I would like to show that a bounded linear operator $T: \ell^2(\mathbb{N}) \rightarrow \ell^1(\mathbb{N})$ cannot be surjective.  If we assume surjectivity, then by the Open Mapping Theorem, $T$ is an open map, i.e. for every open $O \subset \ell^2(\mathbb{N})$, its image $T(O)$ is open in $\ell^1(\mathbb{N})$.   Is it then possible to contradict Baire's Theorem by constructing a sequence of open dense subsets of $\ell^1(\mathbb{N})$, whose intersection is not dense?  Any help pointing me into the right direction would be appreciated.
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The second part of an exercise asks that I show:     There does not exist a surjective bounded linear map from $\ell^2(\mathbf{N})$ to $\ell^1(\mathbf{N})$.     The first part was to show that these two spaces are separable in their respective $L^p$ norms.    I've really hit a wall on this one.  I know, in fact, that there does exist a surjective bounded linear map from $\ell^1(\mathbf{N})$ to $\ell^2(\mathbf{N})$.  So my intuition says that if there were also one going the other way, then the two spaces would have to be isomorphic, which is false .  But I don't know of any such theorem that I could apply to "the category of Banach spaces".    My only other idea is to argue by contradiction using the open mapping theorem, but nothing is coming to me...  My guess is that separability is also going to be important, given the first part of the exercise.  -Thanks.
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My professor has stated the following lemma:  Let $U$ and $V$ be two vector spaces and $B$ a basis for $U$. If $f:B\rightarrow V$ is a map, then there exists a unique linear map $A: U \rightarrow V$ such that $A(x)=f(x)\ \forall\ x\in B$  How can I prove this lemma? (This is not a homework problem.)
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Let $\Omega$ be a bounded open subset of $\mathbb{C}$ and $f:\Omega\rightarrow\Omega$ be holomorphic in $\Omega$. Prove that if there exists a point $z_0$ in $\Omega$ such that $$f(z_0)=z_0~~~~\text{and}~~~~f'(z_0)=1$$ then $f$ is linear.  Please give some hints. Thanks in advance!
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For a real Banach space $X$ let $A:X\rightarrow X^*$ be a positive operator in the sense that $(Ax)(x)\geq 0$ for all $x\in X$. Show that $A$ is bounded.   I don't know how to do that, maybe it's an application of the closed graph theorem?
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If $X$ is a Banach space, and $T:X \to X$ is a bounded linear operator with norm &lt; $1$, then $I-T$ has a bounded inverse defined by $(I-T)^{-1} = \sum_{n=0}^\infty T^n$.   Thinking in terms of a converse, if $T$ is any bounded linear operator defined on $X$, then does the existence of a bounded inverse $S=(I-T)^{-1}$ imply that $S$ can be represented as $S=\sum_{n=0}^\infty T^n$?
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This is a variation of the  questioned some time ago.  For a complex Hilbert space $H$ let $T: H \rightarrow H$ be a bounded operator. We call that $T$ is a trace-class operator if the following sum  $$\sum_{i}{\langle |T|e_{i}, e_{i} \rangle} &lt; \infty$$  converges, where $|T| = (T T^{*})^{\frac{1}{2}}$ is the absolute value of the operator.  Assume that  $$\sum_{i}{\langle Te_{i}, e_{i} \rangle}$$  converges for any basis in the space $H$. How to prove that if the aformentioned property holds then the operator is a trace class operator?  The progress on the problem is the following: Given an arbitrary bounded operator $T: H \rightarrow H$, one can use the following decomposition $$T = \big( \frac{T + T^{*}}{2} \big) ^{*} + i \big( \frac{T^{*} - T}{2i} \big) ^{*}$$  The latter line gives the decomposition $$T = A + i B$$ where $A, B$ are normal operators.  For the normal operators we can apply the spectral theorem that proposes that $T$ is unitary equivalent to  $$(UT U^{-1})(f(x)) = g(x) f(x)$$  where  $$U: H \rightarrow L^{2}(X, \mu)$$ $$g \in L^{\infty}(X, \mu)$$  Though this decomposition classify the operator in a broad sence, i see no direct way to conclude the statement. Are the any hints that may extend the previous argument? If not, are there any ways to conclude the statement?
