Give $C^1([a,b])$ the sup metric induced from $(C^0([a,b]),||.||)$. I want to know if $C^1([a,b])$ is complete. My thinking is to show that $C^1([a,b])$ is a closed subset of $C^0([a,b])$.  Let $(f_n)$ be a sequence of $C^1([a,b])$ functions where $f_n\to f$ (sup-norm convergence). Since differentiability implies continuity, $(f_n)\in C^0([a,b])$, where sup convergence is the same as uniform convergence. The sequence $(f'_n)$ is also in $C^0([a,b])$.   I know that If $f_n$ is a sequence of differentiable functions with $f_n\to f$ uniformly and $f'_n\to g$ uniformly, then $f$ is differentiable and $f=g$.  So if $(f'_n)$ converges, it converges to $f'$. Since $(f'_n)\in C^0([a,b])$, $f'\in C^0([a,b])$, so $f\in C^1([a,b])$. So if I can show $f'_n$ converges in $C^0[a,b]$, then $C^1([a,b])$ is closed and the proof is done. But this only works if must $(f'_n)$ converge. Is this true, and if so why?
28274
The space $ C^1[0,1] $- the space of all continuously differentiable functions on $ [0,1]$ is not a Banach space with respect to the sup norm,$ \|.\|_{\infty} $ since the uniform limit of a continuously differentiable function need not be differentiable.  How can I illustate this statement using a counter example? Can I use $ f_{n}=\frac1 n \sin nx $ as a counter example?  Also, is $ C^1[0,1] $ is same as the space given by $ X=\{f\in C^1[0,1]:f(0)=0 \}$. Can I use the same example to show that this is not a Banach space?  More help is appreciated! Thanks!
23505
Let $-\infty\lt a\lt b \lt \infty$. Let $C[a,b]$ is space of continuous functions and a function $\|.\|:C[a,b]\to R$ is given by $\|f\|:=\int_a^b |f(x)|d(x)$  a). Show that $\|\cdot\|$ is a norm in $C[a,b]$  b). Show that $C[a,b]$ with this norm is not a Banach space.  Well the 1st question is kind of easy. Non-negativity, Homogenity, and Triangle inequality is easily proven. At the 2nd question I have a problem.   I know the $$C[0,1]$$ space with Norm $$\|f\| = \int_0^1|f(x)|d(x)$$ is not complete that means not Banach-Space. But how do I prove the generality?   Does anyone has a suggestion?  Thanks
111370
Let $A$ be a closed and bounded subset of $\mathbb{R}^{n}$. Given that $f: A \to A$ satisfies $d(f(x),f(y)) &lt; d(x,y)$ for all $x,y \in A$, $x\neq y$ where $d$ is the usual Euclidean metric. Show that $f$ has a unique fixed point.    My observations:    Since $\mathbb{R^{n}}$ is complete, then $(A,d)$ is complete since $A$ is closed; Since $A$ is bounded, then $A \subset B_{R}$ for some $R &gt; 0$; Banach Fixed Point Theorem guarantees a unique fixed point if $f$ is contraction on $A$.   Please correct me if I am wrong. My quandry here is showing that $f$ is a contraction, i.e.  $$ d(f(x),f(y)) \leqslant k\cdot d(x,y), \hspace{3mm} k \in (0,1) $$  I'm looking for some hints or tips. Thank you.
120961
The question is two parts. Suppose $B$ denotes the closed unit ball in $\mathbb{R}^n$ and $U$ is an open set containing the ball. Let $f$ be a $C^1$, rank $n$ function $U \to \mathbb{R}^n$ such that $||f(x) - x|| &lt; 1/2$ for $x \in U$. First, show that $||f||^2$ attains a minimum in the interior of $B$. Second, show that $f$ actually hits $0$.  It is not so hard to see that the second claim implies the first. This is because $||f||^2$ cannot possibly be any smaller than $0$, so if $f(x) = 0$ for some $x$, this is as low as it goes. It is a minimum since $f$ is rank $n$, and thus a local homeomorphism by the inverse function theorem. This implies it is locally injective, and thus a local minimum. It must happen in the ball, since the inequality given says in fact it has to happen in the open ball of radius $1/2$.  It is a common theme among the writers of these questions that the first part be a special case of the second, which is often times provable without the first. Thus I'd like to find an argument which establishes this fact directly.  Any hints?
