Let $A_{n}:=\{f \in  C([0,1]): \exists x \in [0,1], \forall y \in [0,1] \operatorname{so that} |f(x)-f(y)|\leq n |x-y|\}$  I have already shown that $A_{n}$ is closed w.r.t $(C([0,1]), \vert \vert \cdot \vert \vert_{\infty})$ but now I want to show that $A_{n}$ is nowhere dense, and I am given the hint of considering the linear function $s:[0,1] \to [0,C]$ with slope $\pm(n + D)$ where $D$ and $C$ are some constants $&gt; 0$.   $(\overline{A_{n}})^{c}=(A_{n})^{c}:=\{f \in C([0,1]): \forall x \in [0,1], \exists y\in[0,1] \operatorname{so that} |f(x)-f(y)|&gt; n|x-y|\}$  Let $g \in C([0,1])$ and $ \epsilon &gt; 0$: since $g$ is continuous for all $x \in [0,1]$ then there exists $\delta_{x} &gt; 0$ so that for all $|x-y|&lt; \delta_{x}\Rightarrow |g(x)-g(y)|&lt;\epsilon$  $\vert \vert g - f \vert \vert_{\infty}=\sup_{x \in [0,1]}|g(x)-f(x)|$  I am unsure on how to use the linear function $s$. Any ideas?
121460
Let $E_n$ be $$E_n:=\{f\in C[0,1]\mid \text{exist } x_f\in[0,1] \text{ such that } |f(x)-f(x_f)|\leq n|x-x_f|,\, \forall x\in[0,1]\}.$$  How show that $E_n$ is nowhere dense, that is, $\mathrm{int}\  \overline{E_n} = \emptyset$?  Can someone help me?
45963
1) Construct a continuous function $f$ on $\mathbb{R}$ that is integrable on $\mathbb{R}$ but $\displaystyle\limsup_{x \to \infty} f(x)  = \infty$.  I took the function that is equal to $n$ on $[n, n + 1/n^{3})$ and made it continuous by saying that $f$ is the line segment joining $n$ and $n+1$ on $[n + 1/n^{3}, n+1)$. But I am failing to prove this integrable. For this, $\lim f(x) = \infty$ but how do you prove in general, if limit does not exist that $\limsup$ is infinity?  2) Prove that if $f$ is uniformly continuous and integrable on $\mathbb{R}$ we have $\displaystyle\lim_{|x| \to \infty} f(x) = 0$.  Any help is appreciated.  Thanks
21777
This is from Apostol's Mathematical Analysis.  Let $f$ be a function defined in $[0, 1]$ such that for each real number $y$ either there is no value of $x \in [0, 1]$ such that $f(x) = y$ or there are exactly two such values of $x$. Prove that a) $f$ must be discontinuous on $[0, 1]$. b) $f$ has an infinite number of discontinuities in $[0, 1]$.  Long back I had tried to solve the problem. I reasoned for a) as follows: Suppose that $f$ is continuous and let $m, M$ be its minimum and maximum values on $[0, 1]$. Clearly if $m = M$ then $f$ is constant and clearly for every $x \in [0, 1]$ we have $f(x) = m = M$ which contradicts given property of $f$. So we must have $m &lt; M$. Clearly there are now four distinct points $a, b, c, d$ in $[0, 1]$ such that $f(a) = f(b) = m, f(c) = f(d) = M$. By checking possible linear orderings of $a, b, c, d$ I could find a find a value of $y \in (m, M)$ for which there exist more than 2 values of $x$ such that $f(x) = y$. However this approach looks very clumsy and I suspect much simpler and easier proof is possible. Part b) seemed bit complex and my only guess was that probably $f$ was discontinuous on each subinterval of $[0, 1]$ but I could not prove this claim (may be the claim is wrong!). Perhaps there is a simpler solution to this part of the problem. Apostol's also asks to provide an example of such a function. Any hints on the overall problem will be appreciated.  Update: After looking at Marc's answer, I think I have found a way to show that there are infinitely many discontinuities. Earlier the solution was a part of the question itself but it made the question look quite lengthy hence I moved it to a separate answer.