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Let $V,W$ be normed vector spaces, and $L(V,W)$ be the space of bounded linear operators. Usually I would only see the statement "If $W$ is Banach, then $L(V,W)$ is Banach.". But  writes that there is a converse: "If $L(V,W)$ is Banach, and if $V$ is non-trivial, then $W$ is Banach". This is pretty interesting since I never seen a converse before. I was wondering if anyone has a nice proof.  I tried reversing the proof for the usual direction, but the inequalities can't be reversed. I also tried to start with a Cauchy sequence in $W$, and construct linear operators (using Hahn-Banach to control the operator norms), but alas I cannot say much about the distance between these operators (much less say that it is Cauchy)
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So I've got a bijection. It clearly has an inverse, but how exactly do I prove that the inverse is a linear map as well?     Suppose that the linear map $T:U\to V$ is a bijection. So $T$ has an inverse map $T^{-1}:V\to U$. Prove that $T^{-1}$ is a linear map.
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On a finite dimensional vector space, the answer is yes (because surjective linear map must be an isomorphism). Does this extend to infinite dimensional vector space? In other words, for any linear surjection $T:V\rightarrow V$, AC guarantees the existence of right inverse $R:V\rightarrow V$. Must $R$ be linear?  How about $T:V\rightarrow W$ linear surjection in general?
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I want to prove that the geometric Hahn Banach theorem implies the analytic one.   Edit:  To avoid confusion I will state the vesion of H.B theorems im familiar with:  Analytic H.B:   Let $X$ be a linear space(over $\Bbb R$) and $Y\subset X$ a subspace. Let $p:X \to \Bbb R$ be a sub-additive function. Suppose $f:Y\to \Bbb R$ is a linear map s.t $f(y) \le p(y)$ for all $y\in Y$ then there exists an extention $g:X\to \Bbb R$ of $f$ s.t. $g(x)\le p(x) $ for all $x\in X$.  Geometric H.B:  Let $X$ be a linear space. $K \subset X$ convex s.t. each point in $K$ is an internal point. Let $D$ be a plane disjoint from $K$ then there exists hyperplane that contains $D$ and disjoint from $K$.  In my problem $X$ is NOT a normed space so there are no open sets.   Given $X$ a linear space and $Y\subset X$ a subspace $p:X\to \Bbb R$ sub-additive, and $f:Y\to \Bbb R$ linear s.t $f(y)\le p(y)$ for $y\in Y$ we need to extend $f$ to $g:X\to \Bbb R $ and $g(x)\le p(x)$ fo r all $x\in X$  So, we look at $X \times \Bbb R $and define $K = \{(x,t) : t&gt;p(x)\}$.  The fact that $K$ is convex is easy. How can I show directly that every point in $K$ is internal? (can't say that $K$ is open).  Now after showing that, we can look at $Graph(f)$ and observe that $Graph(f)\cap K = \emptyset$.  So by the geometric H.B theorem we have a hyperplane (which is a maximal subspace in this case because $(0,0)\in Graph(f)$ ) M containing $Graph(f) $ and disjoint from $K$.  Now my problem is to show that each $x\in X$ has a unique $t\in \Bbb R$ s.t. $(x,t) \in M$.  (I need it in order to extend $f$).  I know that if we take $v_0\notin M$ then each $v \in X \times \Bbb R$ has a unique representation as $v = \alpha v_0 + m$ for $m \in M$ , this is because $M$ is a maximal subspace. not sure if that helps.  Thanks for helping!
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This is a question I'm struggling with for a while:  Let $H$ be a Hilber space. Let  $T,S: H\rightarrow H$ be linear operators (not neccessarily bounded) such that for every $x,y\in H$: $\langle Tx,y\rangle=\langle x,Sy \rangle$.  Prove $T,S$ are bounded.    What I did so far:  First attempt:  $||Tx||^2=\langle Tx,Tx\rangle=\langle x,STx\rangle\le||x||*||STx||\le||x||*||S||*||Tx||$  I'm not entirly sure I'm allowed to do the last inequallity, since $||S||$ might be $\infty$.  It follows $||Tx||\le ||x||*||S||$  and analogly  $||Sx||\le ||x||*||T||$  From the first I'd get $||T||\le ||S||$ and from the second $||S||\le||T||$. Therefore $||S||=||T||$. I don't know where to go from here.  Second Attempt: assume  $T$ isn't bounded, therefore so is $S$, therefore there are series $(x_n),(y_n)\subset H$ such that $||x_n||=||y_n||=1$ and $\lim ||Tx_n||=\infty,\lim ||Sy_n||=\infty$. I'll keep it going soon, but it didn't go so well..    I'd love some guide. Not neccessarily the whole solution, but something that would tell me what i'm missing.  Thanks :)
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I would like an example of a map $f:H\rightarrow R$, where $H$ is a (infinite dimensional) Hilbert space, and $R$ is the real numbers, such that $f$ is continuous, but $f$ is not bounded on the close unit ball $\{ x\in H : \|x\| \leq 1\}$.  Actually, $H$ could be replaced by any Banach space (but not just a normed space-- that's too easy).  My motivation is that if $f$ is linear, this is impossible; but I have next to no intuition about non-linear functions.  Edit: Here's an example for $c_0$ which is even differentiable (disclaimer: I found it here: ).  Define $f:c_0\rightarrow F$ (where F is your field, real or complex) by $$ f(x) = \sum_{n=1}^\infty x_n^n \qquad (x=(x_n)). $$  You can estimate the sum by a geometric progression, so it does converge.  A bit of checking shows that f is Frechet differentible (so certainly continuous).  But $f(1,1,\cdots,1,0,\cdots)=n$ (if there are $n$ ones) so $f$ is not bounded on the closed unit ball.  What I don't immediately see is how to adapt this to $\ell^2$, say.