100129
i am struggeling with the following problem: let $f:X\to Y$ be a continuous mapping bet ween two metric spaces.   Does it hold that $f^{-1}(\overline{A})=\overline{f^{-1}(A)}$ and $f^{-1}(int(A))=int(f^{-1}(A))$, where the overline denotes the closure of a set and $int()$ the interior of a set.  One direction is obvious: $f^{-1}(A)\subset f^{-1}(\overline{A})$ and by continuity of $f$ the set$f^{-1}(\overline{A})$ is closed. So we proved $\overline{f^{-1}(A)}\subset f^{-1}(\overline{A}$. Any idea or reference where I can find the proof or an counter example?
25618
Let $K$ be a compact metric space. Let $\{f_n\}_{n=1}^\infty$ be a sequence of continuous functions on $K$ such that $f_n$ converges to a function $f$ pointwise on $K$.   on Walt. Rudin's book Principles of mathematical analysis, 7.13, if we assume   (1). $f$ is continuous;   (2). $f_n(x)\geq f_{n+1}(x)$ for all $x\in K$ and all $n$;   then it is proved that $f_n$ converges to $f$ uniformly on $K$.   Is there counterexample satisfying (1) but not (2)? And satisfying (2) but not (1)?
19604
Working on the following problem from Munkres:      Let $(X, d_{X})$ and $(Y, d_Y)$ be metric spaces; let $Y$ be complete.   Let $A \subset X$. Show that if $f:A \to Y$ is uniformly continuous,   then $f$ can be uniquely extended t a continuous function $g:  \overline{A} \to Y$, and $g$ is uniformly continuous.   If there exists such an extension, it must be unique, and $f(x) = \lim_{n \to \infty}f(x_n)$ for $x_n$ a sequence in $A$ converging to $x \in \overline{A}$. To show this limit exists, let $x_n \to x$, then $f(x_n)$ is Cauchy, since uniformly continuous functions take Cauchy sequences to Cauchy sequences. Then $f(x_n)$ converges by completeness, and I claim it converges to $f(x)$.  Now, to show $g$ is uniformly continuous, I kind of brute-forced it, and I was wondering if anyone had a more elegant method. (I consider the proof up to now pretty elegant.) What I did was: Suppose for contradiction that $\exists \epsilon &gt;0$ s.t. $\forall \delta &gt;0\,\, \exists x_\delta, y_\delta \in \overline{A}$ s.t. $d(x_\delta, y_\delta)&lt; \delta$ and $d(g(x_\delta), g(y_\delta))&gt; \epsilon$. Let $\delta := \delta_{\epsilon}/2$, where $\delta_\epsilon$ is taken from the uniform continuity of $f$. Now $d(g(\cdot), g(\cdot))$ is continuous...in particular, it is continuous in each variable separately. Therefore we can vary $x_{\delta}$ and $y_{\delta}$ a little bit, to get $x^*,\, y^* \in A$ such that $d(x^*, y^*)&lt;\delta_{\epsilon}$ and $d(f(x^*), f(y^*))&gt;\epsilon$, a contradiction.   Does anyone know a more elegant method to show the uniform continuity?
44605
Let $f:[0,1]\rightarrow [0,1]$ be a homeomorphism. Show that , there exists a sequence of polynomials  $$(P_n(x))_n$$ such that $P_n(x)$ converge uniformly to $f$ on $[0,1]$ and every $P_n(x)$ is a homeomorphism from $[0,1]$ to itself.   I think that if we put additional condition that f be $C^1$ then $f'$ will be non-negative and continuous. Moreover, Weierstrass theorem says that there will be exist non-negative sequence of polynomials $(Q_n(x))_n$ such that it converges uniformly to $f'$ on $[0.1]$. Now it seems that $$P_n(x)=\int_{\mathbb{0}}^{x}Q_n(t)\:dt$$ works .  Is this true ? Is there other ways?   Thanks
81241
