Let $f:\mathbb{R}\to (0,1)$, $f=t\sin\left(\frac1t\right)$. I think this function is uniformly continuous, but I don't see how to prove this formally. I can observe that if $s,t\in (0,1)$ and $|s-t|&lt;\delta$, then the smaller $\delta$ we take, the smaller the quantity $$\left|t\sin\left(\frac1t\right)-s\sin\left(\frac1s\right)\right|\mbox{  (*)}$$ becomes. So we can take any $\varepsilon&gt;0$ and any $s,t$, such that (*) is less than $\varepsilon$, and there will necessarily be a $\delta&gt;0$ such that $|s-t|&lt;\delta$.  I would appreciate some suggestions on what approach to take.
122632
A student showed me the following exercise:   Is $h(x) = x \sin\left(\frac{1}{x}\right)$ uniformly continuous on $(0,1)$?  Admittedly, it has been a while since I have looked at problems involving uniform continuity and so I am not sure how to go about this problem.  My guess is that $h(x)$ is uniformly continuous on this interval but I am not sure how to go about picking an appropriate $\delta &gt; 0$ such that $|x-c|&lt; \delta$ implies $|h(x) - h(c)| &lt; \epsilon$ for $x,c \in (0,1)$.   The student and I looked at $|x \sin(x) - c \sin(c)| \leq |x \sin(x)| + |c \sin(c)| \leq |x| + |c|$ although this feel like the wrong path to go down.  I appreciate any help! Thanks in advance!
46294
I'm told that a function defined on an interval $[a,b]$ or $(a,b)$ is uniformly continuous if for each $\epsilon\gt 0$ there exists a $\delta\gt 0$ such that $|x-t|\lt \delta$ implies that $|f(x)-f(t)|\lt \epsilon$. Then it gives a little note saying that $\delta$ cannot depend on $x$, it can only depend on $\epsilon$.   With ordinary continuity, the $\delta$ can depend on both $x$ and $\epsilon$. I'm just a little lost on why $|x-t|\lt \delta$ implies $|f(x)-f(t)|\lt \epsilon$, and how $\delta$ can't depend on $x$ but only $\epsilon$.
104503
$\sin \frac{1}{x}$ is not uniformly continuous on $[1, \infty]$  I was trying to come up with a counter example using sequence, but I could not find one.
58784
Determine whether the function$$f(x)={\sin x \over x}$$is uniformly continuous in $\mathbb{R}$.  I am using the definition that for $\epsilon&gt;0$ there exists $\delta&gt;0$ such that $|f(x)-f(y)|&lt;\epsilon$ whenever $|x-y|&lt;\delta$.
109196
Some days ago I answered a question that asked to find  $$\mathop {\lim }\limits_{x \to 0} {x^\alpha }\int\limits_x^1 {\frac{{f\left( t \right)}}{{{t^{\alpha  + 1}}}}dt} $$  given that $f$ is continuous in $[0,1]$  I proceeded as follows:  $$\eqalign{   &amp; t = x\cdot u  \cr    &amp; dt = x\cdot du \cr} $$  So this is produces:  $$\mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right)}}{{{u^{\alpha  + 1}}}}du} $$  I then thought: "Well, if $f$ is continuous in the closed interval, then it is also uniformly continuous, so I can assume   $$\mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right)}}{{{u^{\alpha  + 1}}}}du}  = f\left( 0 \right)\int\limits_1^\infty  {\frac{{du}}{{{u^{\alpha  + 1}}}}}  = \frac{1}{\alpha }f\left( 0 \right)$$  This turned out to be true. However, I wasn't very comfortable with such "move". So now I'm thinking, one can put  $$\mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right) - f\left( 0 \right)}}{{{u^{\alpha  + 1}}}}du}  + \mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{f\left( 0 \right)}}{{{u^{\alpha  + 1}}}}du} $$  And then   $$\left| {\int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right) - f\left( 0 \right)}}{{{u^{\alpha  + 1}}}}du} } \right| &lt; \int\limits_1^{\frac{1}{x}} {\frac{{\left| {f\left( {xu} \right) - f\left( 0 \right)} \right|}}{{{u^{\alpha  + 1}}}}du}  &lt; \epsilon \frac{{1 - {x^\alpha }}}{\alpha }$$  However, this is still insufficient since I need to adress the behaviour of the upper limit too. Can someone show me how to adress both behaviours simultaneously?    