Let $(V, \|\cdot\|)$ be a Banach space over $\mathbf R$ and $W$ be a closed subspace of $V$. We know that $V/W$ becomes a normed linear space under the quotient norm $\|\cdot\|_q$ defined as $\|v+W\|_q=\inf_{w\in W} \|v+w\|$ for all $v\in V$.     Suppose $v\in V$ and we have $\|{v+W}\|_q=\lambda$. Does there necessarily exist a $w\in W$ such that $\|{v+w}\|=\lambda$.   Clearly, the above is true if $\lambda=0$ or if $V$ is finite dimensional (because we can use the compactness of $S^n$).  But I am getting nowhere with the statement in block quotes.  We get a sequence $(w_n)$ in $W$ such that $\|v+w_n\|&gt;\lambda+1/n$ for all $n$. But this is weaker than saying $\|v+W\|_q=\lambda$.
123984
Let $E$ be a Banach space and $F$ a closed subspace. It is well known that the quotient space $E/F$ is also a Banach space with respect to the norm  $$ \left\Vert x+F\right\Vert_{E/F}=\inf\{\left\Vert y\right\Vert_E\mid y\in x+F\}. $$ Unfortunately in a set of lecture notes on (Lie) group representations (material for our study group) the author accidentally used here $\min$ instead of $\inf$. Probably a mostly harmless booboo, because at that point it was only needed to get a Banach space structure on the quotient, and we will probably be concentrating on Hilbert spaces anyway, where the problem does not arise.   Namely from Rudin's Functional Analysis I could not find a proof that the minimum should always be attained. Except in the case of a Hilbert space, where an application of parallelogram law (the sum of the squared norms of the two diagonals of a parallelogram equals that of the four sides) allows us to find a Cauchy sequence among a sequence of vectors $(y_n)\subset x+F$ such that  $$\lim_{n\to\infty}\left\Vert y_n\right\Vert_E=\left\Vert x+F\right\Vert_{E/F}.$$  But anyway, the suspicion was left that the infimum is there for a reason (other than conveniently allowing us to sweep this detail under the rug at that point of the development of theory), so in the interest of serving our study group I had to come up with a specific example, where the minimum is not achieved. It's been 25 years since I really had to exercise the Banach space gland in my brain, so it has shrunk to size of a raisin. Searching this site did help, because I found . There we have $E=C([0,1])$, the space of continuous real functions on $[0,1]$ equipped with the sup-norm. If we denote by $\Lambda$ the continuous functional $$ \Lambda: E\to\mathbb{R},f\mapsto\int_0^{1/2}f-\int_{1/2}^1f $$ and let $F=\ker\Lambda$, then the answer to the linked question proves that there is no minimum sup-norm function in the coset $\Lambda^{-1}(1)$.   So I have a (counter)example, and the main question has evolved to:     When can we use minimum in place of infimum in the definition of the quotient space norm?   My thinking:   It seems to me that the answer is affirmative, if $F$ has a complement, i.e. we can write $E=F\oplus F'$ as a direct sum of two closed subspaces such that the norm on $E$ is equivalent to the sum of the norms on $F$ and $F'$-components. But the first point also raises the suspicion that the question may be a bit ill-defined (and uninteresting) in the sense that the answer might depend on the choice of the norm $\left\Vert\cdot\right\Vert_E$ among the set of equivalent norms. However, if we, for example, perturb the sup-norm of $C([0,1])$ in the above example by multiplying the functions with a fixed positive definite function before taking the sup-norm, the argument seems to survive, so may be replacing the norm with an equivalent one is irrelevant?   So to satisfy my curiosity I also welcome "your favorite example" (one with a finite-dimensional $F$ would be nice to see), where we absolutely need the infimum here. Bits about any sufficient or necessary conditions for the minimum to be sufficient or (as a last resort :-) pointers to relevant literature are, of course, also appreciated.
