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17

A compact set must be bounded. Otherwise we can take $\{ x_n \}_{n=1}^\infty$ such that $\| x_n \| \geq n$. This will have no convergent subsequence, which we can prove by showing that it has no Cauchy subsequence. A compact set must be closed. Otherwise we can pick a sequence which converges to a point in the closure which is not in the set. This will ...


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 ...


14

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 ...


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 ...


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.


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.


10

Let me provide a detailed proof: Assume that $v_1,\ldots, v_n\in S$ are orthonormal functions, i.e., $\int_0^1 v_iv_j\,dx=\delta_{ij}$, and for a fixed $a=(a_1,\ldots,a_n)\in\mathbb R^n$ define $\varPhi_a :\mathbb R^n\to \mathbb R$ as $$ \varPhi_a(x)=\sum_{j=1}^n a_jv_j(x). $$ Then $$ \|\varPhi_a\|_{L^2}^2=\sum_{j,k=1}^n\int_0^1a_ja_kv_jv_k ...


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

Suppose that $$f(x)=E(x)+O(x),$$ for all $x$, where $E(x)$ is even and $O(x)$ is odd. Then for all $x$, $$f(-x)=E(-x)+O(-x)=E(x)-O(x).\tag{2}$$ Add. We get $f(x)+f(-x)=2E(x)$, and therefore $E(x)=\frac{f(x)+f(-x)}{2}$.


10

Preduals of $\ell_1$ are very interesting creatures. The question is sligthly ill-posed but let me comment on that anyway. We have quite a lot isometric preduals of $\ell_1$. Indeed, for any countably infinite ordinal number $\alpha$ (endowed with the order topology) we have $C_0(\alpha)^* \cong \ell_1$. This essentially follows from the ...


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's the way I like to think about it. I'll start with the finite dimensional space $\Bbb{R}^n$ because it looks like that's where you are, but I'll give an analogy for infinite dimensional spaces as well. The quantity $z^Tx$ represents a linear functional on $\Bbb{R}^n$, that is a linear function which eats a vector and spits out a real number: $$ ...


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 ...


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

By the hypothesis the Neumann series $\sum\limits_{n=0}^\infty T^n$ is absolutely convergent so convergent and we have $$(I-T)\sum\limits_{n=0}^\infty T^n=\left(\sum\limits_{n=0}^\infty T^n\right)(I-T)=I$$ so $I-T$ is invertible and its inverse is $\sum\limits_{n=0}^\infty T^n$, moreover we have $$\left\|\sum\limits_{n=0}^\infty ...


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

Here is an outline for proving your statement. $\ell^p$ is a space of (some) functions $\mathbb N\to\mathbb R$, and so is $L^p$ (with the choices indicated by OP). (I chose to work with the reals, but this choice is irrelevant.) The counting measure gives nonzero measure to all nonempty sets, so each equivalence class in $L^p$ only consists of one ...


8

The answer is: Not in general. For example, $\Omega=(0,2\pi)$, $u_n(x)=\sin nx$. Then $u_n\rightharpoonup 0$, since $$ \int_0^{2\pi} f(x)\,\sin nx\,dx\to 0=u, $$ for all $f\in L^2[0,2\pi]$. Meanwhile $$ v_n(x)=u_n^2(x)=\sin^2 nx=\frac{1}{2}-\frac{\cos (2nx)}{2}\rightharpoonup \frac{1}{2}=v, $$ and $u^2\ne v$. Note. If $\{u_n\}$ is bounded and ...


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

Literally, the claim is wrong, since the space has dimension $2^{\aleph_0}$, and there are Banach spaces with the same dimension (e.g. the $\ell^p(\mathbb{N})$ spaces). Since the rationals are dense in $[a,b]$, there is a linear injection of $C^\infty([a,b])$ into $\mathbb{R}^{\mathbb{Q}\cap [a,b]}$, and the latter space has cardinality ...


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

We know that, $L_\infty$ is not separable, so neither does its dual $L_\infty^*$. It is remains to recall that $L_1$ is separable.


7

No, consider $J\oplus_2 J^*$, where $J$ is a James space


7

This is the proof that I know. But it's not easy, but it is well known in the literature. $L^1$ has a property called cotype 2, that is, there is a constant $C>0$ such that for any $x_1,\dots,x_n \in L^1$ $$ E \left\|\sum_{k=1}^n \epsilon_k x_k \right\|_1 \ge C \left(\sum_{k=1}^n \|x_k\|_1^2\right)^{1/2} ,$$ where $\epsilon_k$ is a sequence of ...


7

Yes, this can be done without making the $C^*$-algebra concrete in $B(H)$. All we need is that $z=0$ iff $z^*z=0$, which follows from the $C^*$-identity $\|z^*z\|=\|z\|^2$. Lemma Let $a\in A$. The following assertions are equivalent. 1 - $p=a^*a$ is idempotent, i.e. $p^2=p$. 2 - $aa^*a=a$ 3 - $a^*aa^*=a^*$ 4 - $q=aa^*$ is idempotent. Proof ...


7

Note that $f'$ (and thus $|f'|$) is Riemann-integrable on $[0,1].$ Then, using the condition $$\int_0^1 f(x) \,dx = 0$$ we see $$\int_0^1 f(x)^2\, dx = \int_0^1 f(x)(f(x) - f(0))\, dx = \int_0^1 f(x)\int_0^xf'(t)\,dt\, dx \le \int_0^1 |f(x)| \int_0^x|f'(t)|\,dt\, dx.$$Similarly, $$\int_0^1 f(x)^2\, dx = \int_0^1 f(x)(f(x) - f(1))\, dx \le \int_0^1 |f(x)|\, ...


7

Another way to solve this problem is the following: first, let's prove a more general theorem. Assume that $f:\mathbb{R}\to\mathbb{R}$ is a bounded continuous function. Consider the problem $$-\Delta u=f(u),\ u\in H_0^1(D)\tag{1}$$ Let $F(x)=\int_0^x f(s)ds$ and $I:H_0^1(D)\to\mathbb{R}$ the energy functional associated with $(1)$, i.e. $$I(u)=\int_D ...


7

Note that you don't need uniform continuity, merely continuity (of course, these are equivalent since $[0,1]$ is compact) of $f$ on $[0,1]$ to apply Weiertrass' approximation theorem. You are on the right way. Approximate $f$ by a polynomial $p$ uniformly over $[0,1]$, so that $$\lVert f-p\rVert_\infty<\varepsilon$$ Since $p$ is a polynomial, ...



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