Let the space $(C[0,1],\|\cdot\|_{L^1})$ where    $$ \forall f \in C([0,1]); \|f\|_{L^1}= \int^1_0 |f(x)| \, dx  $$ show that $(C[0,1],\|\cdot\|_{L^1})$ is not a Banach Space    Def of Banach space is a complete norm space.   def of complete if it is a cauchy seq in norm space then it converges in the norm space.     Guessing that a seq that might work would be something like a trig function that would be cauchy but it will not converge in the space (either by not converging or not being continous) or somhow fail to be a norm space. Not sure what route to take at this point. Will try to play with obvious cauchy seq and see what happens.   A hint would be appreciated thanks.
120632
Is $C[0,1]$ a Banach space with respect to the norm $\|f\| = \int\limits_0^1|f(t)| \, dt$?   People keep telling me it is, but lets consider: $f_n(x) = x^n$. This function defines a Cauchy sequence, yet the limit clearly isn't a continuous function!
119839
Define norm as $\int |f|$ (Riemann integral) on  $\mathcal R^1[0,1]$, the space of riemann integrable functions on $[0,1]$ with identification $f=g$ iff $\int |f-g|=0$.  Let $\{ r_1,r_2,\cdots \}$ be the rationals in $[0,1]$, and let $f_n=1_{\{r_1,\cdots,r_n\}}$. Then $f_n$ is a cauchy sequence in $\mathcal R^1[0,1]$. I want to show that there is no $f\in \mathcal R^1[0,1]$ such that $f_n$ converges to $f$ in norm. How can I show it?  Obviously the pointwise limit $f=1_\mathbb{Q}$ is not contained in $\mathcal R^1[0,1]$, but can I use this fact? I think that there can be other candidates, since convergence in norm and pointwise convergence are different.
88024
I am trying to understand better Cauchy sequence in the book that I am reading he is trying to provide counter argument that in general metric spaces we can find sequence that are Cauchy but aren't convergent for example $\frac{1}{n}$ is Cauchy but doesn't converge in the space T = (0,1].  I am trying to prove that the sequence is Cauchy to begin with if we consider $n \geq m &gt; 0$ Consider $|a_n - a_m| = \frac{m - n}{mn} &lt; \frac{m}{mn} = 1/n &lt; \epsilon$. So if we choose N = $\frac{1}{\epsilon}$ that should work since for $n, m &gt; N = \frac{1}{\epsilon}$   we have the following:  $|a_n - a_m| = \frac{m - n}{mn} &lt; \frac{m}{mn} = \frac{1}{n} &lt; \epsilon$.  what do you guys think any the proof is good or is there anything to make it better ? It is always good to learn from the extra bit of details as it makes my brain understand the subject more !
27676
Let $(X, \|\cdot\|_X)$ be a normed vector space and $(X^{\ast},\|\cdot\|_{X^{\ast}})$ its dual space. I have to prove, that  $$ \forall x\in X:\quad \|x\|_X = \sup_{T\in X^{\ast}}\{|T(x)| : \|T\|_{X^{\ast}}=1\}$$  How can I show this using Hahn-Banach theorem?
15667
i have this question :  in an example of the compact embedding, the autor gives a demonstration of : the sobolev space $W^{1,1}(\mathbb{R}^n)$ is not compactly embedded in $L^1(\mathbb{R}^n)$  So let $F\in D(\mathbb{R}^n)$(=the space of smooth functions with a compact support in $\mathbb{R}^n)$ ., not identically equal to zero and $\{x_n\}$ a sequence such that lim $x_n=+\infty$ when $n\rightarrow \infty$. so $F_n(x)=F(x-x_n)$ is bounded in $W^{1,1}(\mathbb{R}^n)$ and it converge a.e. to 0.  so if it converge strongly in $L^1$ we will have :$||F_n||_{L^1}=||F||_{L^1}=0$, an this is a contradiction .  my question is : where is the contradiction and how to prove that the embedding is compact in "this case or in normed (Banach) spaces (general case)"?  thank you very much.
90252
Let $(V,|\cdot|)$ be a normed space. Let $x, y\in V.$ If $|x|,|y|\geq 1$, then does it follow that   $$\left|\frac{x}{|x|}-\frac{y}{|y|}\right|\leq |x-y|\;?$$   Actually I got a hint in form of a picture, but I failed to use it.
16703
Problem Statement  Let $(X,\|\cdot \|)$ be a Banach space and $Y$, $Z$ closed subspaces of $X$. If every $x \in X$ can be uniquely represented as $x=y+z$ for $y \in Y$ and $z \in Z$ then show that there exists $c$ such that $\|y\|\leq C\|x\|$ and $\|z\| \leq C\|x\|$  Attempt  If $L:X \rightarrow Y$ given by $L(x)=L(y+z)=y,$ then we will be done if we can show that $L$ is a bounded linear operator. Alternatively, by adding the two inequalities, we can show that the norm $\|\cdot\|$ is equivalent to $\| \cdot \|_{Y\times Z}$, where $\|(y,z)\|=\|y\|+\|z\|$. I attempted the first approach via the closed graph theorem but I am unable to make any headway.