29696
I'm trying to solve this previous qual problem from my univeristy: Let $L_n$ be the continuous linear functions on $L^\infty(\mathbb{R})$ given by $$L_n(\phi)=\frac{1}{n!}\int_0^\infty x^ne^{-x}\phi(x)\,dx.$$  Prove that $L_n$ has no subsequence that converges in the weak* topology of $L^\infty(\mathbb{R})^\ast$. So far, I managed to see that $\frac{1}{n!}x^ne^{-x}$ converges uniformly on compact sets to 0, hence the weak* limit of any subsequence of $L_n$ must take compactly supported $L^\infty$ functions to 0. I've also tried to construct a test function directly for any subsequence but so far I have yet to come up with such a construction. Can anyone pleas help me? Some hints would be very much appreciated! Thanks!
42031
I'm looking for proof verification/help for the title question. Here is what I have now: Let $f:X \rightarrow Y$ be continuous with $X$ limit point compact. Let $V$ be an infinite subset of $f(X)$. Then we also have that $f^{-1}(V)$ is also an infinite subset of $X$. Since $X$ is limit point compact there is $x \in X$ such that $x$ is a limit point of $f^{-1}(V)$. Then $x \in \overline{f^{-1}(V)}$ by definition of the closure. Hence since $f$ is continuous, $$ f(x) \in f(\overline{f^{-1}(V)}) \subseteq \overline{f(f^{-1}(V))} \subseteq \overline{V} $$ Then since $f(x) \in \overline{V}$, it is either a limit point of $V$ or an element of $V$. However since is infinite, any neighborhood of $f(x)$ will contain infinitely many points of $V$ in either case. Thus $V$ has a limit point and $f(X)$ is limit point compact.
52254
Let $f$ be an entire function and let for each $a\in \mathbb R$, there exists at least one coefficient $c_n$ in $f(z)=\sum\limits_{n=0}^{\infty}c_n(z-a)^n$, which is zero. Then    $f^{(n)}(0)=0$ for infinitely many $n\geq 0$ $f^{(n)}(0)=0$ for every $n\geq 0$ $f^{(2n+1)}(0)=0$ for every $n\geq 0$ There exists $k\geq 0$ such that $f^{(n)}(0)=0$ for all $n\geq k$   We know that $c_n=\frac{f^{(n)}(a)}{n!}$ for all $n\in \{0,1,2\ldots\}$. Thus for $a=0$, $c_n=\frac{f^{(n)}(0)}{n!}$. By hypothersis, atleast one $c_n=0$. After that I could not do anything. Please help!
20661
Let $V$ be a vector space with $\dim V=n$ . If $\ker f\subset \ker g$ where $f,g $ are non-zero linear functionals then show that $f=cg$ for some $c\in F$.  Now let $\mathcal B=\{v_1,v_2,\ldots ,v_n\}$ be a basis of $V$, since $f,g$ are non-zero linear functionals then $\exists v_i\in \mathcal B $ such that $g(v_i)\neq 0\implies f(v_i)\neq 0$ Take $i=1$ without any loss of generality so take $g(v_1)\neq 0,f(v_1)\neq 0$. Now take $c=\dfrac{f(v_1)}{g(v_1)}$ Then we need to show that $(f-cg)(v_i)=0\forall i$ Now $(f-cg)(v_1)=0$ How to show that $(f-cg)(v_i)=0\forall i\ge 2$ Can someone please help? Note::Another Question Why do we need the dimension of the vector space to be finite?
77075
Let $f$ be an entire function and $L$ a line in $\mathbb{C}$ such that $f(\mathbb{C})\cap L=\emptyset$. Show that $f$ is constant function.  If $f$ is not constant  then $f(\mathbb{C})$ is dense set in $\mathbb{C}$, but how can I use that line does not intersect the image set?