Would this work?  Let $P$ be the statement that $$\mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right)}}{{{u^{\alpha  + 1}}}}du}  = \frac{{f\left( 0 \right)}}{\alpha }$$  Then $P$ is true if and only if  $$ \forall \epsilon  &gt; 0\exists \delta  &gt; 0$$  Such that if $$\left| x \right| &lt; \delta $$ then $$ \left| {\int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right)}}{{{u^{\alpha  + 1}}}}du}  - \frac{{f\left( 0 \right)}}{\alpha }} \right| &lt; \epsilon $$  But then   $$\left| {\int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right) - f\left( 0 \right)}}{{{u^{\alpha  + 1}}}}du}  + \int\limits_1^{\frac{1}{x}} {\frac{{f\left( 0 \right)}}{{{u^{\alpha  + 1}}}}du - \frac{{f\left( 0 \right)}}{\alpha }} } \right| &lt; $$  $$\left| {\int\limits_1^{\frac{1}{\delta }} {\frac{{f\left( {xu} \right) - f\left( 0 \right)}}{{{u^{\alpha  + 1}}}}du}  + \int\limits_1^{\frac{1}{\delta }} {\frac{{f\left( 0 \right)}}{{{u^{\alpha  + 1}}}}du - \frac{{f\left( 0 \right)}}{\alpha }} } \right| \leqslant $$  $$\varepsilon \frac{{1 - {\delta ^\alpha }}}{\alpha } + f\left( 0 \right)\frac{{1 - {\delta ^\alpha }}}{\alpha } - \frac{{f\left( 0 \right)}}{\alpha } &lt; $$  And since  $$\frac{{1 - {\delta ^\alpha }}}{\alpha } &lt; \frac{1}{\alpha }$$ $$\epsilon \frac{{1 - {\delta ^\alpha }}}{\alpha } + f\left( 0 \right)\frac{{1 - {\delta ^\alpha }}}{\alpha } - \frac{{f\left( 0 \right)}}{\alpha } &lt; \frac{\epsilon }{\alpha } &lt; \epsilon $$
32010
This should be a very easy question but it might just be that I'm confusing myself. So we have the definition of a function $f$ on $S$ being continuous at $x_0$:     For any $\epsilon$>0, there exists $\delta&gt;0$ such that: whenever $|x-x_0|&lt;\delta$, we have $|f(x)-f(x_0)|&lt;\epsilon$   And I assume the negation is     There exists $\epsilon$>0 such that for all $\delta&gt;0$, $|x-x_0|&lt;\delta$ yet $|f(x)-f(x_0)|\ge \epsilon$.   Now I want to show that the function $f(x)=\sin(\frac{1}{x})$ together with $f(0)=0$ cannot be made into a continuous function at $x=0$. So I need to show that there exists $\epsilon&gt;0$ such that for all $\delta&gt;0$, $|x|&lt;\delta$ yet $|f(x)|\ge\epsilon$.  Let $\epsilon = \frac{1}{2}$. Then no matter what $\delta$ we choose, let $|x|&lt;\frac{1}{2}$. It is certainly possible that $|f(x)|\ge \frac{1}{2}$, because, well, $\frac{1}{x}$ can really take on arbitrarily large value as $x$ is small.  Now, what confuses me is that, as $x$ gets small, $f(x)$ can certainly be greater than $\frac{1}{2}$ for infinitely many times, but it will be less than that infinitely many times, too. But I suppose it doesn't really matter. So I think there's something wrong with my negation but I couldn't figure out where.  Update: The correct version can be found . Watch for Lemma 4.6
51248
