Show $f(x):=\sqrt{x}$ is uniformly continuous on $[0,1]$.  What I did:  Need to show that $\forall \varepsilon&gt;0: \exists\delta&gt;0$ such that $\forall x,y\in(0,1): |x-y|&lt;\delta\Rightarrow|f(x)-f(y)|&lt;\varepsilon$  Choose $\delta=\varepsilon^2$.  then for $x,y\in(0,1)$ with $|x - y| &lt; \delta$,  $|f(x)-f(y)|=|\sqrt{x}-\sqrt{y}|$  $|\sqrt{x}-\sqrt{y}|\le|\sqrt{x}+\sqrt{y}|$  $|\sqrt{x}-\sqrt{y}|^2\le|\sqrt{x}-\sqrt{y}||\sqrt{x}+\sqrt{y}| =|x-y|&lt; \varepsilon^2$  $|\sqrt{x}-\sqrt{y}|\le\sqrt{|x-y|}&lt;\sqrt{\varepsilon^2}=\varepsilon$
29138
Prove that the function $\sqrt x$ is uniformly continuous on $\{x\in \mathbb{R} | x \ge 0\}$.  To show uniformly continuity I must show for a given $\epsilon &gt; 0$ there exists a $\delta&gt;0$ such that for all $x_1, x_2 \in \mathbb{R}$ we have $|x_1 - x_2| &lt; \delta$ implies that $|f(x_1) - f(x_2)|&lt; \epsilon.$  What I did was $\left|\sqrt x - \sqrt x_0\right| = \left|\frac{(\sqrt x - \sqrt x_0)(\sqrt x + \sqrt x_0)}{(\sqrt x + \sqrt x_0)}\right| = \left|\frac{x - x_0}{\sqrt x + \sqrt x_0}\right| &lt; \frac{\delta}{\sqrt x + \sqrt x_0}$   but I found some proof online that made $\delta = \epsilon^2$ where I don't understand how they got? So, in order for $\delta =\epsilon^2$ then $\sqrt x + \sqrt x_0$ must $\le$ $\epsilon$ then $\frac{\delta}{\sqrt x + \sqrt x_0} \le \frac{\delta}{\epsilon} = \epsilon$. But then why would $\epsilon \le \sqrt x + \sqrt x_0? $ Ah, I think I understand it now just by typing this out and from an earlier hint by Michael Hardy .
52003
For $\alpha &gt;1$, $x^\alpha$ is not uniformly continous over $\mathbb R$ , essentially because $\forall \delta&gt;0, \lim_n (n+\delta)^{\alpha}-n^{\alpha}=\infty$  I don't have a clear intuition about the behavior of uniformly continuous functions over an unbounded interval.      Is is safe to say that if $f$ is uniformly continuous over $\mathbb R$ then $\displaystyle \forall \alpha &gt;1, f=o_{\infty}(x^{\alpha})$ ?
69272
I would like to show that the function $\frac{1}{x}$ is not uniformly continuous on $(0,1)$ using two approaches.   First Approach:   We have the fact that if a function $f$ is uniformly continuous on an open interval $(a,b)$, then the function $f$ is bounded on $(a,b)$. By using its contrapositive, since $\frac{1}{x}$ is not bounded on $(0,1)$, it is not uniformly continuous.  Second Approach:  Proof by definition. Note that the definition of non-uniform continuity is:  There exists an $\epsilon_0&gt;0$ such that for all $\delta&gt;0$, there exists $x,y \in (0,1)$ such that $|x-y|&lt;\delta$ but $|\frac{1}{x} - \frac{1}{y}| \geq \epsilon_0$.  Assume that $\delta &lt; 1$. Let $x = y + \frac{\delta}{2} \in (0,1), y &lt; \frac{\delta}{2} \in (0,1), \epsilon_0 = 1$. Then we have   $$|\frac{1}{x} - \frac{1}{y}| = |\frac{1}{y + \frac{\delta}{2}} - \frac{1}{y}| = \frac{\frac{\delta}{2}}{|y(y + \frac{\delta}{2})|} &gt; \frac{\frac{\delta}{2}}{(\frac{\delta}{2})(\delta)} = \frac{1}{\delta} &gt; 1 = \epsilon_0 $$  If $\delta \geq 1$, then let $\delta^{\prime} = \frac{1}{\delta}$ so that $\delta^{\prime} \leq 1$. The rest proceeds the same as above.   Question: Are my two approaches correct?