7683
Let $V$ and $W$ be vector spaces over a field $F$. Let $L(V,W) = \{T:V\to W : T \text{ is linear} \}$, that is, $L(V,W)$ is the collection of all linear functions from $V$ to $W$. For $S,T \in L(V,W)$ and $a \in F$, define new functions $(S+T):V \to W$ and $ aT :V \to W$ by $(S+T)(x) = S(x)+T(x)$ ; $(aT)(x) = a(T(x))$.   Prove that $S+T$ belongs to $L(V,W)$ and $aT$ belongs to $L(V,W)$.
122695
In the 4th book of Stein and Shakarchi, the statement of the Hahn-Banach theorem is as follows:     Suppose $V_0$ is a linear subspace of $V$ and $p$ is a real sub-linear function on $V$, and that we are given a linear functional $l_0$ on $V_0$ that satisfies $$l_0(v)\leq p(v) ~\forall ~v\in V_0$$   Then $l_0$ can be extended to a linear functional $l$ on $V$ that satisfies    $$l(v) \leq p(v)~ \forall ~ v \in V$$   The proof given is as follows:     We have already proved that if $l_0$ can be extended to a subspace W, then it can also be extended to the subspace $W+ w$ for any vector $w$. We well order all the vectors in $V$ that do not belong to $V_0$, and denote this ordering by $&lt;$. Among these vectors, we call a vector $v$ extendable if the linear functional $l_0$ has an extension of the kind desired to the subspace spanned by $V_0.v$ and all the vectors $&lt;v$. What we want to prove is in effect that all vectors not in $V_0$ are extendable. Assume the contrary, then because of the well-ordering we can find the smallest $v_1$ that is not extendable. Now if $V_0'$ is the space spanned by $V_0$ and all the vectors $&lt;v_1$,then by assumption $l_0$ extends to $V_0'$. The previous step, with $V_0'$ in place of $V_0$ allows us then to extend $l_0$ to the subspace spanned by $V_0'$ and $v_1$, reaching a contradiction. This proves the theorem.   Firstly, how does proving that all vectors not in $V_0$ are extendable imply that an extension exists for the whole space? We know that for any $v \notin V_0$ there exists a valid extension, say $l_v$ for the space spanned by $V_0,v$ and all vectors $&lt;v$. Say we define $l(v):=l_v(v)$. Now we take an arbitrary $w&gt;v$. Couldn't it happen that $l_w(v) \ne l_v(v)$. In this case how can one define $l$? Will it help if we change the definition of "extendable" to something like: A vector $v \notin V_0$ is called extendable if $l_0$ has an extension of the kind desired to the relevant subspace which is equal to the already defined functionals $l_w~ \forall w&lt;v$ in the relevant subspaces? Also, how does one get the statement :"Now if $V_0'$ is the space spanned by $V_0$ and all the vectors $&lt;v_1$,then by assumption $l_0$ extends to $V_0'$."
