I found an example of a function $f: \mathbb{R}\to\mathbb{R}$ that is continuous and bounded, but is not uniformly continuous. It is $\sin(x^2)$. I think it's not uniformly continuous because the derivative is bigger and bigger as $x$ increases. But I don't know how to prove this is uniformly continuous. Is $\sin(x^2)$ uniformly continuous then? if it isn't, can you guys think of any other examples? thanks
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If I want to prove that the function$\ f(x)=\sin(x^2)$ is not uniformly continuous on the domain $\mathbb{R}$, I need to show that: $\exists\varepsilon&gt;0$ $\forall\delta&gt;0$  $\exists{x,y}\in\mathbb{R}\ : |x-y|&lt;\delta$ and $|\sin(x^2) - \sin(y^2)|\geq\varepsilon$. So let's take $\varepsilon = 1$. Then I want $|\sin(x^2)-\sin(y^2)|\ge1$. That's the case if $\sin(x^2)=0$ and $\sin(y^2)=\pm1$. Thus $x^2=n\pi$ and $y^2=n\pi + \frac{1}{2}\pi$. Now I'm stuck on expressing x and y, which I want to express in $\delta$, to ensure that $|x-y|&lt;\delta$. Thanks in advance for any help.
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Let $f(x) = \frac{1}{x}$ for $x &gt; 0$ and take our set at which the function act on $(0,1]$. This function is continuous but not uniformly continuous on $A$.  To prove this consider $\epsilon = 10$,and suppose we have $\delta$, $0 &lt; \delta &lt; 1$. Taking $x = \delta$ and $p = \frac{\delta}{11}$, we get  $|x - p| &lt; \delta$ and $|f(x) - f(p)| = \frac{11}{\delta} - \frac{1}{\delta} &gt; 10$.   In the following example from my book shows this is continuous but not uniform continuous but the example is missing correct ? since we also need to find counter example for $\delta \geq 1$ right ? If someone could present that example too that would be nice.  Also my question is different since the possible duplicate above shows very different question to the question I am asking here since I am asking why we must consider $\delta \geq 1$ so please read carefully my question.
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I want to prove or disprove that $f(x)=\sin(x\sin x)$ for all $x\in (0,\infty)$ is uniformly continuous.  We know that $g(x)=x\sin x$ is not uniformly continuous on $(0,\infty)$. But what about $f$? I am not getting any hint. Any help is appreciated.
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Can we know from the graph of a function that whether the function is uniformly continuous or not? If the set on which the function is defined is not compact, so what one can say in that case?
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Give an example of a function $f:\mathbb{R}^2\to\mathbb{R}$ such that $f'_u(0,0)$ exists in all directions $\|u\| = 1$, but $f$ is not differentiable at $(0,0)$. You have to show that your example satisfy the above requirement.
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I need to prove that if $f: (0,1) \rightarrow \mathbb{R}$ is Uniformly continuous then it is  bounded.   Thank you.
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I must show that if $f$ is continuous on $(0,1)$ and $$\lim_{x \to 0}f(x) = + \infty$$ then $f$ is not uniformly continuous. I am not sure if we have to use epsilon- delta proofs or how to show this.
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How would I go about showing that if $f$ is uniformly differentiable then $f'$ is continuous.  my attempt: A differentiable function $f:[a,b]\to \Bbb R$ is said to be uniformly differentiable on $[a,b]$ if $\forall \epsilon&gt;0 \exists \delta&gt;0:\forall x,y\in[a,b]$ we have $$0&lt;|x-y|&lt;\delta\implies |\frac{f(x)-f(y)}{x-y}-f'(y)|&lt;\epsilon$$  I don't know where to go from here, any help would be highly appreciated.
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I'm trying to find an example of a continuous function $f:[1,\infty) \to \Bbb R $ such that  $$f(x) &gt;0 $$  $$\int_{1}^{\infty} f(x) \ dx \ \text{converges and } \int_{1} ^{\infty} f(x)^2 dx \ \text{diverges}.$$
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I have to prove that the function given below is continuous in $\mathbb R^2$:  $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$f(x,y)$= \begin{cases}  e^{-\text(\frac{1}{|x-y|})} &amp; \text{if $x\neq y$} \\ 0 &amp; \text{if $x=y$} \end{cases}   I have just started learning continuity in two/more dimensions ,so please help how to prove that the function is continuous on any path I take in 2 dimensions...
