinuous function the domain of definition of the subdifferential is dense in the domain of $f$, $$ \mathrm{dom}(f) = \overline{ D(\partial f)}. $$ This implies that there is $x\in X$ such that $\partial f(x)\ne\emptyset$, as for proper convex and lower semicontinuous $f$ it holds $\mathrm{dom}(f) \ne ...


1

Suppose $E$ is a spectral measure on $\mathbb{C}$. Choose any unit vector $x \in \mathcal{H}$, and look at the probability measure $\mu_{x}(S)=\|E(S)x\|^{2}$. If there are countably many disjoint Borel subsets $\{ S_{j}\}_{j=1}^{\infty}$ with $\mu_{x}(S_{j})\ne 0$, then you can construct a Borel function $f$ such that $\int |f|^{2}d\mu_{x} =\infty$. Then ...


1

Suppose $H$ has a countably dense subset $\{ s_{n}\}_{n=1}^{\infty}$ of non-zero vectors. Using induction we may discard each $s_{n}$ which is a linear combination of the previous vectors in the sequence in order to arrive at a new sequence $\{ t_{n}\}_{n=1}^{\infty}$ of vectors for which the $\{ t_{1},t_{2},\cdots,t_{k}\}$ is linearly independent for each ...


1

Suppose that $a\ne0$, and consider first the case of continuous functions $f\in L^2([0,1])$, in this case $(e^{at}\phi(t))'= e^{at}f(t)$. So, since $\phi(0)=0$, $$e^a\phi(1)=\int_0^1e^{at}f(t)dt\tag{1}$$ Now, integrating $\phi'+a\phi=f$ on $[0,1]$, we get $$\phi(1)+a\,l(f)=\int_0^1f(t)dt\tag{2}$$ Combining $(1)$ and $(2)$ we obtain $$ ...


1

Let $F$ be the scalar field. Since $f\ne 0$, there is a $v\in H$ such that $f(v)\ne 0$; let $\alpha=f(v)$. For surjectivity let $\beta\in F$ and consider $$f\left(\frac{\beta}{\alpha}v\right)\;.$$ For the rest, note that if $\dim H>1$, then $\ker f\ne(0)$ (why?), so pick any non-zero $u\in\ker f$ and consider $f(u+v)$.


1

Another example of a vector space of functions is $P_2(t)$, the set of all polynomials with real coefficients, in the variable $t$, of degree at most $2$. That is $$P_2(t)=\{a+bt+ct^2:a,b,c\in\mathbb{R}\}$$ This space is isomorphic with $\mathbb{R}^3$, via $\phi:a+bt+ct^2\to (a,b,c)$.


1

First $\|A\|^{2} \le \|A^{\star}A\|$ holds because $$ \|Ax\|^{2}=(Ax,Ax)=(A^{\star}Ax,x) \le \|A^{\star}Ax\|\|x\| \le \|A^{\star}A\|\|x\|^{2}. $$ Consequently, $\|A\|^{2}\le \|A^{\star}A\|\le \|A^{\star}\|\|A\|$ gives $\|A\|\le \|A^{\star}\|$. Replacing $A$ by $A^{\star}$ in this inequality gives $\|A^{\star}\|\le\|A\|$; hence, $\|A\|=\|A^{\star}\|$. ...


1

If $H$ is finite dimensional then $A$ must be finite - else keep adding points to $A$ to maintain the $\|x-y\|$ property until $A$ is maximal. Then the set of open balls of radius $\epsilon$ around elements of $A$ is an open cover of the unit sphere. If $H$ is infinite-dimensional we should be able to use the Riesz Lemma to construct a counter example for ...


1

Not true. Take $f(x,y)=x+iy$ and $X$ the space spanned by $f(x)$ and $\overline{f}(x,y)=x-iy$ over $\mathbb C$. Now, every element o $X$ is of the form $u=af+b\overline{f}$. Define the inner product $$ \langle u_1,u_2\rangle=a_1\overline a_2+4 b_1\overline b_2. $$ Then $$ \|\,\overline f\|=2\|\,f\|. $$



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