So I understand the definition of uniform continuity, but wanted some suggestions to prove that a function is or isn't uniformly continuous. I have looked ahead and have seen that if a function is continuous on a compact domain, then the function is also uniformly continuous. But for the purposes of the examples below, I cannot use compactness, or the fact that a function is uniformly continuous if it is continuous and has a bounded derivative. Here's one example:  prove $f(x)= \sqrt{x}$ is uniformly continuous on $(0, \infty)$. I have seen this question asked before on this forum, and the respondents used the traditional definition. Ie. they found a delta such that if $|x-y|&lt;\delta$, then $|\sqrt{x}-\sqrt{y}|&lt; \epsilon$. I was wondering if I could try this in a slightly different way. Suppose I prove that $f$ is continuous at $x_0$, and the choice of $\delta$ doesn't depend on $x_0$. After some scratch work, I proved that $f(x)$ is continuous at $x_0$ provided that $|x-x_0|&lt; \min\{1,(\sqrt{x_0+1} + \sqrt{x_0})*\epsilon\}$. Since $x_0$ is always greater than $0$, we can say that $\delta= \min\{1, \epsilon\}$. Because I proved that our choice of $\delta$ doesn't depend on $x_0$, is this a valid proof that $f(x)$ is uniformly continuous? Here's another example: I want to prove that $f(x)=x^2$ is not uniformly continuous on $(0, \infty)$. After some work, I proved that $f(x)$ is continuous at $x_0$ (in dom $f$), provided that $\delta&lt; min \{1, \frac {\epsilon}{ (2x_0+1)}\}$. Here, since our choice of delta does depend on $x_0$, we can say the function is not uniformly continuous. So am I approaching these problems correctly? I realize that I have not been using the definition of uniformly continuity. (that if $f$ is uniformly continuous, then if $|x-y|&lt;\delta$, then$|f(x)-f(y)|&lt; \epsilon$. Also, sorry about the formatting. I realize I should learn proper formatting soon!. Also, I have read the following post regarding this topic:
11949
Put precisely, if $f:\mathbb R\to\mathbb R$ is continuous with  $$ \lim_{s\to 0^+}s^{-1}(f(t+s)-f(t-s))=0 \kern20mm (*) $$ for all $t\in\mathbb R$ , is $f$ then constant?  I guess it is but I cannot see the proof. Obviously, if instead of ($*$) we had $\lim_{s\to 0^+}s^{-1}(f(t+s)-f(t))=0$ , i.e. the right-handed derivative zero, then the assertion could be proved indirectly by a simple compactness argument applied to closed intervals $[a,b]$ with $f(a)\not=f(b)$ . However, I cannot see how this argument could be modified to get the above result. Is there possibly a reference to this, or is the result even false?
21229
Please help me out with this problem:  Let $f_n(x) : [0,1] \to \mathbb R$ be a sequence of continuous functions convergent at every $x \in [0,1]$ to a continuous function $f: [0,1]  \to \mathbb R$ . Does $f_n$ converge uniformly to $f$?  My first solution: I read this example in  posted earlier and I think it works: Take a sequence $f_n:[0,1] \to \mathbb R$ such that $f_n$ increases linearly from $0$ to $1$ on the interval $\left[ 0,\frac{1}{n} \right]$, decreases linearly from $1$ to $0$ on the intreval $\left[ \frac{1}{n}, \frac{2}{n} \right]$, and is $0$ on $\left[ \frac{2}{n},1 \right]$. Then $f_{n} \to 0$ nonuniformly.  Is this answer true? Also, if anyone can give some examples of such function with detailed proof, I will very appreciative?
102154
I have been doing some excercises on total variation when the following questions came up to my mind:  (1) Let $f$ be continuous on the interval $[0,1]$ and be of bounded variation. Is it true that its total vatiation function $TV(f_{[0,x]})$ is uniformly continuous? i.e. Is it true that for $\forall\space\epsilon&gt;0$, $\exists\space\delta&gt;0$, such that for arbitrary interval $[a,b]$ with $|b-a|&lt;\delta$, we have $TV(f_{[a,b]})&lt;\epsilon$?  (2) Alternatively, if it is not uniformly continuous, can I say that for $\forall\space{}x\in[0,1]\text{ and } \forall\space\epsilon&gt;0$ , $\exists\space\text{ nondegenerate interval }I \text{ such that }x\in{}I\text{ and } TV(f_{I})&lt;\epsilon$?  Thank you!