107421
Give an example of a continuous function $f : [0,  ∞) \mapsto [0, ∞)$ such that  $\int_{0}^{\infty}f(x)dx$ exists but  $f$ is unbounded.   I have been thinking about this. And I have come to the conclusion that I will need to construct a function, $f$, such that $f$ is a sequence of triangles of increasing height, but decreasing base. I obviously need $f$ such that both the height of the triangles, and sum of the bases tends to infinity. But I also need that the $\sum (\text{height} \times \text{base}) \leq \infty $
88187
Let $$f\colon [0,1]\rightarrow [0, \infty )$$  be  continuous. Suppose that  $$\int_{0}^{x} f(t)\,\mathrm dt \ge {f(x)} \quad\text{for  all }x\in[0,1].$$  Then   $A.$ No  such  function  exists.  $B.$ There  are  infinitely  many  such  functions.  $C.$ There is  only  one  such  function  $D.$ There  are  exactly  two  functions.  Now  I  was  thinking,  since $\int_{0}^{x} f(t) dt$  is  the  area  under  the  graph  of  $f(t)$  from $0$  to  $x$ so it  can  be  easily  made  greater  than  the  value  of  $f(t)$  at  one  point, namely  $x$.  So  there   will  be  infinitely  many  such  functions  satisfying  this  condition.   Is  my  reasoning  correct or  not?  Thanks.
51262
I was trying to do some of these questions to check my understanding about the topic, but I'm not sure if they're correct. Here are my answers.   1) Suppose $f$ is continuous on $[0,1]$. Must there be a nondegenerate closed subinterval $[a,b]$ of $[0,1]$ for which the restriction of $f$ to $[a,b]$ is of bounded variation?   No, because $f(x)$ might be a constant function. In this case, $f(x)$ will not be the difference of two increasing functions...and so will not be of bounded variation.  2) Let $f$ be the Dirichlet function, the characteristic function of the rationals in $[0,1]$. Is $f$ of bounded variation on $[0,1]$?   No, because we cannot express it as the difference of two increasing functions.  3) Let $f$ be a step function on $[a,b]$. Find a formula for its total variation.  If we partition $[a,b]$ in such a way that for each $[x_{i-1}, x_i]$ (which is part of the partition), $x_{i-1}$ and $x_i$ each correspond to the midpoint of a step. I know this is not a formal answer, but I'm just trying to check my understanding.
2881
Let $S = [0,1) \cup [2,3]$ and $f\colon S \rightarrow \mathbb R$ be such that $f(S)$ is connected . Which of the following are true:  a) $f$ is discontinuous exactly at one point.  b) $f$ is discontinuous exactly at two points.  c) $f$ is discontinuous at infinite many points.  d) $f$ is continous.  I think $f$ is continuous . We need to show that $\lim_{x \to a^+}f(x) = f(a)$ for any $a \in S$. Since $f(x)$ is increasing, $\lim_{x \to a^+}f(x) \leq f(a)$. Suppose $\lim_{x \to a^+}f(x) &lt; f(a)$. Then there is $l$ such that $\lim_{x \to a^+}f(x) &lt; l &lt;f(a)$, which contradicts connectedness of $f(S)$. Similarly we can show that $\lim_{x \to a^-}f(x) = f(a)$.  I would be thankful if anyone would check my solution and correct me if there are any errors.
101987
let $x=(x(1),x(2),\ldots,x(n),\ldots)\in c_0$  So for any $\varepsilon&gt;0$ there exists a $n_0\in\mathbb{N}$ such that $|x(n)|&lt;\varepsilon_2$ for all $n\geq n_0$.  Now for all $n\geq n_0$ ,$||x_n−x||_\infty=\sup{|x(m)|}_{m\geq n_0}\leq \varepsilon_2&lt;\varepsilon$  i have a doubt that regarding the last line. What makes it possible?
44547
I am trying to solve this exercise from Royden's 3rd edition.     The question is as follows: Let $f$ be a real-valued function defined for all real numbers. Show that the set of points at which $f$ is continuous is a $G_{\delta}$.    Let $$A_n = \{y : \text{there is a }~\delta_y \gt 0 : |f(s)-f(t)|\lt 1/n ~ \text{whenever}~ s,t \in (y-\delta, y+\delta)\}\;.$$     Then by the definition of open sets, $A_n$ is open.     To complete the proof, I need help in showing that $f$ is continuous at say $x$ if and only if $x\in \cap A_n$.     If $f$ is continuous at $x$, the there is a $\delta \gt 0$ such that $|f(x) - f(a)| \lt 1/n$ whenever, $x\in (a-\delta, a+\delta)$. so $x \in A_n$ son it must be in $\cap A_n$.  Thanks.