56832
Let $h(x) = x$ for all rational numbers $x$ and $h(x) = 0$ for irrational numbers. I'm trying to show that $h$ is continuous at $x=0$ and at no any other point.  Defintion of continuity at $x_0$: $(\forall x)(\forall \varepsilon &gt; 0)(\exists \delta &gt; 0)$ such that $|x-x_0| &lt; \delta \implies |h(x) - h(x_0)| &lt; \varepsilon$.  For this part of the question I took $\delta = \epsilon$. Then $|x - 0| &lt; \delta \implies |h(x)| &lt; \varepsilon$ because if $x$ is rational, then $x &lt; \varepsilon$ and if $x$ is irrational then $|x| &lt; \varepsilon$.   Defintion of discontinuity at $x_0$: $(\exists x)(\exists \varepsilon &gt; 0)(\forall \delta &gt; 0)$ such that $|x-x_0| &lt; \delta \implies |h(x) - h(x_0)| \geq \varepsilon$.  I tried to divide the problem into two cases:  1) If $x_0$ is rational, then $|h(x) - h(x_0)| = |h(x) - x_0|$. Then pick $x$ to be a irrational number. Then I have $|h(x) - x_0| = |x_0|$. And take $\varepsilon = \frac{|x_0|}{2}$.  2) Now if $x_0$ is irrational $|h(x) - h(x_0)| = |h(x)|$. Now here is where I was unsure of myself: If I pick $x$ to be a rational number, then $|h(x)| = |x|$. I was not sure about choosing $\varepsilon$ so that $|x| \geq \varepsilon$. Would it be valid to take $\varepsilon = \frac{|x|}{2}$? (I mean $|x|$ here and not $|x_0|$)  Could I get some feedback? Thanks!  Edit To finish off the last part, let $\varepsilon = \frac{|x_0|}{2}$. Now I want to choose an $x$ in the rational numbers. I also need to satisfy $x_0 - \delta &lt; x$ (since $|x_0 - x| &lt; \delta$) and $|x| \geq \frac{|x_0|}{2}$. So now I pick $x \in (|x_0|-\delta, |x_0|) \cap (\frac{|x_0|}{2}, |x_0|)$.
117576
Let $f\colon(a,b) \to \mathbb R$ be differentiable and suppose that there exists an $M&gt;0$ such that $|f&#39;(x)| \leq M$ for all $x$ in $(a,b)$. Prove that $f$  is uniformly continuous.
118193
I want a $\epsilon-\delta$ proof for the continuity of $f(x)=x^n$ around a point $a$ on the domain of this function.  Here's my sketch:  $$|f(x) - f(a)| = |x^n - a^n| = |x-a||x^{n-1} + x^{x-2}a  +\dotsb + xa^{n-2} + a^{n-1}|$$  For every pair $x^{n - j}a^{j-1} + x^{j-1}a^{n-j}$ we can write as   \begin{multline} x^{j-1}a^{j-1}\bigl((x^{n-2j +1} - a^{n-2j+1}) + 2a^{n-2j+1} \bigl)\\ = x^{j-1}a^{j-1}\bigl((x - a)(x^{n-2j} + x^{n-2j -1}a +\dotsb + xa^{n-2j-1} + a^{n-2j}) + 2a^{n-2j+1} \bigl) \end{multline}  for every $j = 1,2,..., n$   Since $|x-a|&lt;\delta$:  $$\bigl|x^{j-1}a^{j-1}\bigl((x - a)(x^{n-2j} + x^{n-2j -1}a +\dotsb + xa^{n-2j-1} + a^{n-2j}) + 2a^{n-2j+1} \bigl)| &lt; |x^{j-1}a^{j-1}| \bigl( \delta (x^{n-2j} + x^{n-2j -1}a +\dotsb + xa^{n-2j-1} + a^{n-2j}) + |2a^{n-2j+1}| \bigl)$$   From now on I can't think of anything that would help me to conclude, I can't find any pattern that would help me summarize it, I would be glad if you help me.  Thanks.
