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25

No, such a bijection from the unit interval $I$ to the unit square $S$ cannot exist. Since $I$ is compact and $S$ is Hausdorff, a continuous bijection would be a homeomorphism. But in $I$ there are only two non-cut-points, whereas in $S$ each point is a non-cut-point.


17

$\Bbb R[x]$, the polynomials in one variable. All the continuous functions from $\Bbb R$ to itself. All the differentiable functions from $\Bbb R$ to itself. Generally we can talk about other families of functions which are closed under addition and scalar multiplication. All the infinite sequences over $\Bbb R$. And many many others.


14

Fredholm theory, originally developed for studying (systems of) differential equations, had been around for several decades when Gel'fand et al., working in the '50s, noticed that the Fredholm index was homotopy invariant, i.e., if $F_t$, $a \leq t \leq b$ is a continuous path of Fredholm operators, then $\operatorname{Index}(F_t)$ is constant in $t$, and so ...


13

If $V$ is a Banach space we call $V'$ the dual space (see continuous dual space on wikipedia), i.e. the space of linear continuous functionals $\xi \colon V\to \mathbb R$. Then it is well known that there exists a natural injection $$ J \colon V \to V'' $$ defined by $$ J(v)(\xi) = \xi(v) $$ for all $\xi \in V'$. We know that $J$ is an isometry, in ...


12

Take $$f\left(x\right)=\begin{cases} x^{-1/2}, & x\in\left(0,1\right]\\ 0, & \text{otherwise} \end{cases} $$ This function is obviously in $L^1$; note also that $f\ast f$ is $0$ on $(-\infty,0] \cup [2,+\infty)$, $\pi$ on $(0,1]$, and it decays from $\pi$ to $0$ continuously on $[1,2]$. Therefore, $f \ast f(x)$ is everywhere defined, in the sense ...


12

I submit this counterexample which, in my opinion, proves that the statement is false. In the vein of this MathOverflow post by Gerald Edgar, let $V$ denote the real vector space of all polynomials of one variable and let $$\lVert P\rVert=\max_{x\in[0, 1]} \lvert P(x)\rvert,\qquad \forall P\in V.$$ Moreover, let $$W=\{a_0+a_2x^2+a_4x^4+\dots+a_{2k}x^{2k}\ ...


12

First, a general remark. A common way to prove that some operator $T$ is not compact is to exhibit an infinite-dimensional subspace $M$ on which $T$ has a lower bound: that is, there exists $c>0$ such that $$\|Tx\|\ge c\|x\|,\quad \forall\ x\in M \tag{1}$$ If (1) holds, then the image of unit ball under $T$ contains a ball of radius $c$ in the ...


12

In short the list of reasons is: 1) The modern trend in math today is to develop non-commutative analogues of well known theories. Vaguely speaking operator spaces are normed spaces over "non-commutative" scalars, in fact over matricies. 2) There were several long standing problems, that were solved via methods of operator space theory. As soon as a ...


11

We can finish the argument as follows. (Note: We'll assume that the limit in question exists for $f$ and establish that it's equal to $f(1)$. Technically, we should prove that this limit exists as Peter Tamaroff notes below (thanks!). A minor modification of the following argument simultaneously establishes the existence of the limit and its value but we'll ...


11

The problem is that an element of a Hamel basis might be an infinite linear combination of the other basis elements. Essentially, linear dependence changes definition.


11

The reason is that we would like to define a norm by the following formula: $$ \|f\|_p:=\left(\int_X |f|^p d\mu \right)^{1/p}. $$ Therefore, we need to have triangle inequality (Minkowski's Inequality) which is available only for $p\geq 1$. See more details in Chapter $6$ of Folland's book: Real Analysis.


10

For any $f \in L^1$ we have $$\lim_{\omega \to \infty} \underbrace{\int_{\mathbb{R}} f(x) \cdot e^{-\imath \, x \omega} \, dx}_{=:\hat{f}(\omega)} = 0, \tag{1}$$ this result is known as Riemann-Lebesgue lemma. As $f \in L^1$ we can choose a sequence $(f_n)_{n \in \mathbb{N}}$ of simple functions such that $f_n \stackrel{L^1}{\to} f$. Since ...


10

Your question is equivalent to the following: does every pre-Hilbert space (= inner product space) $M$ admit an orthonormal basis? It turns out that the answer is "no". A counter-example can be found in N. Bourbaki's Topological Vector Spaces, Exercise V.2.2. I tried to solve the exercise, but I'm not quite sure that I understood it. Let ...


10

Here is a more elementary method than you proposed: First, note that if $f$ is continuous on $[0,1]$, then it is necessarily bounded on $[0,1]$; say $\lvert f(x)\rvert\leq M$ for all $x\in[0,1]$. If we define $\delta_n:=\frac{1}{\sqrt{n}}$, then $$ \left\lvert n\int_0^{1-\delta_n}f(x)x^n\,dx\right\rvert\leq ...


