First I constructed the negation to Lipschitz continuity:   $ \forall L &gt; 0, \exists x, y \in [a,b]\ |f(x) - f(y)| &gt; L|x - y| $  For $f(x) = \sqrt x$, notice $$|f(x) - f(y)| = |\sqrt x -\sqrt y| \cdot\frac{|\sqrt x + \sqrt y |}{|\sqrt x + \sqrt y |} = \frac{| x - y|}{|\sqrt x + \sqrt y|} \ge L |x - y| .$$  The problem is $I = [0,1]$ with $ \frac{1}{|\sqrt x + \sqrt y|}$ assuming values between $ \left(\frac{1}{2},\infty\right)$ and  $ \frac{1}{|\sqrt x + \sqrt y|} \ge L$. So for sufficiently large $L$, the desired inequality for a function not being Lipschitz continuous cannot hold. Can someone explain the issue?
121377
Show that f(x)=$\sqrt{x}$ is uniformly continuous, but not Lipschitz continuous.  I can prove that it's uniformly continuous. But why is it not Lipschits? How do I check the definition?
92704
This came up as a question in my last  real analysis test. Speaking to my teacher after the test, he said he'd forget to mention that the function should be continuous, and in fact there is an easy couterexample for his argument when the function is not continuous. Still after thinking for a long time I was unable to come up with a function with uncoutable strict local minima or with a proof that these don't exist.
106543
I don't know how to prove this example. Let $y(t):\mathbb{R}^+\to \mathbb{R}$ be a nonincreasing function. Assume that there is a constant $A&gt;0$ such that \begin{equation} \int_{s}^{+\infty}y(t)dt \leq Ay(s), \quad 0\leq s&lt;+\infty \end{equation} Then $y(t)\leq y(0)e^{1-\frac{t}{A}}$, for all $t&gt;0$.
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 $\lambda \geq 1$.
98812
I tried to use the definition and arrived this far:  $|f(x)-f(y)|=\left|\frac{x}{1+|x|}-\frac{y}{1+|y|}\right|=\frac{|x-y+x|y|-y|x||}{(1+|x|)(1+|y|)}\leq|x-y+x|y|-y|x||$.  Any suggestion for ending the proof?  I also tried to prove that $\frac{x}{1+x}$ is uniformly continuous on $[0,\infty[$ and that $\frac{x}{1-x}$ is uniformly continuous on $[-\infty,0[$, but I wonder if we can use the definition with the function $f(x)=\frac{x}{1+|x|}$ itself.
22291
Is a function defined at a single point continuous?  For example $f:\{0\}\to\{0\}$ defined by $f(x)=\sqrt{x}+\sqrt{-x}$ is a sum of two continuous functions and is therefore continuous, however for $f$ to be continuous, it has to have the limit $\lim\limits_{x_0\to x}f(x_0)=f(x)$ but it dosen't seem to have that limit.
78312
Is there a counterexample?  Proposition: Let $f:\mathbb{R} \longrightarrow \mathbb{R}$ be a differentiable function  such that $f':\mathbb{R} \longrightarrow \mathbb{R}$ is differentiable at $a\in\mathbb{R}$ (possibly differentiable at a single point). Then $f':\mathbb{R} \longrightarrow \mathbb{R}$ is continuous at an interval $(a-\delta,a+\delta)$ for some $\delta&gt;0$.  Any hints would be appreciated.
139
How do I prove $\sqrt{x + y} \le \sqrt{x} + \sqrt{y}$? for $x, y$ positive?  This should be easy, but I'm not seeing how.  A hint would be appreciated.
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.
87890
Is the function $f(x)=|x|^{1/2}$ Lipschitz continuous near $0$? If yes, find a constant for some interval containing $0$.   I think the answer is yes since I can find $L=1$ that satisfies Lipschitz continuity criteria in a interval close to $0$, am I right?
635
My questions are about the reasoning made in the note  (which is in Norwegian).   To prove that $f(x)=\frac{1}{x}$ is not uniformly continuous, the authors use the following "result" (Sats 2.12 in the note, which I translate below):  Result 2.12: If for every $h&gt;0$ we have that $|f(x+h)-f(x)|$ is unbounded on $I$, then $f$ is not uniformly continuous on $I$.  Proof: The result follows directly from the definition of uniform continuity.   In Example 2.14 (Eksempel 2.14), the authors look at  $$|f(x+h)-f(x)|=\left|\frac{1}{x+h}-\frac{1}{x}\right|=\left|\frac{h}{x(x+h)}\right|.$$ They then claim that the above quantity is not bounded for any $h&gt;0$, since   $$\lim_{x\rightarrow 0}\left|\frac{h}{x(x+h)}\right|=\infty.$$  Question 1: Is it not possible to choose $h=x^2$, thereby obtaining   $$\frac{h}{x(x+h)}=\frac{1}{\frac{x}{h}(x+h)}=\frac{1}{x(1/x+1)}=\frac{1}{x(1/x+1)}=\frac{1}{1+x}\rightarrow1 \text{ as }x\rightarrow 0.$$  Therefore, |f(x+h)-f(x)| is not unbounded, so we cannot use the result "Result 2.12". Is my argumentation correct?? Is it OK to choose $h$ like I have done?   Question 2: However, it seems correct to me that my argumentation is all you need to prove that  $f(x)=\frac{1}{x}$ is not uniformly continuous on (0,1). Here $x_1=x,x_2=x+x^2$, so that $|x_1-x_2|$ can be made arbitrarily small. However $|f(x_1)-f(x_2)|=1$, which stays the same regardless of how small we make $|x_1-x_2|$. Is this correct?