103524
I have to prove that $f(x)=\log(1+x^2)$ is Uniform continuous in $[0,\infty)$ (with $\epsilon ,\delta$ formulas...)  I wrote the definition: (what I have to prove):  $\forall \epsilon&gt;0 \quad \exists \delta&gt;0 \quad s.t. \quad \forall x,y\in [0,\infty) : \quad |x-y|&lt;\delta \quad \Rightarrow \quad |f(x)-f(y)|&lt;\epsilon $  So I tried developing $|f(x)-f(y)| = |\log(1+x^2)-\log(1+y^2)| = |\log(\frac{1+x^2}{1+y^2})| $...   now this is where I try to make it bigger and simplify the expression so i can choose the right $\delta$ which depends on the $\epsilon$, and then say that if the simplified bigger expression is still smaller than $\epsilon$ then of course the original $|f(x)-f(y)|$ is smaller than $\epsilon$      but what can I do with this expression? any algebraic tricks?
68193
Let $f : [a, +\infty) \rightarrow \mathbb{R}$ be continuous, such that   $\lim_{x \rightarrow +\infty} f(x) = l \in \mathbb{R}$. Show that f is   uniformly continuous.   I want to split the domain - one in which we have continuity, and the other in which the values of f are approaching $l$. This means there are theree possibilities for our $x$ and $y$ using the uniformly continuous definition: x and y are in the first part, x and y are in the second, x is in the first and y in the second.   How would I do this proof?
2015
Suppose that $f$ is continuous at $x_0$ and $f$ satisfies $f(x)+f(y)=f(x+y)$. Then how can we prove that $f$ is continuous at $x$ for all $x$? I seems to have problem doing anything with it. Thanks in advance.
100251
A. $f$ is bounded      B. $f$ may not be uniformly continuous      C. $f$ is uniformly continuous      D. $f$ is unbounded.   Let $f(x)=\sqrt x$. Then $g(x)=x$ is uniformly continuous and unbounded. Hence option A can not be true.  Let $f(x)=1$. Then $g(x)=1$ is uniformly continuous and bounded. Hence the option D can not be true.  How should I check uniform continuity?  EDIT:  Since $g$ is uniformly continuous, hence for a given $\epsilon \gt 0$, $\exists \delta \gt 0$ depending only upon $\epsilon$ such that $|g(x)-g(y)| \lt \epsilon$ whenever $|x-y| \lt \delta$  $\forall x,y$.  $\Rightarrow |\sqrt g(x) -\sqrt g(y)||\sqrt g(x)+\sqrt g(y)|\lt \epsilon$ whenever $|x-y| \lt \delta$.  $\Rightarrow |f(x)-f(y)||f(x)+f(y)|\lt \epsilon$ whenever $|x-y| \lt \delta$.  What should be my next step now? How can I use the continuity of $f$?
76305
$f:X\rightarrow Y $ is continuous iff every open ball in $Y$ has open pre-image. It is between topological spaces.  $f:X\rightarrow Y$ is uniformly continuous if $\forall \epsilon &gt;0,\forall x \in X, |x-y|&lt;\delta \implies |f(x)-f(y)|&lt;\epsilon$.  My question is what is the difference between uniform continuity and continuity of a map.
90509
I think I don't understand how it works.. I found some proofs.. okay, let's see:  Well I'd like to show that the function,  $$f(x) = \begin{cases} 0 &amp; x \not\in \mathbb{Q}\\ 1 &amp; x \in \mathbb{Q} \end{cases}$$  is discontinuous.   Now with epsilon-delta-definition:  Let's choose an $\varepsilon &lt; 1$, for example $\varepsilon := 1/2$. And: $\delta &gt; 0$ .  I have to show that $|f(x) - f(x_0)| &gt; \varepsilon$  So if $x_0 \not\in \mathbb{Q}$ let $x$ be rational in $(x_0- \delta, x_0+ \delta)$  if $x_0 \in \mathbb{Q}$ let $x$ be irrational in $(x_0- \delta, x_0 + \delta)$  In the end it is $|f(x) - f(x_0)| = 1 &gt; 1/2$.  Why is $|f(x) - f(x_0)| = 1 $ ?
84077
Please help me to prove the following;  Let $ \ f:[0,1]^2\longrightarrow\mathbb{R}$ be such that: (i) $\ f(x,\cdot)$ is measurable for each fixed $x\in[0,1]$; (ii) $\ f(\cdot,y)$ is continuous for each fixed $y\in[0,1]$. Show that $\ f$ is measurable.