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I have the following definition:   Let ($X$,$\mathcal{T}$) and ($X'$, $\mathcal{T'}$) be topological spaces. A surjection $q: X \longrightarrow X'$ is a quotient mapping if $$U'\in \mathcal{T'} \Longleftrightarrow q^{-1}\left( U'\right) \in \mathcal{T} \quad \text{i.e. if } \mathcal{T'}=\{ U' \subset X' : q^{-1}\left( U' \right) \in \mathcal{T} \}$$  and the properties:   $q$ is a bijective quotient mapping $\Leftrightarrow$ $q$ is a homeomorphism  In general, $q$ quotient $\not \Rightarrow q$ open.  If $U \in \mathcal{T}, q(U)\subset X'$ is open if $q^{-1}\left( q\left( U \right) \right) \in \mathcal{T}$ but not in general.   I could not find an example of quotient mapping for which $q^{-1}\left( q\left( U \right) \right)$ is not open. I would understand the idea better if you could show me one.
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Is there some simple characterization of weak convergence of sequences in the space $\ell_\infty$? If yes, is there some similar claim for nets?   I was only able to come up with a characterization of sequential weak convergence using limit along ultrafilters, which I will describe below. I wonder whether there is some insight into this characterization. (E.g. whether there is some simple reformulations which does not use ultralimits.) Moreover, I do not know whether at least this characterization works for nets, too.  We have the following result describing weak convergence in $C(K)$, see e.g. Corollary 3.138,  in Banach Space Theory by Fabian, Habala et al. (It is a consequence of Rainwater theorem and characterization of extreme points of unit ball in $C(K)^*$.)     Let $K$ be a compact topological space. Let $\{f_n\}$ be a bounded sequence in $C(K)$ and $f\in C(K)$. Then, if $f_n\to f$ pointwise, we have $f_n\overset{w}\to f$.   Moreover we have isometric isomorphism between $\ell_\infty$ and $C(\beta\mathbb N)$, which is described e.g. in the Wikipedia article on  or in  of Carothers' book A short course on Banach space theory. This isomorphism assigns to each bounded sequence $(x_n)$ the continuous function $\overline x$ on $\beta\mathbb N$ defined by $$\overline x(\mathscr U) = \operatorname{\mathscr U-lim} x_n,$$ where $\operatorname{\mathscr U-lim} x_n$ denotes the  of $x_n$ w.r.t the ultrafilter $\mathscr U$.  Combining the above results we get the following characterization:      Let $f^{(n)},f\in\ell_\infty$. The sequence $f^{(n)}$ converges to $f$ weakly if and only if for every ultrafilter $\mathscr U$    $$\lim_{n\to\infty} \operatorname{\mathscr U-lim} f^{(n)}= \operatorname{\mathscr U-lim} f.$$   (The above claim for principal ultrafilters is just a pointwise convergence. But in the above claim the equality is required for free ultrafilters, too.)
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Let $T\colon L^2[0,1] \to L^2[0,1]$ be a bounded linear map of Hilbert spaces such that if $f\in L^2[0,1]$ is continuous then so is $Tf$. Show that there is a constant $C$ such that $$\sup_{x\in[0,1]}|Tf(x)| \leq C \sup_{x\in[0,1]}|f(x)|$$  I have no idea how to go about. Please at least give hints.
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I am wondering how to prove that a non-zero degree map from $A \to A$ is surjective. For example, identifying $S^1 \subset \mathbb{C}$, we can take $f:S^1 \to S^1$ via $f(z) = z^k$ with $k\neq 0$. This is a map of degree $k$. How can I show that it is surjective?  Note that I am interested in general topological spaces, not the circle example in specific.
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Suppose $X$ and $Y$ are topological vector spaces, $\dim (Y)&lt;\infty$, and $f:X\to Y$ is linear and surjective. Prove that $f$ is open and if the null space $N$ of $f$ is closed, then $f$ is also continuous.  The second assertion is easier since the quotient space $X/N$ has the universal mapping property.