Let $Tf(x)=\int_{[0,x]}fdm$ for $0 \leq x \leq 1$.  Prove that $C_0((0,1])=\{f \in C[0,1] : f(0)=0\}$ is a closed subspace of $C[0,1]$    I need a bit of help with this one. I was thinking of proving that the complement is open, or maybe showing that given any convergent sequence in $C_0((0,1])$ its limit has to be in $C_0((0,1])$.... Could use some help though!! THanks!
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?
104103
I'm trying to prove that proper subspaces of $ R^n $ are closed and have empty interiors.  To prove that they are closed I'm trying to use the fact that inverse images of closed sets are closed sets by continuous functions. So I'm trying to find a linear map such that a proper subspace of $ R^n $ is its Kernel but I don't know how to express it so that it would work on all dimensions.  Regarding the fact that they have empty interiors, I really don't know what to do.  Thanks in advance.
82153
I am trying to prove that under the $L^2$ norm,  $C([0,1])$ does not give rise to a  complete metric space. To do this I am trying to find a Cauchy Sequence which does not converge in $C([0,1])$. As a template (on $C([a,b])$) I am led to believe the following is Cauchy: $$f_n (x) = \begin{cases}  1 &amp;\mbox{if } 0 \leq x \leq \frac{1}{2} \\  1 - 2n(x-\frac{1}{2}) &amp; \mbox{if } \frac{1}{2}\leq x \leq \frac{1}{2n} + \frac{1}{2}\\ 0 &amp; \mbox{if } \frac{1}{2n} + \frac{1}{2} \leq x \leq 1  \end{cases}  $$ But I am struggling to show this is Cauchy, I have tried integrating from 0 to 1 but this is giving a very nasty integral and I was wondering if anyone has a better method?
108229
The set $C[0,1]$ is the set of all continuous functions $f:[0,1]\to \mathbb{R}$. Show that $C[0,1]$ is a vector space. Note: for $f,g$ (as elements of) $C[0,1]$, we define $(f+g)(x) = f(x)+g(x)$ and for a scalar $c,$ $(cf)(x) = cf(x)$.  The first step in showing that something is a vector space is to show that vector $u$ + vector $v$ is in $V$.   I don't know what I'm supposed to assign as $u$ and $v$, though. Please help. Thanks.
44045
Let $X$ be a banach space, and let $A$, and $B$ be closed linear subspaces. Assume that $$\inf\{\|x-y\|\mid x\in A, y\in B, \|x\|=\|y\|=1\}&gt;0$$  I want to show that $A+B$ is closed.   I was thinking of doing something like, let $z$ be a limit point of $A+B$, then there exist $z_n\in A+B$ such that $z_n\rightarrow z$, each $z_n$ can be written as $a_n+b_n$. Then I wanted to do something along the lines of determining whether $a_n$ and $b_n$ have a limit (if they do, then call them $a$ and $b$ and then $a\in A$, $b\in B$, and $z=a+b\in A+B$), but I can't seem to be able to use the condition given.   I am studying for a qual, so you can go ahead and either give a solution or sketch it.
14317
I'm trying to solve this question:  Let $$M=\{f\in C([0,1])\ | \ f(1)=0\}.  $$  Is $M$ closed in $(C[0,1]),d_1)$?  At first, I assumed that $M$ was closed, so I tried to prove that if there is a sequence $f_1,f_2,f_3,...\in M,$and $\ f_n\rightarrow f\,$ with respect to $d_1$, then $f \in M$.    However, I got stuck and checked out the answers, and the answer I was given was just a counterexample of a sequence $(f_n)_n$ where $f_n \in M,\ f_n\rightarrow f$ wrt $d_1$, and $f\notin M$.    I understand that it's a lot easier to give a counterexample, but unfortunately I suck at analysis and cannot tell whether a set is closed or not upon inspection or come up with a counterexample right away.  Are there other ways of proof, like proof by contradiction, to show that this set is not closed?
81852
This question is in relation to the space $C(I)$, $I = [a, b]$. Define $P_n =\{ a_0+\dots+a_nx^n \mid a_i \in \mathbb{R}\}$ (any or all $a_i$ could be zero); clearly $P_n \subset C(I)$. The norm I'm using is $\lVert f\rVert_I = \sup_I |f(x)|$. Is $P_n$ closed under $\lVert\cdot\rVert_I$?   I am almost sure the answer is "yes", but I can't seem to prove it. My first instinct was to biject  $P_n$ to $\mathbb{R}^{n+1}$, using coefficients as coordinates, and prove that sequences of degree-$n$ polynomials converge to degree-$n$ polynomials, but I can't prove that the metric $\lVert\cdot\rVert_I$ is equivalent to the standard metric on $\mathbb{R}^{n+1}$. Next, intuitively given a function $f \in C(I), \notin P_n$ I should be able to find some $\epsilon &gt; 0$ such that there are no low-degree polynomials "nearby", then that function was not a limit point of $P_n$ so $P_n$ is closed. Again I have no idea how to prove this. What should I do?