121831
Let $V$ be a real vector space. Suppose that $\Vert \cdot \Vert_1$ and $\Vert \cdot \Vert_2$ are two norms on $V$ which are equivalent.  I suspect the following to be true.  Let $(x_n)_{n=0}^\infty$ be a sequence in $V$. Then $(x_n)_n$ is a Cauchy sequence w.r.t.  $\Vert \cdot \Vert_1$ if and only if $(x_n)_n$ is a Cauchy sequence w.r.t. $\Vert \cdot \Vert_2$.  Is it true?   I'm looking for a good way to explain why "equivalence" of norms is the right way of "comparing" norms without mentioning anything from topology besides "convergence".
29181
Given the normed space $C^1[0,1]$ of differentiable functions with continuous derivatives on $[0,1]$. The norm is defined as $$\|x\| = \max_{[0,1]} |x(t)|$$ I'd like to prove that the given normed space is not a Banach space. In the attempt of solving this problem, I have thought about the possibility to construct a Cauchy sequence in the space which doesn't converge in the space. However until now I haven't got any idea. Then I proceeded to think about constructing an equivalent norm to the given one where the space can be easily proven to be not a Banach space. However, I have got nothing either. Now I'm stuck without a clue. Please give me a hint to a correct direction. Any help is greatly appreciated.
83662
Let $\{f_k\}_{k=1}^\infty\in L^{loc}_1(\mathbb R^n)$ be a sequence of real value functions such that  $\mbox{supp}(f_k)\subseteq\{|x|\leq k^{-1}\}$, $\quad \int f_k(x)dx=1$ $\quad for\quad all \quad k=1,2........$  show that the sequence $\{f_k^2\}_{k=1}^\infty$ does not converge in $\mathscr{D}^\prime(\mathbb R^n)$ $as \quad k\rightarrow \infty$
60535
Let $C$ be the Banach space of all complex continuous functions on $[0,1]$ endowed with the supremum norm. Let $B$ be the closed unit ball of $C$. Does there exist a $\Lambda\in C^*$ for which $\Lambda (B)$ is open? Note that $C$ is a locally convex space and $B$ is convex, balanced and closed set, by the separation theorem of convex sets,we know there exists $\Lambda\in C^*$ such that $|\Lambda f|\leq 1$ for all $f\in B$ and $|\Lambda f|&gt;1$ for $f\in B^c$. But how about this when we consider some concrete space above?
80977
Let $l^2$ be the space of square summable sequences with the inner product $\langle x,y\rangle=\sum_\limits{i=1}^\infty x_iy_i$.   (a) show that $l^2$ is H Hilbert space.    To show that it's a Hilbert space I need to show that the space is complete. For that I need to construct a Cauchy sequence and show it converges with respect to the norm. However, I find it confusing to construct a Cauchy sequence of sequences?
110984
Suppose $E$ is a normed space of all continuous functions on $[0,1]$ with norm $\int_{0}^{1}|f(t)|dt$. Prove that $E$ is not complete   I know that we must do is to find a Cauchy sequence of continous functions that doesn't converge in $E$, but I can't find that sequence. Can any one help me? Thanks.
57796
$C[a,b]$ is a normed vector space of all continuous complex valued functions on $[a,b]$, with supremum norm   $$\|f\|_\infty=\sup_{t\in [a,b]}|f(t)|.$$  The metric induced by the norm is  $$d(f,g)=\|f-g\|_\infty = \sup_{t\in [a,b]}|f(t)-g(t)|.$$  Show that $C[a,b]$ is a complete metric space under this metric induced by the supremum norm (i.e. show $C[a,b]$ is a Banach space).
103174
Define $C_1^1[a,b]$ to be the space of continuously differentiable functions on $[a,b]$, with norm $$\|f\|_1=\left(\int_a^b[|f|^2+|f'|^2]\mathsf dx\right)^{1/2}.$$ Show that this is a proper definition of a norm. Is this normed space complete?  I'm stuck on the triangle inequality part. I'm having a hard time showing that $$\|f+g\|_1\leq \|f\|_1+\|g\|_1.$$ Any hints or solutions are greatly appreciated.
17404
Consider the space $C^1([0,1])$ and the function $d:C([0,1])\times C([0,1]) \to \mathbb R$ defined as $d(f,g)=|f(0)-g(0)|+sup_{x \in [0,1]}|f'(x)-g'(x)|$. Decide whether the metrics $d$ and $d_{\infty}$ are topologically equivalent in $C^1([0,1])$ (where $d_{\infty}=sup_{x \in [0,1]}|f(x)-g(x)|)$  My attempt at a solution:  If two metrics are topologically equivalent, then they have the same convergent sequences. Honestly, I couldn't do anything. I am trying to define a sequence of functions $\{f_n\}_{n \in \mathbb N}$ such that $f_n \to f$ in, for instance, $(C^1([0,1]),d)$ but $f_n \not \to f$ in $(C^1([0,1]),d_{\infty})$. Could it be this two metrics are topologically equivalent? If this is the case, how could I prove it? If not, I would appreciate any hint to find an adequate sequence of functions that works for what I am trying to prove.
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.