121665
Let $V$ and $W$ be vector spaces (say over the reals). There is a linear injection $V^* \otimes W^* \to (V \otimes W)^*$ which sends $\sum_i f_i \otimes g_i \in V^* \otimes W^*$ to the unique functional in $(V \otimes W)^*$ sending $v \otimes w \mapsto \sum_i f_i(v) \cdot g_i(w)$ for all $(v,w) \in V \times W$. In the finite dimensional case it is easy to see this is an isomorphism by comparing dimensions on both sides. I'm aware that this is not an isomorphism in the infinite-dimensional case (see Relation with the dual space in the  on tensor products) but I must admit that I'm not sure why. Cardinal arithmetic does not seem help here and, even if it did, I would be much happier to see an explicit example of a functional in $(V \otimes W)^*$ outside the range of this map. I'm having trouble cooking one up myself.  Added: Let me elaborate on my comment about cardinal arithmetic. Suppose that $X$ and $Y$ are infinite dimensional vector spaces over a field $k$. I'm confident that $\dim (X \otimes Y) = \dim X \cdot \dim Y$ holds. It is clear that $|X^*| = |k|^{\dim X}$ since we may identify a functional on $X$ with a function from a basis for $X$ to $k$. Also I think I've convinced myself that if $\dim X \geq |k|$ then in fact $|X| = \dim X$ ie. the cardinality and dimension of a vector space agree when the dimension is larger than the cardinality of the ground field. Consequently, assuming say that $|k| \leq \dim X \leq \dim Y$ it seems we have $$ \dim(X \otimes Y)^* = |k|^{ \dim X \cdot \dim Y} = |k|^{ \dim Y} = \dim Y^* = \dim X^* \cdot \dim Y^* = \dim (X^* \otimes Y^*)$$  which implies that in fact $(X \otimes Y)^*$ and $X^* \otimes Y^*$ are isomorphic but, rather frustratingly, the obvious map does not do the job. Did I make a mistake here? If not, is it generally true (ie without making assumptions about $|k|$) that there exists some isomorphism $(X \otimes Y)^* \to X^* \otimes Y^*$?
84189
There is a conjecture: "The only subspace of a normed vector space $V$ that has a non-empty interior, is $V$ itself." (here, the topology is the obvious set of all open sets generated by the metric $||\cdot||$).  I have a proof for the case $V$ is finite dimensional. Because, let $V$ have dimension $n$ and a subspace $S$ of $V$ have dimension $m &lt; n$. Let $\{v_1,v_2,\ldots,v_m\}$ be a basis for $S$, extended to the basis $\{v_1,\ldots,v_m,\ldots,v_n\}$ of $V$. Now, suppose that $p = b_1v_1+\cdots+b_mv_m$ is an interior point of $S$. Now, consider the norm $N(a_1v_1 +\cdots+a_nv_n)=\max(|a_1|,\ldots,|a_n|)$. Then, there is an $r&gt;0$, such that $||x-p||&lt;r$ and $x$ is in $V$ $\implies$ $x$ is in $S$, since on a finite dim. space, all norms are equivalent. Let $m&lt;k\leq n$,and chose $v$ in $V$ as:  $$v=\left(b_1+\frac{r}{2}\right)v_1+\left(b_2+\frac{r}{2}\right)v_2+\cdots+\left(b_m+\frac{r}{m}\right)v_m+\frac{r}{2}v_k$$  Then, $N(v-p)\leq\frac{r}{2}&lt;r$, so $v$ is in $S$ and by the subspace property of $S$, $v_k$ is in $S$ too, a contradiction to $m&lt;n$.    I have primarily 2 questions:  (1) Is there a simpler method to proof the conjecture for the finite dimensional case?  (2) Is the conjecture true for the infinite dimensional case?  Sorry, if the question admits a very trivial answer. The motivation behind my question , is the fact that an open interval in $\mathbb{R}$ is not open in $\mathbb{R}\times\mathbb{R}$, etc.