10

Unfortunately, taking $\mathcal{H}=H^2(\mathbb{R}^3)$ won't work. Indeed, the Laplacian of an $H^2$ function needs not be $H^2$, but the Hamiltonian should be an operator from $\mathcal{H}$ to $\mathcal{H}$. The only mathematically sound way out is taking $\mathcal{H}=L^2(\mathbb{R}^3)$ and considering the Laplacian as an unbounded operator, meaning that ...


9

By the Cauchy Schwarz Inequality, for any integrable function $f(x)$: $\displaystyle\left(\int_a^b f(x) \cdot f(x)^2\,dx\right)^2 \le \left(\int_a^b f(x)^2\,dx\right) \left(\int_a^b (f(x)^2)^2\,dx\right)$ $\displaystyle\left(\int_a^b f(x)^3\,dx\right)^2 \le \left(\int_a^b f(x)^2\,dx\right) \left(\int_a^b f(x)^4\,dx\right)$ But by the given conditions, we ...


9

Suppose that $R^2=T$. Then $\ker R\subseteq\ker T=\Bbb Re_0$, where $e_0=\langle 1,0,0,\ldots\rangle$. Clearly $\ker R$ is non-trivial, so $\ker R=\ker T$. Moreover, $T$ is surjective, so $R$ must also be surjective. In particular, $e_0=Rx$ for some $x\in\ell^1(\Bbb N)\setminus\ker T$, and therefore $R^2x=Re_0=0\ne Tx$. Now suppose that $R^2=S$, and let ...


9

$\mathbb Q(\sqrt 2)$ has two norms which are equivalent on $\mathbb Q$ derived from the two embeddings into $\mathbb R$


9

Hint: Consider what happens to the connected $[0,1]$ if the point $\frac12$ is removed. What happens to $[0,1]\times[0,1]$ when $f(\frac12)$ is removed?


8

The space of continuous functions of compact support on a locally compact space, say $\mathbb{R}$. The space of compactly supported smooth functions on $\mathbb{R}^{n}$. The space of square summable complex sequences, commonly known as $l_{2}$. This is the prototype of all separable Hilbert spaces. The space of all bounded sequences. The set of all linear ...


8

Let $e \in C_0(X)$ be the unit. For each $x \in X%$ there is a $f \in C_0(X)$ with $f(x) \ne 0$. As $e(x)f(x) = f(x)$ we must have $e(x) = 1$. So $e$ is the constant function $e=1$. But $1$ "vanishes at infinity" only if $X$ as compact: By definition there is a compact $K$ such that $1 = |1(x)| \le \frac 12$ outside $K$, i. e. on $X \setminus K$, so we must ...


8

You can apply a theorem of Grothendieck to the closure of $S$ in $L^2$ which is (as you show) contained in $C([0,1]) \subseteq L^\infty$. Grothendieck's theorem says that every closed subspace of $L^p(\mu)$ (where $\mu$ is a probability measure on some measurable space and $0<p<\infty$) which is contained in $L^\infty(\mu)$ is finite dimensional. A ...


8

Even worse. You can define a $L^p$-distance but the space is not locally convex. See "Examples of spaces lacking local convexity" in http://en.wikipedia.org/wiki/Locally_convex_topological_vector_space.


8

You do not need the axiom of choice to prove "There exists a normed space with a discontinuous linear functional". Consider the space $c_{00}$ of all sequences of real numbers with only finitely many nonzero terms, with the supremum norm. The functional $f(x) = \sum_{n=1}^\infty n x(n)$ is discontinuous. Everything in this argument is completely explicit ...


8

This is the operator-valued version of the fact that the map $$x\mapsto \frac{1-ix}{1+ix}$$ transforms the real line onto the unit circle. In operator terms, the real line becomes the set of self-adjoint operators. And the unit circle is the group of unitary operators. So, let $T$ be self-adjoint. Then $(I\pm iT)^*=(I\mp iT)$, which implies ...


8

Maybe a good point to start is this useful corollary of Baire Cathegory Theorem the cardinality of an Hamel base of a Banach Space can be finite or uncountable. It can't be countable The proof is a delightful application of Baire theorem. Now to give an explicit example, we can consider the space $\ell^2 $ which has the standard base $ M:=$ $\lbrace ...


8

Let $T\in J$ where $J$ is a closed two-sided ideal in $B(H)$. Consider its polar decomposistion $T=U|T|$, where $|T|=(T^*T)^{1/2}$. Clearly $T^*T\in J$, so its square root $|T|$ is in $J$ too (granted, we use some small bit of functional calculus there). Therefore, $T^*=|T|U^*\in J$ as well. Edit: As commented by Martin Argerami, this proof works for norm ...


8

First, note that $$\int_0^1 x^n f(x)dx\to 0$$ since $f$ is bounded, so we can prove that $$(n+1)\int_0^1 x^n f(x)dx\to f(1)$$ But note $$\left( {n + 1} \right)\int_0^1 {x^n}f (1)dx = f(1).$$ so it suffices to consider the case $f(1)=0$. THM Suppose that $f:[0,1]\to \Bbb R$ is continuous and $f(1)=0$. Then $$\mathop {\lim }\limits_{n \to \infty } ...



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