I need to proof that $(C_0([0,\infty)), \ \|\cdot\|_\infty)$ is complete. I think that I did it correctly (or atleast am thinking in the right direction), but I still struggle with writing it down correctly. I used that $(C_b([0,\infty)), \|\cdot\|_\infty)$ is complete, so therefore as $C_0[0,\infty) \subset C_b[0,\infty) $, I wanted to use that if $A$ is a closed subset of a complete space $(M,d)$, that $(A,d)$ is complete. Thus, it came down to showing that $C_0[0,\infty)$ is closed.  Let $f_n(k)$ be a sequence of elements out of $C_0[0,\infty)$ that converges to an element $f$ out of $C_b[0,\infty)$. I need to proof that $f(k) \in C_0[0,\infty)$.   Per the definition of convergence, we know that $\|f_n(k) - f(k) \|_\infty \rightarrow 0$ whenever $n \rightarrow \infty$. We use this to find $ |f_n(k)-f(k)| \leq\|f_n(k) - f(k)\|_\infty  \rightarrow 0$.  Then we consider $|f(k)| = |f(k) - f_n(k) +f_n(k)| \leq |f(k)-f_n(k)| + |f_n(k)|$. If we take limits on both sides, we get: $\lim_{k\to \infty} |f(k)| \leq \lim_{k\to \infty} |f(k)-f_n(k)| + \lim_{k\to \infty} |f_n(k)| = 0 + 0 = 0$ (the first term follows from what we have previously stated and the second term follows from the fact that $f_n(k)$ is an element of $C_0$. Thus, $f \in C_0$ and therefore $C_0$ is closed and therefore it is complete.  My questions are the following;   What exactly does it mean to be a sequence of functions? I used $f_n(k)$ and $f(k)$ in my proof because I saw that it was used in another exercise, but I'm not sure what it means to write that $k$ down. Is the structure of my proof sound? Is there a better way to proof such statements?   Thanks for reading,  K. Kamal
32597
Let $(X,d)$ be a metric space and $A\subset X$ and $x\in X$. Then  $x\to d(x,A)$ is a uniformly continuous function.  If $\partial A=\{x\in X\,:\,d(x,A)=0\}\cap\{x\in X\,:\,d(x,X-A)=0\}$, then $\partial A$ is closed for any $A\subset X$.  If $A,B$ are subsets of $X$  then $d(A, B)=d(B,A)$. If the function is uniformly continuous then $d(x,y)&lt;\delta $ implies $d(d(x,A)-d(y,A))&lt;\epsilon$ , I can not handle the last expression. Difficulty continues for 3rd choice also. At least give me some hints. And for the 2nd choice I think it is true, as it is intersection of two closed sets. As finite intersection of closed sets closed $\partial A$ closed. And logic for former two sets closed is that the sets are preimage of closed set (singleton $\{0\}$ is closed as $\mathbb R$ is $T_1$) under continuous map. I am not very much sure about my ideas and want a verification.
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.
74489
Let $(X,d)$ be a metric space and let $S$ be the set of Cauchy sequences in $S$. Define "~" in $S$ by "$\{s_k\}$ ~ $\{t_k\}$" to mean that $d(s_k,t_k) \to 0$ as $k \to \infty$. Let $X'$ denote the set of equivalence classes of $S$ and let $s'$ denote the equivalence class of $s = \{s_k\}$.   I know $\phi(s',t') = \lim\limits_{k \to \infty} d(s_k,t_k)$ is a metric on $X'$.   Show $(X',\phi)$ is complete.  For $x \in X$, define $x'$ to be the equivalence class of the constant sequence ${x,x,...}$. Show that the function $x \to x'$ is an isometry of $X$ onto a dense subset of $X$.   If anyone can share a proof, or steer me in the right direction, maybe tell me what it is that I need to prove specifically, that would be great.
61739
Let $f_1 : [-1, 1] \rightarrow \mathbb{R};\: f_1(0) = 0 $ be a continuously differentiable function and $\lambda &gt; 1$. Consider the sequence of functions defined inductively by $f_k(x) := \lambda f_{k-1}\left(\dfrac{x}{\lambda}\right);\: k \geq 2;\: x\in [-1, 1]$. Find the pointwise limit of the sequence of functions $(f_n)$.   I figured out that $f_k(0)=0 \:\forall\: k\implies f_k\rightarrow 0 $ when $ x=0$. Also $f_k$ is continuously differentiable $\forall \:k$.   $\Bigg($Since $\quad\displaystyle\lim_{h\rightarrow0}\dfrac{f_2(x+h)-f_2(x)}{h}=\lambda\lim_{h\rightarrow0}\dfrac{f_1\left(\frac{x+h}{\lambda}\right)-f_1\left(\frac{x}{\lambda}\right)}{h}.\: $Since$ f_1 $is continuously differentiable we see $ f_2 $ is differentiable and its derivative is continuous and hence $f_k$. $\Bigg)$.  