120404
Let $ c_0 = \{ x = (x_n)_{n \in \mathbb N} \in l^\infty : \lim_{n \to \infty} x_n = 0\}$. Show that $c_0$ is a Banach space with the norm $\rVert \cdot \lVert_\infty$  I am capable of showing the space where the limit of $x_n$ exists is normed linear space but am having trouble with showing that the limit of Cauchy sequences must converge to 0.   Let $(x^{(n)})_{n \in \mathbb N}$ be a Cauchy sequence in $c_0$ such that $x^{n} = (x^n_1, x^n_2,...)$. Fix $k \in \mathbb N$ consider the sequence $(x^n_k)_{n \in \mathbb N}$ in $\mathbb F$. For any $n,m \in \mathbb N$  $\lvert x^n_k - x^m_k \rvert \le \sup_{k \in \mathbb N} \lvert x^n_k - x^m_k \rvert = \lVert x^n - x^m \rVert_\infty \lt \epsilon $   (1)  Thus $x^n_k$ is Cauchy in $\mathbb F$ and so has limit $y_k$ such that $y = (y_1,y_2,...)$ and y is the limit of $x^n$  To show that such a y exists we look at the value of $\lvert y_n - y_m \rvert \le \lvert y_n - x^N_n \rvert + \lvert x^N_n - x^N_m \rvert + \lvert x^N_m - y_m \rvert \lt \epsilon$ for all $n,m \ge N$     (2)  The middle expression on RHS of (2)  is $\lt \epsilon/3$ by (1)  The other two are also $\lt \epsilon/3$ follow from $x^N_k$ being Cauchy and converging to $y_k$  This shows that $\lim_{n \to \infty} y_n$ exists but we still have not shown that $y \in c_0$.  I know that to show y tends to 0 i should show that $\lvert y_k \rvert \lt \epsilon$ for $k \ge N$  This is where I am stuck. Perhaps $\lvert y_k \rvert = \lvert \lim_{n \to \infty} x^n_k \rvert$ and then we can take the limit function outside the absolute value sign by continuity? Then we might say due to it being a Cauchy sequence $x^n_k \lt \epsilon$. I know this last bit isn't at all convincing so I could do with some help.
90957
I want to prove the following theorem (no idea whether it has a name):  Let $V = \mathbb{R}^n$ or $\mathbb{C}^n$ and $\|\cdot\|$ be a norm on $V$. Then, there exist $C_1, C_2 &gt; 0$ such that for all $x \in V$:  $$C_1 \|x\|_\infty \leq \|x\| \leq C_2 \|x\|_\infty$$  I first let $x \neq 0$ (otherwise it would be trivial). Then, I divided by $\|x\|_\infty$ and normalized the vector $x$ such that $\|x\|_\infty = 1$. That left me with  $$C_1 \leq \|x\| \leq C_2$$  but I don't see how this could help me or how I could possibly limit an unknown norm. How can I proceed? Or is this the wrong way anyway? Thanks for any answers.
31280
My question is:  Let $ V $ be a vector space (over $ \mathbb K\in\{\mathbb{R}, \mathbb{C}\} $), $ X,Y\subseteq V $ two subspaces equipped with norms $ \|\cdot\|_X, \|\cdot\|_Y $ such that $ (X,\|\cdot\|_X) $ and $(Y,\|\cdot\|_Y)$ are Banach spaces and $ D\subseteq X\cap Y$. If $ D $ is dense in $ (X,\|\cdot\|_X) $ and $(Y,\|\cdot\|_Y)$, is $ D $ also dense in $ X\cap Y $ equipped with $ \|\cdot\|:=\|\cdot\|_X + \|\cdot\|_Y $?  At first sight, it seemed very clear to me that this should be true. But I even fail to answer the following (possibly) easier question:  Let $ X $ be a vector space over $ \mathbb K $ equipped with two norms $ \|\cdot\|_1, \|\cdot\|_2 $ such that $ (X,\|\cdot\|_1) $ and $(X,\|\cdot\|_1)$ are Banach spaces and $ D\subseteq X$. If $ D $ is dense in $ (X,\|\cdot\|_1) $ and $(X,\|\cdot\|_2)$, is $ D $ also dense in $ X $ equipped with $ \|\cdot\|:=\|\cdot\|_1 + \|\cdot\|_2 $?  The answer is yes, if $ \|\cdot\|_1, \|\cdot\|_2 $ are equivalent, so I tried to think about a counterexample using nonequivalent norms on a specific space and I also found a nice paper about nonisomorphic complete norms () but it didn't helped me so far to construct anything useful for my question.  Thanks for your help!