31155
I have a non homework related question from a text and require a nice clear proof/disproof please  Is it true that a subset that is closed in a closed subspace of a topological space is closed in the whole space?  my ideas:  if $H$ is the subset of the topological space $X$  if the subset is closed in the closed subspace, the complement is open in the subspace, which means the complement is of form $U\cap H$ for some $U$ open in $X$  if the subspace is closed the complement is open which means complement of $H=U$ for some open $U$ in $X$  kind thanks
74618
Consider the vector spaces $C([0, 1])$ and $C^1([0, 1])$ with norm, \begin{align*} \displaystyle \|f\|_\infty=\sup_{x\in [0, 1]}|f(x)|, \end{align*} and let $T:C^1([0, 1])\to C([0, 1])$ the operator given by, \begin{align*} \displaystyle Tf=f'. \end{align*} Is $T$ defined this way an open mapping? If it's, how to prove it?
29942
Consider $C([0, 1])$, the linear space of continuous complex-valued functions on the interval $[0, 1]$, with the norm  $$\displaystyle\lVert f\rVert_1 = \int_0^1 \lvert f(x)\rvert \,dx.$$  I have to show that $C([0, 1])$ is not complete with respect to this norm. I have found the following example from a book.  Let $f_n \in C[0,1]$ be given by  $$f_n(x) := \begin{cases}  0   &amp; \text{if $0 \le x \le \frac1{2}$}\\  n(x-\frac{1}{2}) &amp; \text{if $\frac {1}{2} &lt; x \le \frac {1}{2} + \frac {1}{n}$}\\  1              &amp; \text{if $ \frac {1}{2} + \frac {1}{n} &lt;x \leq 1 $} \end{cases}$$  How to prove that $f_n$  is a Cauchy sequence with respect to $\lVert \cdot\rVert_1$?   If I use basic definition then I have to prove that $\lVert f_n - f_m\rVert_1 &lt; \epsilon$ $\forall n, m &gt; N$. But I am finding it difficult to prove this.  Please help me to understand how to prove that $f_n$ is Cauchy sequence in $C([0, 1])$.  Thanks
103108
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.
92721
Question: Let $E$ be a normed space. Let $G$ be a closed subspace of $E$ and let $F$ be a finite dimensional subspace of $E$. Show that $F+G$ is a subspace of $E$ and is closed.  I'm having trouble in showing $F+G$ to be closed. I know that $F$ is itself closed and complete, as it is a finite dimensional subspace of a normed space, and that if $F$ were compact that $G+F$ would be closed. I also know that the closed unit ball of any finite dimensional normed space is compact.  I tried two methods. One was to take a convergent sequence $(x_n)_{n \in \mathbb N}$ in $G+F$. Then we can write $x_n = f_n + g_n$ where $f_n$ and $g_n$ are sequences in $F$ and $G$ respectively, and I attempted to find a way to force the individual components $f_n$ and $g_n$ inside the unit ball which would enable me to say that they had convergent subsequences. I couldn't see how to do this, however.   The other thought was to try and show that $F$ is compact, but I don't see a way to do this as I can't imagine it to be simply true without some other conditions on $F$.  Are one of these methods the right way to go? Or should I go another direction? I would appreciate any help I can get, although I would prefer not to be presented with a full proof so that I can do some work for myself. Thanks!
11246
Define $c_0$ to be the space of real sequences in $\ell^\infty$ that converge to $0$ and $\ell^\infty$ be the set of all bounded sequences. Show that $c_0$ is closed in $\ell^\infty$.   I am reading in the solution in  and I don't understand how we define $$x^{k}_n - \omega_n$$  Because $(x_n^k) = ( \{x^1_1,x_2^1,\dots \}, \{x_1^2, x_2^2,\dots \}, \dots)$ and $(\omega_n) = (\omega_1, \omega_2, \dots)$  The objects in $(x_n^k)$ are sequences (sets) and the objects in $\omega_n$ are real numbers, how can you subtract a number from a set?