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Is $C^\infty(\mathbb{R})\subset C_b(\mathbb{R})$ dense? I.e. is any continuous bounded function $f:\mathbb{R}\to\mathbb{R}$ the uniform limit of smooth functions?   On any bounded interval this is true since by Stone-Weierstrass polynomials are smooth and dense in the continuous functions. I guess the result is still true for $\mathbb{R}$ but I don't know how to prove it. Since I don't have any integrability conditions on $f$ the technique of convoluting with mollifiers is probably not applicable.
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Determine whether the function$$f(x)={\sin x \over x}$$is uniformly continuous in $\mathbb{R}$.  I am using the definition that for $\epsilon&gt;0$ there exists $\delta&gt;0$ such that $|f(x)-f(y)|&lt;\epsilon$ whenever $|x-y|&lt;\delta$.
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I am thinking about this since couple hour. Is a continuous and bounded function $f:\mathbb R\to\mathbb R$ uniform continuous too? I didn't found a counter example and thus I tried to prove this like this: let $\epsilon&gt;0$ and $a\in\mathbb R$. By continuity, there is a $\delta_a&gt;0$ such that $$d(f(x),f(y))&lt;d(f(x),f(a))+d(f(y),f(a))&lt;\epsilon$$ if $d(x,y)\leq d(x,a)+d(y,a)&lt;\delta_a.$ Then, if $\inf_a\delta_a&gt;0$ I set $\delta=\inf_a\delta_a$ and we are done since $d(f(x),f(y))&lt;\epsilon$ if $d(x,y)&lt;\delta,$ but I didn't success to prove that $\delta\neq 0$.
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I was trying to find a series of differentiable functions that converge uniformly to a non differentiable function, to demonstrate the fact that uniform convergence doesn't imply anything about the differentiability of the function the sequence converges to.  I came up with this example -  $f_n(x)=\left\{  \begin{array}{c} |x| &amp;|x|\gt\frac 1 n \\ \frac x n &amp;|x|\leq\frac1 n \end{array} \right. $  The sequence of functions converges uniformly to $|x|$, and all $f_n$ are differentiable, but $|x|$ isn't differentiable at $x=0$.  Does this seem right?    Edit- people here helped me understand that my functions are not continuous at $x=\frac 1n$, which I forgot to check.  So a correct example will be:  $f_n(x)=\left\{  \begin{array}{c} |x| &amp;|x|\gt\frac 1 n \\ \frac {nx^2}2+ \frac1{2n} &amp;|x|\leq\frac1 n \end{array} \right. $  Which is continuous at $|x|=\frac1n$ and differentiable there.
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For , $(x,y)\in \mathbb R^2$ with $(x,y)\not =(0,0)$ , let $\theta=\theta(x,y)$ be the unique real number such that $-\pi&lt;\theta \le\pi$ and $(x,y)=(r\cos \theta , r\sin \theta)$ , where $r=\sqrt{x^2+y^2}$. Then the resulting function $\theta :\mathbb R^2 \setminus\{(0,0)\}\to \mathbb R$ is (A) differentiable. (B) continuous but not differentiable. (C) bounded but not continuous. (D) neither bounded nor continuous.  Here , $\theta=\arctan\left(\frac{y}{x}\right)$. So it is differentiable in $\mathbb R^2 \setminus \{(0,0)\}$. So option (A) is correct. Am I correct ?
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I understand the formal definition of uniform continuity of a function, and how it is different from standard continuity.  My question is: Is there an intuitive way to classify a function on $\mathbb{R}$ as being uniformly continuous, just like there is for regular continuity? I mean, that for a "nice" function $f:\mathbb{R} \to \mathbb{R}$, it is usually easy to tell if it is continuous on an interval by looking at or thinking of the graph of the function on the interval, or on all of $\mathbb{R}$. Can I do the same for uniform continuity? Can I tell that a function is uniformly continuous just by what it looks like? Ideally, this intuition would fit with the Heine-Cantor theorem for compact sets on $\mathbb{R}$.
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I am working on a homework problem from Avner Friedman's Advanced Calculus (#1 page 68) which asks     Suppose that $f(x)$ is a continuous function on the interval $[0,\infty)$. Prove that if $\lim_{x\to\infty} f(x)$ exists (as a real number), then $f(x)$ is uniformly continuous on this interval.   Intuitively, this argument makes sense to me. Since the limit of $f(x)$ exists on $[0,\infty)$, we will be able to find a $\delta &gt; |x_0 - x_1|$ and this implies that, for any $\epsilon&gt;0$, we have $\epsilon &gt; |f(x_0) - f(x_1)|$ (independent of the points chosen). I am aware that the condition of uniform continuity requires that $\delta$ can only be a function of $\epsilon$, not $x$.   What information does the existence of a real-valued limit provide that implies $f(x)$ is uniformly continuous on this interval?