23933
Let $(X,\|\;\|)$ be a normed vector space over $K$ and let $A,B\subseteq X$ be s.t. $A$ is compact and $B$ is closed. I want to prove that:  $$(\alpha A+\beta B)\;\text{ is closed }\;\forall \alpha,\beta\in K$$ So, given our function $\odot$ as in , my attempt goes like this: Let $\alpha,\beta\in K$. If $\alpha=\beta=0$, clearly $$\alpha A+\beta B=\{0\}\text{ is closed}$$ Now, since $\odot$ is continuous and $A$ is compact, we have that $\;\forall \alpha\neq0,\;\alpha A=\odot(\alpha,A)\;$ is compact and thus closed.  Then, let $\;y\in \beta B$ so,  $$ y=\beta b_y $$ for some $b_y\in B$. But, since $B$ is closed, there exists $\{b_n\}_{n\in\Bbb N}\subset B$ s.t. $b_n\xrightarrow[n\to\infty]{}b_y$ So let $y_n=\beta b_n\;\;\forall n\in\Bbb N$. Thus, $\{y_n\}_{n\in\Bbb N}\subset \beta B$ and $\|y_n-y\|=\|\beta b_n-\beta b_y\|=|\beta|\|b_n-b_y\|\xrightarrow[n\to\infty]{}0$ Thus, we get that  $$\forall y\in\beta B\;\exists\;\{y_n\}_{n\in\Bbb N}\subset \beta B \text{  s.t.  } y_n\xrightarrow[n\to\infty]{}y$$ so $\beta B$ is closed $\;\forall \beta\in K$. Now, $$ \text{Let }x\in (\alpha A+\beta B)\Rightarrow\;\exists\;a^*\in\alpha A\;\land\;b^*\in\beta B\text{  s.t.  } x=a^*+b^* $$ so, since $\alpha A$ and $\beta B$ are closed, $$ \exists\;\{a_n^*\}_{n\in\Bbb N}\subset\alpha A\text{  s.t.  }a_n^*\xrightarrow[n\to\infty]{}a^*\\ \exists\;\{b_n^*\}_{n\in\Bbb N}\subset\beta B\text{  s.t.  }b_n^*\xrightarrow[n\to\infty]{}b^* $$ so, taking $x_n=a_n^*+b_n^*\;\forall n\in\Bbb N$ is clear that $x_n\in (\alpha A+\beta B)\;\forall n\in\Bbb N$ and $$ \|x_n-x\|=\|a_n^*+b_n^*-a^*-b^*\|\le\|a_n^*-a^*\|+\|b_n^*-b^*\|\xrightarrow[n\to\infty]{}0 $$ so $x_n\xrightarrow[n\to\infty]{}x$ and, since this happens $\;\forall x\in (\alpha A+\beta B)$, thereby $(\alpha A+\beta B)$ is closed $\;\forall \alpha ,\beta\in K$  What do you think? Did I miss something?
75296
Note that (1) all norms on finite dimensional vector space are equivalent.  And here (2) norms on vector space are equivalent iff they have same topology.   Question : How can we prove (2) ? Please recommend reference  Thank you in anticipation
79183
Suppose that we have a Hilbert Space $H$ and $M$ is a closed subspace of $H$. Let $T\colon H\rightarrow M$ be the orthogonal projection onto $M$.   I have to show that $T$ is compact iff $M$ is finite dimensional.  So if we assume that $M$ is finite dimensional then $\overline{T(B(0,1))}$ is a closed bounded set in a finite dim vector normed space and so it is compact. Which gives that $T$ is compact.  But I am unsure how to prove that if $T$ is compact then $M$ is finite dimensional?  Thanks for any help
56555
I am trying to prove that given a subspace $\mathbf{W}$ in $\mathbb{R^{n}}$, the subspace and its orthogonal complement 'cover' whole of $\mathbb{R^{n}}$ through '+' where we define $\mathbf{W}$+$\mathbf{W^{\perp}}$ as linear combinations of vectors both in the subspace and in its orthogonal complement. It seems intuitively right, and I can prove that the sum of their dimensions adds up to n, but I am not sure how to prove the question I am looking at. Thanks!