88696
Suppose the function $f:[0,\delta) \to \mathbb{R}$ is continuous, differentiable in $(0,\delta)$ and $f(0)=0$.  If the limit $\displaystyle \lim_{x \to 0+}\frac{f(x)}{f'(x)}= L$ exists, then is it always the case that $L = 0$.  This seems to be true both for functions with well-behaved derivatives such as $f(x) = x,$  $$\lim_{x \to 0+} \frac{f(x)}{f'(x)}=\lim_{x \to 0+} \frac{x}{1}=0,$$  as well as functions with "bad" derivatives such as   $$f(x) = \begin{cases}x \ln x &amp;\mbox{if }x&gt;0 ,\\0  &amp;\mbox{if } x=0,  \end{cases},$$  where  $$\lim_{x \to 0+} \frac{f(x)}{f'(x)}=\lim_{x \to 0+} \frac{x \ln x }{1+ \ln x}=0.$$  Is there a simple proof or counterexample?
65761
Let $a&lt;b$ be real numbers. Let $f:[a,b]\to\mathbb{R}$ be a continuous non-negative function. Suppose $\int_{[a,b]}f=0,\text{ then }f(x)=0 \forall x\in[a,b]$    Proof: Suppose for the sake of contradiction $\exists x_0\in[a,b] \text{ such that } f(x_0)= D&gt; 0$  Since the function is continuous at $x_0$, we have $$\forall \epsilon&gt;0, \exists \delta&gt;0 \text{ such that } |f(x)-f(x_0)|\leq\epsilon\text{ whenever } x\in[a,b]\cap(x_0-\delta,x_0+\delta)$$ choose $\delta=\delta_1$ such that $\epsilon=D/2$, then we can say: $$-D/2 \leq f(x)-f(x_0)\leq D/2$$ $$-D/2 \leq f(x) - D$$ $$f(x)\geq D/2&gt;0$$ This implies that $\int_{[a,b]\cap(x_0-\delta_1,x_0+\delta_1)}f &gt;0$  Now since $\int_{[a,b]} f = 0$. we have  $$\int_{[a,b]}f=\int_{[a,x_0-\delta_1]}f+ \int_{(x_0-\delta_1, x_0+\delta_1)} f + \int_{[x_0+\delta_1,b]}f=0$$ $$\int_{[a,x_0-\delta_1]}f+\int_{[x_0+\delta_1,b]}f&lt;-D/2,$$ but this is a contradiction since $f(x)\geq 0$ and hence $\int_{[a,x_0-\delta_1]}f+\int_{[x_0+\delta_1,b]}f\geq 0$  Is my proof correct?
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:
17745
I found this problem on an old FB post and have just been humbled by it.  Let $f(x)$ be continuous on $[0,n]$ such that $n &gt; 0$ and $f(x) + f(n - x)$ does not vanish on $[0,n]$ then evaluate  $$ \int_0^n \frac{f(x)}{f(x) + f(n-x)}dx $$  Any clues/hints would be appreciated.
75860
I got stuck on this bit from Spivak's Calculus:  When proving that $f$ does not approach $l$ at $a$, be sure to negate the definition correctly: If it is not true that   for every $\varepsilon&gt;0$ there is some $\delta&gt;0$ such that, for all $x$, if $0&lt;|x-a|&lt;\delta$, then $|f(x)-l|&lt;\varepsilon,$   then   there is some $\varepsilon&gt;0$ such that for every $\delta&gt;0$ there is some $x$ which satisfies $0&lt;|x-a|&lt;\delta$ but not $|f(x)-l|&lt;\varepsilon.$    I don't understand why the negation of the top sentence is given by the bottom sentence. Thank you so much! EDIT My problem is not why &quot;continuous functions are those that let you interchange limits&quot;. It's more just a question about how you can negate sentences in general.
101691
Let $A\subset \mathbb{R}^n$ be a nonempty convex and closed set, and let $f:A\to\mathbb{R}$ be a lower semicontinuous and convex function.   Does this imply that $f$ is continuous on $A$?  I know that in the (relative) interior of $A$, $f$ is locally lipchitz so my question is about the points on the boundary.  Of course, I consider the continuity of $f$ with respect to $A$ and not the continuity of the extended function $\tilde{f}:\mathbb{R}^{n}\to\mathbb{R}\cup\left\{+\infty\right\}$ defined by $\tilde{f}(x):=f(x)$ for $x\in A$ and $\tilde f(x):=+\infty$ for $x\notin A$, which is obviously not upper semicontinuous on the boundary of $A$.  If the answer to my question is negative a counterexample would be welcome. Thanks!