44777
Prove that if $f:(0,\infty)→\mathbb{R}$ satisfying $f(xy)=f(x)+f(y)$, and if $f$ is continuous at $x=1$, then $f$ is continuous for $x&gt;0$.   I let $x=1$ and I find that $f(x)=f(x)+f(1)$ which implies that $f(1)=0$. So, $\lim_{x\to1}f(x)=0$, but how can I use this to prove continuity of $f$ for every $x \in \mathbb R$?   Any help would appreciated. Thanks
9106
This is a homework problem for my real analysis class. It is problem 13.1 from Mattuck's Introdution to Analysis. The question reads:       Suppose F(x) is continuous on some open interval I and c is a maximum point inside this interval. Is it true that f(x) must be increasing immediately to the left of c and decreasing immediately to the right of c? Proof or counterexample. (Note: A constant function is considered to be both increasing and decreasing.)      My friend and I have been attempting to find a counterexample. We figured a function that begins oscillating infinitely many times near c but converges to c would work. We're having difficulty constructing such a function out of elementary functions so we think we will have to define it in another way, piece-wise using receding intervals. I have no clue how this would look. Any advice would be greatly appreciated.
106020
Prove that the function $x \mapsto \dfrac 1{1+ x^2}$ is uniformly continuous on $\mathbb{R}$.  Attempt: By definition a function $f: E →\Bbb R$ is uniformly continuous iff for every $ε &gt; 0$, there is a $δ &gt; 0$ such that $|x-a| &lt; δ$ and $x,a$ are elements of $E$ implies $|f(x) - f(a)| &lt; ε.$ Then suppose $x, a$ are elements of $\Bbb R. $ Now \begin{align} |f(x) - f(a)|  &amp;= \left|\frac1{1 + x^2} - \frac1{1 + a^2}\right| \\&amp;= \left|  \frac{a^2 - x^2}{(1 + x^2)(1 + a^2)}\right| \\&amp;= |x - a| \frac{|x + a|}{(1 + x^2)(1 + a^2)} \\&amp;≤ |x - a| \frac{|x| + |a|}{(1 + x^2)(1 + a^2)} \\&amp;= |x - a| \left[\frac{|x|}{(1 + x^2)(1 + a^2)} + \frac{|a|}{(1 + x^2)(1 + a^2)}\right] \end{align} I don't know how to simplify more. Can someone please help me finish? Thank very much.
49362
I am to show the continuity of this function with the help of $\epsilon$-$\delta$ argument.   The function is: $g: \Bbb{R} \rightarrow \Bbb{R}$, $x \mapsto x^2$.   Given the $\epsilon$-$\delta$ definition of limit, I tried as follows:   We must have: $|f(x)-f(x_0)|&lt;\epsilon$, so then $|x^2-x_0^2|=|(x-x_0)(x+x_0)|=|(x-x_0)||(x+x_0)|$, so $|x-x_0|&lt;\frac{\epsilon}{|x+x_0|}$. So now I will choose $\delta=\frac{\epsilon}{|x+x_0|}$.   Is it enough? I am writing this sentences like a machine, but I am not understanding intuitively.
39349
Show that the bounded function $f(x) = \sin(x^2)$ is continuous, but not uniformly continuous, on the interval $[0,\infty)$.  The problem should be solved using the following definitions: Definition of Continuous Function Let $X$ and $Y$ be metric spaces. We sat that a function $f:X\rightarrow Y$ is continuous if given any open set $V \subseteq Y$, its inverse image $f^{-1}(V)$ is an open subset of $X$. Definition of Uniformly Continuous Function Let $(X,d_x)$ and $(Y,d_y)$ be metric spaces. If $E \subseteq X$, then we say that a function $f: X\rightarrow Y$ is uniformly continuous on $E$ if for every $\epsilon &gt; 0$ there exists a $\delta &gt; 0$ such that for all $x,y \in E$ we have $$d_x(x,y) &lt; \delta \longrightarrow d_y(f(x),f(y)) &lt; \epsilon$$ For the continuity part it is clear that $f^{-1}(a,b), (a,b) \subseteq (-1,1) = \{x:f(x) \in (a,b)\}$ which appears to be an open set however I doubt this is sufficient as a proof since it holds for any function $f$. Not sure how to go about proving $f$ not uniformly continuous. Seems to be difficult to prove from the definition alone.