94728
How to show that quotient space $X/Y$ is complete when $X$ is Banach space, and $Y$ is a closed subspace of $X$?   Here's my attempt: Given a Cauchy sequence $\{q_n\}_{n \in \mathbb{N}}$ in $X/Y$, each $q_n$ is an equivalence class induced by $Y$, I want to find a representative $x_n$ in $q_n$ so that the induced sequence $\{x_n\}_{n \in \mathbb{N}}$ is also a Cauchy sequence in $X$. But I don't know how to construct such sequence.
123661
Prove that the set    $$ W= \{f \in C([0, 1]):f(0)=0\}$$   is a closed linear subspace of $C([0,1]).$   Here $C([0, 1])$ is the space of continuous functions equipped with the uniform norm  $||f||_\infty = \sup_{x\in[0,1]}|f(x)|$.   I need some help in how to proceed  with this problem.   First, I want to prove that $W$ is, indeed, a subset of $C([0,1])$. For this, we have that for any two functions $f,g\in W$, $f+g\in W$, since $f(0) + g(0) = 0$. Then, also for any $\lambda \in \mathbb{R}$, we have that $\lambda f \in W$, since $\lambda f(0) = 0$. Finally, we also have $0 \in W$. So $W$ is a linear subspace of $C([0, 1])$.  My problem is how to prove that $W$ is closed. Based on the definition of a closed set, I wanted to show that a sequence of continuous functions $f_n$ converges uniformly to $f$ and that this $f$ is also in $W$. I started with something like this:  Let $f_n$ be a sequence of continuous functions such that $f_n(0) = 0$ for any $n\in \mathbb{N}$. We need to prove that $f_n \to f$, uniformly. i.e $$ \|f_n - f\|_\infty = \sup_{x\in[0,1]} |f_n(x) - f(x)|\to 0$$ as $n\to \infty$.   How can I proceed?
117110
I  found a question about a property of the . However the question was not properly answered (it used a projection argument which might not be true in a general Banach space).  I was wondering whether the following (weaker) statement holds:  Let $X$ be a Banach space and suppose $A,B,C_0\subset X$ are bounded, closed, convex and non-empty subset. Do we then have $$A+C_0=B+C_0\implies A=B?$$
17097
I am sure the answer to this is (kind of) well known. I've searched the web and the site for a proof and found nothing, and if this is a duplicate, I'm sorry.  The following question was given in a contest I took part. I had an approach but it didn't solve the problem.     Consider $V$ a linear subspace of the real vector space $\mathcal{M}_n(\Bbb{R})$ ($n\times n$ real entries matrices) such that $V$ contains only singular matrices (i.e matrices with determinant equal to $0$). What is the maximal dimension of $V$?   A quick guess would be $n^2-n$ since if we consider $W$ the set of $n\times n$ real matrices with last line equal to $0$ then this space has dimension $n^2-n$ and it is a linear space of singular matrices.  Now the only thing there is to prove is that if $V$ is a subspace of $\mathcal{M}_n(\Bbb{R})$ of dimension $k &gt; n^2-n$ then $V$ contains a non-singular matrix. The official proof was unsatisfactory for me, because it was a combinatorial one, and seemed to have few things in common with linear algebra. I was hoping for a pure linear algebra proof.  My approach was to search for a permutation matrix in $V$, but I used some 'false theorem' in between, which I am ashamed to post here.
22048
Put   $$ l^{\infty} = \{ (x_n) \subseteq \mathbb{C} : \forall j \; \;\ \;|x_j| \leq C(x)\} $$  I want to show that $c_0$, the space of all sequences of scalars that converges to $0$ is closed subspace of $l^{\infty}$.  MY atttemt: Take arbitrary sequence $(x_n) \in c_0$. We know $x_n \to 0$. but $0 \in c_0$. Therefore, every sequence in $c_0$ converges to an element in $c_0$. Hence, $c_0$ must be closed  Is this correct? thanks