# Tag Info

1

Sorry to drag up such an old thread, but concerning the first question: For $E$ a Banach space, a theorem of Edward Odell states that the following assertions are equivalent: $\ \ \$1) $E$ contains no copy of $\ell_1$. $\ \ \$2) Every completely continuous operator on $E$ (to some Banach space) is compact. This result is referenced in Corollary 5 of ...

1

If you let $\mathcal{F} = \{ f_n | n\in \mathbb{N} \}$ then this set is not necessarily closed, as you say. However, it has compact-closure and so the sequence $f_n$ converges to a point in $\overline{\mathcal{F}}$ though not necessarily to a point in $\mathcal{F}$ itself.

3

Since $A^2$ is compact, it has an eigenvalue $\alpha$. As $A^2$ is self-adjoint, $\alpha$ is real. Let $\lambda\in\mathbb C$ with $\lambda^2=\alpha$. As $\alpha$ is an eigenvalue for $A^2$, we have that there exists nonzero $v$ with $(A^2-\alpha I)v=0$. Then $$0=(A^2-\alpha I)v=(A-\lambda I)(A+\lambda I)v=0.$$ If $(A+\lambda I)v=0$, then $-\lambda$ is ...

3

For this to hold it is crucial that $H$ is a complex Hilbert space since on $\mathbb{R}^2$ a rotation $T$ by $\pi/2$ has no eigenvalues, yet $T^2 = -I$ is compact and self-adjoint. The spectral theorem for compact self-adjoint operators yields an eigenvector $w$ of $A^2$ with real eigenvalue $\lambda$. Without loss of generality we can rescale $A$ ...

1

This is the first thought that I had. I do not see how it leads to what you want but maybe you have an idea with it: Since $A^2$ is a self-adjoint compact operator either $||A||^2$ or $-||A||^2$ is an eigenvalue for $A^2$.

2

$\alpha$ implies that $I - K$ is bijective, hence injective, giving in turn "not $\beta$". Not $\beta$ says that $I - K$ is injective, hence you get the bijectivity from 4. This gives $\alpha$.

3

Ad (1): Gram–Schmidt can readily be applied to any countable collection of non-zero vectors. Just remember how Gram–Schmidt orthonormalisation works: you simply replace $v_{k+1}$ with its orthogonal projection onto $\{v_1,\dotsc,v_k\}^\perp$, and then normalise (if non-zero). Now, suppose that $N(1-K)$ is infinite-dimensional, and that $\{v_k\}$ is a ...

2

Let $T_n :\ell_2 \to \ell_2 ,$ $T_n (x) =\left(x_1 ,\frac{x_2}{2} , \frac{x_3 }{3} ,...,\frac{x_n }{n} ,0,0,0,...\right) .$ Then $T_n$ is finite rank operator because $\mbox{dim} (T(\ell_2 )) =n<\infty .$ Moreover $$||T_n -T ||_{\ell_2 \to \ell_2} =\sup_{||x||_2 =1 } ||(T_n -T)(x) ||_2 \leq \sqrt{\sum_{j=n+1}^{\infty} \frac{1}{j^2}}\to 0.$$ Hence $T$ is ...

1

As continuous functions map compact sets into compact sets, composition of a bounded operator with a compact operator is compact. So, if $T$ is compact and invertible, so are $I_X=T^{-1}T$, $I_Y=TT^{-1}$. Now if $I_X$ is compact, it maps the unit ball into a compact set, itself. So the unit ball of $X$ is compact, which implies that $X$ is ...

1

If $T$ is invertible then the image of the open unit ball in $X$ is open in $Y$ (it is the preimage of the ball under $T^{-1}$). If $Y$ is infinite-dimensional, the closure of this image cannot be compact (by Riesz's lemma, as you mentioned). Therefore $Y$ must be finite-dimensional. From there $X$ clearly must be finite-dimensional as well.

1

Note that $T$ is a multiplication operator (multiplication by the identity function). It is not hard to check that the spectrum of a multiplication operator is the closure of the range of the corresponding function. That is, the spectrum of $T$ is $[0,1]$. So $T$ is not compact, as the spectrum of a compact operator can only accumulate at $0$.

2

Do you know that if $T$ is compact and $f_n \rightharpoonup f$ then $Tf_n \to Tf$? There exists a sequence such that $f_n \rightharpoonup 0$, $\|f_n\|_2 = 1$ and $\mbox{supp} f_n \subset (\frac{1}{2},1)$. To construct this sequence, you could just take an orthonormal basis of $L^2(\frac{1}{2},1)$, extend it by $0$ to $(0,1)$ and re-normalize (if you are not ...

1

You can choose $0<\delta\leq\varepsilon$ for $x,z\in [0,1]$, $|x-z|<\delta \Rightarrow |f(x,y)-f(z,y)|\leq\varepsilon$, $y\in [0,1]$, because $f$ is uniformly continuous in $[0,1]\times [0,1]$. So in your case, $$|Tv_n(x)-Tv_n(z)|\leq c\varepsilon+Mc\delta\leq c(M+1)\varepsilon.$$

2

Your professor did not make a mistake. A diagonalization argument will show that if $X^*$ is separable, then any bounded sequence in $X$ has a weakly Cauchy subsequence (c.f. Joeseph Diestel, Sequences and Series in Banach Spaces, pg. 200). A completely continuous operator maps weakly Cauchy sequences to norm convergent sequences. See this post for a ...

1

Ad 1., yes. A subset of a metric space can be precompact (totally bounded) without its closure being compact (which for metric spaces coincides with being sequentially compact) if the space is not complete. The equivalence that every sequence in $A$ has a convergent (in $Y$) subsequence if and only if $\overline{A}$ is (sequentially) compact also holds in ...

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