83708
I have to prove that the function $f:]0,1] \rightarrow \Bbb R$ : $$ f(x) = \begin{cases} \frac1q,  &amp; \text{if $x \in \Bbb Q$  with $ x=\frac{p}q$ for $p,q \in \Bbb N$ coprime} \\ 0, &amp; \text{if $x \notin \Bbb Q $} \end{cases} $$  is continuous in  $x \in \ ]0,1] \backslash \Bbb Q$ (irrational numbers).  I already tried to solve this exercise and I understood how to prove that this function is discontinuous for rational numbers.  I was trying, now, to understand how to prove that for the irrational numbers it is continuous.  I already have the solutions but they are a bit complicated, and I come up using sequences to prove it's continuous: $$\forall x_n \quad x_n\rightarrow a \quad \Rightarrow \quad f(x_n) \rightarrow f(a)$$  I created a sequence $ x_n=\frac1n + a$, and we consider $a$ an irrational number. In order to be continuous $f(x_n)$ has to converge to $f(a)$. In fact:  $$\lim_{n\to\infty} f(x_n)= f(a) = 0 $$  So my proof ends here, but I know that I'm missing something really big because in the solution there is another complicated step using a set and the Epsilon-Delta theorem after what I've showed.  Can someone explain what's wrong and help me with this proof?
69552
$$\lim_{x \to 0} \frac{f(x)}{|x|}=1$$ Then prove that $f$ is not differentiable at $0$. I was given this assignment and I'm not sure about my proof.. it feels like I'm assuming too much and not explaining with enough detail. Is there anything I need to add to make it sufficient? Assume toward contradiction that $f$ is differentiable at $0$. Then by definition we know that $f$ is continuous at $0$. Therefore $\lim _{x\to 0}f\left(x\right)=f\left(0\right)$ We know that $\lim _{x\to 0}\frac{\left(f\left(x\right)\right)}{\left|x\right|}=1$ Therefore we can conclude $\lim _{x\to 0}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \left|x\right|=1\cdot 0=0$ But we can also rewrite this as $\lim _{x\to 0}\left(f\left(x\right)\right)\cdot \frac{\left|x\right|}{\left|x\right|}=f\left(0\right)\cdot 1=0$ Therefore $f(0) = 0$. Since $f$ is differentiable at $0$ we have $\lim _{x\to 0}\frac{\left(f\left(x\right)-f\left(0\right)\right)}{x-0}=\lim _{x\to 0}\frac{\left(f\left(x\right)-0\right)}{x}=\lim \:_{x\to \:0}\frac{\left(f\left(x\right)\right)}{x}\cdot \frac{\left|x\right|}{\left|x\right|}=\lim \:\:_{x\to 0}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \:\frac{\left|x\right|}{x}$ Since the one sided limits of $\frac{f\left(x\right)}{\left|x\right|}$ exist, and are both equal to 1, we have $\lim \:_{x\to 0^+}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \:\:\frac{\left|x\right|}{x}=\lim \:_{x\to \:0^+}\frac{\left(f\left(x\right)\right)}{x}\cdot \:\:\:\frac{x}{x}=1\cdot 1=1$ $\lim \:_{x\to 0^-}\frac{\left(f\left(x\right)\right)}{\left|x\right|}\cdot \:\:\frac{\left|x\right|}{x}=\lim \:_{x\to \:0^-}\frac{\left(f\left(x\right)\right)}{-x}\cdot \:\:\:\frac{-x}{x}=1\cdot -1=-1$ Since the limit doesn't exist, we have a contradiction to the fact $f$ is differentiable at $0$. Therefore $f$ is not differentiable at 0.