99196
I want to prove that all "closed unit ball" of $$ c_0 = \{ \{x_n\}_{n=1}^\infty \in \ell^\infty : \lim_{n\to\infty} x_n = 0\} $$ do not have any extreme point. Would you please help me?  (Extreme Point) Let $X$ be a vector space and $A \subset X$ be convex. We say $x\in A$ is an  if for $x = (1-t)y + tz,\; y,z,\in A, \;t\in(0,1)$ then $y = z = x$.  What I tried is as follows:  Let $B$ be a closed unit ball of $c_0$, that is, $$B = \{\{x_n\}_{n=1}^\infty \in \ell^\infty : \lim_{n\to \infty} x_n = 0 \text{ and } \|x\|_{\ell^\infty}\le 1\}.$$ If there is a extreme point $b = \{b_n\}_{n=1}^\infty\in B$, then we have for $$ b = (1-t)y + tz, \quad y,z\in B,\quad t\in (0,1) $$ implies $$ y = z = b. $$ But I cannot do anymore here. Would you please help me?
25113
Suppose $f$ is a continuous function at $x = c$ in $[a,b]$.  Prove that for any sequence ${x_n}$ in $[a,b]$ converting to $c$, the sequence $\{f(x_n)\}$ converges to $f(c)$.  That is, $$ \lim_{n\to\infty}f(x_n)= f\left(\lim_{n\to\infty}x_n\right)$$    This proof seems simple but there are a few things that I need to know first. If $\{x_n\}$ converges to $c$, is it sufficient to substitute $c$ in for $\lim_{n\to\infty}x_n$?  Also needing some guidance on the structure of this proof.  Thanks!
42395
If $f$ is continuous from $ [0,1]$ to itself and $f(0)=0$, $f(1)=1$ and $f(f(x))=x$ then show that $f(x)=x$ for every $ x $ in the domain of function.  I'd like to prove this statement, my idea goes like below:  Suppose that the converse of the statement holds. then without loss of generality, assume that there exists some $c$ in $[0,1]$ such that $d=f(c)&gt;c$. By symmetry, $f(d)=f(f(c))&lt;f(c)=d$. And $f$ is bijective because $f$ has an inverse function in this domain. So $f$ should be strictly increasing function. But on $[c,d]$, $f$ is decreasing. This is a contradiction.  So it should be that $f(x)=x$ on $[0,1]$.  Is this argument right? And I'd like to know another argument which prove this statement. Thank you for your answer in advance :)
21413
[I really want to apologize if this problem looks a little too long.]  The problem :  This is taken from  and it says...     Prove or disprove the comapctness of the closure of the unit ball of $C^1[0, 1]$ in $C[0, 1]$.   What have I tried?  I think it's pretty clear that the unit ball in $C^1[0,1]$ is $B= \{f \in C^{1}[0,1]: \lVert f \rVert \leq 1\}$. Let us denote $K= \operatorname{cl}_{C[0,1]}(B)$, the closure of $B$ in $C[0,1]$.  I really didn't have much clue how to start working on this problem but finally thought about using the sequential approach. If $a$ and $b$ are two points such that $0&lt;a&lt;b&lt;1$ then I think there can be functions $f_a$ and $f_b$ having the property that $\lVert f_a-f_b \rVert = 1$ which are sufficiently smooth to be in $C^1$.  Now as it is quite tedious to construct these functions explicitly, I scanned a hand-drawn picture of what I think they might look like.     Here $f_a$ and $f_b$ take zero values in almost all of the interval $[0,1]$ and jumps up at $a$ and $b$ respectively. Also, as is clear from the figure, they don't assume non-zero values simultaneously. Now, my argument is clear: however close the two points $a$ and $b$ come to each other (remaining distinct) there will always be functions like $f_a$ and $f_b$. They may get steeper and steeper but will never lose their $C^1$-ness.   So if I consider the sequence $(\frac{1}{n})$ in $[0,1]$, I will get a sequence $(f_{\frac{1}{n}})$ in $B (\subseteq K)$ where for $m \neq n$ we will have $\lVert f_{\frac{1}{m}} - f_{\frac{1}{n}}\rVert = 1$. Hence the sequence $(f_{\frac{1}{n}})$ cannot have a convergent subsequence, thereby proving that $K$ is not sequentally compact and hence not compact.  But I'm not very sure about all these. A friend of mine told me that this same problem has a positive answer and that made me sufficiently confused.  So, here comes my question...     In the above argument, where have I gone wrong?   What is the real answer? And how to prove it?   Thanks a million for reading my extra-long question. And thanks a zillion for any help that you can offer.