61624
Given a function on a closed interval $f\colon I\subset \mathbb{R}\to \mathbb{R}$ with $$f(x+y) \leq f(x) + f(y).$$ Moreover, I know that $f$ is    monotonic increasing continuous on all points except countably many continuous from the right such that the limit from the left at each point exists. $f(nx) \leq nf(x)$ for all $n\in\mathbb{N}$.   Can I say more on this function? In particular, I would like to show that $f(\lambda x) \leq \lambda f(x)$ for all $\lambda \geq 1$ or at least for all rational $\lambda \geq 1$.
23086
Let $u$ be a distribution on $\mathbb{R}^n$ with support = $\left\{0\right\}$. Then there exists $N$ such that $u$ has order $N$. Let $\chi\in C_0^{\infty}(\mathbb{R}^n)$ a smooth function with  $\chi(x) = 1$ for $0\leq |x|\leq 1 $,   $\chi(x)\in [0,1]$ for $1\leq |x|\leq 2$,  $\chi(x)=0$ for $|x|\geq 2$  Denote $ \chi(x/r) = k_r(x)$ for $r\in (0,1]$. For the case $N=0$ I want to show that there exists $c_1$ such that $$|\left\langle u,\phi \right\rangle|\leq c_1 |\phi(0)|  $$ for all $\phi \in C_0^{\infty}(\mathbb{R}^n)$. The idea would be to apply $u$ on $\phi = k_r\phi+(1-k_r)\phi$ and letting $r\to 0$.   So then I guess we can write $$ |\left\langle u,\phi \right\rangle| \leq |\left\langle u,k_r\phi \right\rangle| + |\left\langle u,(1-k_r)\phi \right\rangle |$$ But not really sure how this follows...  Moreover how can i show that $\left\langle u,\phi \right\rangle = \left\langle u,\phi \chi\right\rangle = \left\langle u,\chi\right\rangle\phi(0) $?
83217
Let $G\subset \mathbb{C}$ be a non-empty open set and $f$ be a function holomorphic on $G$. Let $g: G\times G\to \mathbb{C}$ be a function defined as $$g(z,w)= \begin{cases}        \frac{f(z)-f(w)}{z-w}, &amp; z\ne w \\       f'(z), &amp; z=w     \end{cases}$$  I need to prove that $g$ is continuous on $G\times G$.  My approach:  Let $\varepsilon &gt; 0$. Then $\exists \delta_1, \delta_2&gt;0$ such that $\| (z,w)-(z_0,w_0) \|^2\le|z-z_0|+|w-w_0|&lt;\delta_1+\delta_2$ implies that $|f(z)-f(z_0)|&lt;\epsilon$, $|f(w)-f(w_0)|&lt;\epsilon$, since $f$ is continuous on $G$.  Now, $\| g(z,w)-g(z_0,w_0) \|=\left| \frac{f(z)-f(w)}{z-w} - \frac{f(z_0)-f(w_0)}{z_0-w_0} \right|\le \left| \frac{f(z)-f(w)}{z-w}\right| + \left|\frac{f(z_0)-f(w_0)}{z_0-w_0} \right|$  Since $f$ is holomorphic on $G$, it is continuous on $G$, thus $\exists \delta'&gt;0$ such that $| {f(z)-f(w)}|&lt;\varepsilon'|z-w|&lt;\varepsilon'\delta'$ whenever $|z-w|&lt;\delta'$.  And here's where I'm stuck, because it's not clear how to deal with $\left|\frac{f(z_0)-f(w_0)}{z_0-w_0} \right|$, because $z_0, w_0$ are fixed points, so we can't just make them approach each other. Of course, my first thought was to somehow rearrange the inequality in such a way as to obtain $\left| \frac{f(z)-f(z_0)}{z-z_0}\right| + \left|\frac{f(w)-f(w_0)}{w-w_0} \right|$, but I don't see how to do so.  Your help would be appreciated.