114192
Let $f:\mathbb R \rightarrow \mathbb R$ be continuous and compactly supported. How can I prove that $f$ is uniformly continuous ? I was trying to prove it by contradiction but get stuck. My attempt was as follows:  Let $E$ be the compact support of $f$. On $E$ we know that $f$ is uniformly continuous. If we assume that $f$ is not uniformly continuous we know $$ \exists \epsilon &gt; 0 \forall \delta &gt; 0 \exists x,y \in \mathbb R: |x-y|&lt;\delta \wedge |f(x)-f(y)| \geq \epsilon $$ Let $\delta_n := \frac 1n$ and fix this $\epsilon &gt; 0$. Compute a $\delta &gt; 0$ s.t. $\forall x,y \in E:|x-y| &lt; \delta \rightarrow |f(x)-f(y)| &lt;\epsilon$. Let $n \geq N$ s.t. $\frac 1N &lt; \delta$. Then we may assume wlog that $x_n \in E$ and $y_n \in \mathbb R \setminus E$ wehere $x_n,y_n$ are the points corresponding with $\delta_n$. This gives a sequence of points where $|x_n-y_n| \rightarrow 0$ and $x_n \in E$ and $y_n \in \mathbb R \setminus E$.  I now want to use somehow the continuitiy of $f$. How can I do this ? Is this approach a good one ? Can it be more simple ?  New idea: I know that $E$ is compact so $(y_n)_{n=0}^\infty$ has a convergent subsequence $(y_{n_j})_{j=0}^\infty$ whit limit say $y \in E$. Now  $$  |x_{n_j}-y| \leq |x_{n_j}-y_{n_j}|+|y_{n_j}+y| \rightarrow 0 $$ So I can take $x_{n_j}$ close to $y$ to get $|f(x_{n_j})-f(y)| = |f(y)| &lt;\frac \epsilon 2$ by the continuity. I can also take $y_{n_j}$ close to $y$ to get $|f(y_{n_j})-f(y)| &lt;\frac \epsilon 2$. We further have $$  |f(y_{n_j})| \leq |f(y_{n_j})-f(y)| + |f(y)| $$ s.t. $$   |f(y)| \geq |f(y_{n_j})| - |f(y_{n_j})-f(y)| \geq \frac \epsilon 2 $$ because $|f(y_{n_j})| \geq \epsilon$ per construction. So we have $|f(y)| \leq \frac \epsilon 2$ and $|f(y)| &gt; \frac \epsilon 2$ which is a contradiction.    New solution: Let $\epsilon &gt; 0$. Let $\delta_1$ for the uniform continuity on $E$. Further $$  \forall x \in E,\ \exists \delta_x &gt; 0\ \forall y \in \mathbb{R}: |x-y|&lt; \delta_x \rightarrow |f(x)-f(y)| &lt; \epsilon $$ Compute an open finite over of $E$ $$  E \subseteq \bigcup_{i=1}^N B\left(x_i,\frac{\delta_{x_i}}2\right) $$ Write $\delta_i := \delta_{x_i}$. Let $\delta_2 := \min_{i=1,\cdots,N} \frac {\delta_i} 2$ and $\delta := \min(\delta_1,\delta_2)$.   Assume $|x-y|&lt; \delta$. If $x,y \in E$ or $x,y \notin E$ we are done. Otherwise assume $x \in E$ and $y \notin E$. Then $x \in B\left(x_i,\frac{\delta_i}{2}\right)$ for some $i$. Further $$  |y-x_i| \leq |y-x|+|x-x_i|\leq \delta_2 + \frac{\delta_i}2 \leq \delta_i $$ thus $y \in B(x_i,\delta_i)$ which proves the claim.
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 $\epsilon$.
7232
Prove that the function $\sqrt{x}$ is continuous on its domain $[0,\infty)$.  Proof.  Since $\sqrt{0} = 0, $ we consider the function $\sqrt{x}$ on $[a,\infty)$ where $a$ is real number and $s \neq 0.$ Let $\delta=2\sqrt{a}\epsilon.$ Then, $\forall x \in dom,$ and $\left | x-x_0\right | &lt; \delta \Rightarrow \left| \sqrt{x}-\sqrt{x_0}\right| = \left| \frac{x-x_0}{ \sqrt{x}+\sqrt{x_0}} \right| &lt; \left|\frac{\delta}{2\sqrt{a}}\right|=\epsilon.$  Can I do this?
92992
Let $S=\{f:\mathbb{R} \to  \mathbb{R}\}$ that satisfies: $\forall f\in S$, $f'$ exists and $f'$ is continuous and $f(0)=f(1)=0$.  Please prove that $\exists \delta :\forall f\in S$ s.t.  $\int_{0}^{1} (f(x))^2dx\leq \delta \int_{0}^{1} (f'(x))^2dx$. Thanks in advance.