Further since $f_1(0)=0$. So I assumed $f_1(x)=x\cdot g(x)$, where $g(x)$ is continuously differentiable and $g(0)\neq0$. Assume $x\neq0,\:f_2(x)=\lambda\cdot\dfrac{x}{\lambda}g\left(\dfrac{x}{\lambda}\right)=x\cdot g\left(\dfrac{x}{\lambda}\right)\implies f_3(x)=x\cdot g\left(\dfrac{x}{\lambda^2}\right)$ $\implies f_k(x)=x\cdot g\left(\dfrac{x}{\lambda^{k-1}}\right)$.   So taking limit as $k\rightarrow \infty$, we get $\displaystyle \lim_{k\rightarrow\infty}f_k(x)=x\cdot g(0)$. Now I am not sure how to proceed or whether there is something wrong here. Please provide hints or suggestions.  $\rule{17cm}{1pt}$  To sum up (using the arguments made by $\textbf{zhw}$ and $\textbf{Paul Frost}$)  When $\displaystyle x\neq0, \lim_{n\rightarrow\infty}f_{n+1}(x) = \lim_{n\rightarrow\infty}x\,\frac{f_1(x/\lambda^n)}{x/\lambda^n} = \lim_{n\rightarrow\infty}x\,\frac{f_1(x/\lambda^n)-f_1(0)}{x/\lambda^n}=x\:f'(0).$ Thus  $$f_n\rightarrow\begin{cases}0&amp;x=0\\x\:f'_1(0) &amp;x\neq0\end{cases}=f $$
3052
I'm currently reading Evans' PDE book. In it he claims that for $f \in C^2_c(\mathbb{R}^n)$ $$\frac{f(x + he_i) - f(x)}{h} \to \frac{\partial}{\partial x_i}f(x)$$ and $$\frac{\frac{\partial}{\partial x_i}f(x + he_j) - \frac{\partial}{\partial x_i}f(x)}{h} \to \frac{\partial^2}{\partial x_jx_i}f(x)$$ uniformly as $h \to 0$.  My question is why must the convergence be uniform?  Thanks in advance.
69333
A friend of mine gave me the following problem recently:     Let $(a_n)$ be a sequence of real numbers. Suppose that $\sum_{n=1}^\infty a_nb_n$ converges for any sequence $(b_n)$ such that $\lim\limits_{n\to\infty} b_n=0$. Show that $\sum |a_n|&lt;\infty$.   Hint he gave me was to use  (a.k.a. uniform boundedness principle).   I would like to know whether there are different solutions to this, in particular, whether there is an elementary solution. (Using just calculus, without resorting to more advanced results from functional analysis or other areas.)     The argument using Banach-Steinhaus theorem goes as follows:  Let $a=(a_n)$ be a given (fixed) sequence such that $\sum a_nb_n$ converges for each $b=(b_n)\in c_0$.  For every $n$ we can define $f_n\colon c_0\to{\mathbb R}$ as $\sum_{k=1}^n a_nb_n$. We consider $c_0$ with the usual sup-norm, which makes $c_0$ into a Banach space.  Each function $f_n$ is obviously linear.  For any $b\in c_0$ such that $\|b\|=\sup\limits_{n\in\mathbb N} |b_n| \le 1$ we have $$|f_n(b)| \le \sum_{k=0}^n |a_k|.$$ This shows that all functions $f_n$ are bounded linear functionals.   Now if we fix some $b=(b_k)\in c_0$ we have $$\left|\sum_{k=0}^n a_kb_k\right| \le \sum_{k=0}^\infty |a_k|\cdot |b_k|.$$ The sum on the RHS is finite, since it can be expressed as $\sum a_kc_k$, where $c_k=\pm b_k$. (The choice of sign here depends on the signs of $b_k$ and $a_k$.) So we have shown that at each point $b\in c_0$ the values $f_n(b)$ are bounded by the same constant (depending only on $b$).  By Banach-Steinhaus theorem we have that there exists a constant $M$ such that $\|f_n\|&lt;M$ for each $n\in\mathbb N$.  Now it only remains to show that $\|f_n\|=\sum_{k=0}^n |a_k|$. We already know that $\|f_n\|\le\sum_{k=0}^n |a_k|$. The other inequality can be obtained by choosing $b$ in unit ball, such that $b_k$ is either $1$ or $-1$, depending on the sign of $a_k$ for $k=1,2,\dots,n$; and $b_k=0$ for $k&gt;n$.  Now we have that $\sum_{k=0}^n |a_k|&lt;M$ for each $n$, which implies $\sum_{k=0}^n |a_k|\le M$.
96288
Prove that $\{f = \cos(nx) : n=1, ... ,\infty \}$ is a bounded subset (of functions), but not totally bounded, of $C([0, \pi])$ under the supremum norm (or uniform norm, see ).   Note that a totally bounded set has a convergent subsequence in a complete metric space.