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7

Unfortunately, no. However $f'$ is severely constrained in this instance. It can be shown that $f'$ actually satisfies the conclusion of the intermediate value theorem. To give an example of this phenomenon, take $f(x) = x^2 \sin(\frac{1}{x})$ when $x\ne 0$, and $f(0) = 0$. Then $f$ is differentiable everywhere, but its derivative is not continuous at the ...


6

For continuity, the only point of concern is $x=0$ and we need $\lim_{x \uparrow 0} f(x) = 1 = \lim_{x \downarrow 0} f(x) = b$. Hence $f$ is continuous everywhere iff $b=1$. Since $f_-(x)= 2x^2+x+1$ and $f_+(x)= ax+b$ are both smooth everywhere, the only way that $f$ can fail to be differentiable at $x=0$ is if $f_-'(0) \neq f_+'(0)$. Hence we need $f_-'(...


4

On $[0,1]$ we have that $\cos{x} \neq 0$ and also it is continuous. The numerator is continuous as a product of the continuous function $e^{x^2}$ and the composition of the continuous $\sqrt{x}$ with $\sin{x}$(which is continuous) Ratio of continuous functions is continuous.


4

$T= \{x \in \mathbb R \mid (f-g)(x)=0\} = (f-g)^{-1}(\{0\})$, i.e. $T$ is the inverse image under $f-g$ of the singleton $\{0\}$. As $\{0\}$ is closed (a singleton set is closed) and $f-g$ continuous, $T$ is closed.


4

Yes, of course, because the integral of a positive continuous function is positive: $$\int_0^1g(x)dx-\int_0^1f(x)dx=\int_0^1(g(x)-f(x))dx>0.$$


4

The set of points of discontinuity can be everywhere dense; in fact, it can be any countable subset of the plane. Let $D=\{(x_n,y_n):n\in\mathbb N\}$ be a countable subset of $\mathbb R^2$. You can easily contruct a function $h:\mathbb R^2\to[0,1]$ which is discontinuous at $(0,0)$ and continuous everywhere else, and has the property that its restriction ...


3

Recall: a function $f$ is continuous at $c$ if $$\lim_{x \to c^+} f(x) = \lim_{x \to c^-} f(x)$$ If one graphs your proposed $f$ (in red below, with $x=7$ in blue), we can see this is not the case: one limit is is $+\infty$ and the other is $-\infty$: The likely conclusion is there is a typo; I second Michael Hardy's suggestion from the comments in this ...


3

Let $(x_n)$ be a convergent sequence in $T$ with limit $x_0$. You have to show that $x_0 \in T$. To this end use $f(x_n) =g(x_n)$ for all $n$, $x_0 \in [a,b],$ $f(x_n) \to f(x_0)$ and $g(x_n) \to g(x_0).$ Can you proceed ?


3

No. Urysohn's lemma holds if and only if the space is normal. However not every locally compact Hausdorff space is normal. (Check here: https://mathoverflow.net/questions/53300/locally-compact-hausdorff-space-that-is-not-normal) Nevertheless, there is a version of Urysohn's lemma that holds in your case: if $\Omega$ is a locally compact Hausdorff space and ...


3

No, there exist such functions that don't have a minimal period. Set $p_0=1$ and choose $p_n \in \mathbb Q, n=1,2,\ldots$ with $$ p_1 > p_2 > \ldots > p_n > p_{n+1} > \ldots,\quad\lim_{n \to \infty}p_n=1=p_0.$$ Those $p_n, n=0,1,\ldots$ will be the periods of a function that is very close to the final function $f$ that will serve as example....


3

Following your idea, without loss of generality, and towards a contradiction, $f(b)>f(a)$. Define $g(x)=\frac{f(x)-f(a)}{x-a}$ if $x\neq a$ and $g(a)=0.$ Then $g$ is continuous on $[a,b]$. Now, set $g(b)=2\epsilon$ and $c=\inf\{x\in [a,b]:g(x)>\epsilon\}.$ Then, continuity of $g$ implies that $c < b$ and $g(c)=\epsilon.$ (Drawing a picture here ...


2

This is similar to the idea in Kavi Rama Murthy's answer. First, note that for $k = 0, 1, 2, \dots$, you can construct a function $h_k : [-1, 1] \to \mathbb{R}$ which is $C^k$ but is not $(k+1)$ times differentiable at any point: just take repeated antiderivatives (integrals) of a continuous nowhere differentiable function, e.g. the Weierstrass function. By ...


2

You can redefine if the discontinuity is removable i.e. if the limit exists at 7. But here the limit does not exist.


2

$h(x):=f(x)-g(x)$, $h$ is continuous. $T=${$x| h(x)=0$}. Then $T= h^{-1}${$0$} is closed being the inverse image of the closed set {$0$}.


2

Suppose $y\in[a,b]\setminus T$. Then $a:=|f(y)-g(y)|\gt 0$. As $f$ and $g$ are continuous, we can find a $\delta\gt0$ such that $$|f(x)-f(y)|\lt \frac a3,\:|g(x)-g(y)|\lt \frac a3$$whenever $|x-y|\lt\delta$. So if $|x-y|\lt\delta$, by triangle inequaltiy, $$\begin{align*}a=|f(y)-g(y)|&\leq |f(y)-f(x)|+|f(x)-g(x)|+|g(x)-g(y)|\\&\lt\frac a3+|f(x)-g(x)|...


2

Yes. If $A \subseteq \mathbb{R}$ is bounded, then $A$ is contained in some compact set $K \subseteq \mathbb{R}$. The continuous image of a compact set is compact, and thus bounded. So we have $f(A) \subseteq f(K)$ which is bounded, and thus $f(A)$ is as well.


2

So you're considering $X \times \{t\}$ as a subspace of $X \times I$ in the induced topology, and you want to show that $f: X \rightarrow X \times \{t\}, x \mapsto (x,t)$ is continuous. Not only is $f$ continuous; it's a homeomorphism. Since the open sets in $X \times I$ are unions of sets of the form $U \times V$ with $U$ open in $X$ and $V$ open in $I$, ...


2

Write your term as $$\frac{x^3}{y^2}\frac{x+y^2-x}{\sqrt{x}+\sqrt{y^2+x}}$$


2

$|\frac{x^3}{y^2} \left ( \sqrt{y^2 + x} - \sqrt{x} \right )|=|\frac{x^3}{\sqrt{x}+\sqrt{y^2+x}}| \leq |x|^{\frac{5}{2}}$ So the limit of the function at zero is zero.


2

As I said in the comments: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ so we can write this as \begin{align} &3|(x^2 + 1) - (y^2 + 1)| \cdot |(x^2 + 1)^2 + (x^2 + 1)(y^2 + 1) + (y^2 + 1)^2| \\ ={} & 3|x - y| \cdot |x + y| \cdot |(x^2 + 1)^2 + (x^2 + 1)(y^2 + 1) + (y^2 + 1)^2|. \end{align} Then you just need a crude bound on $3|x + y| \cdot |(x^2 + 1)^2 + ...


2

In answer to your first question, yes, this is because of the unboundedness assumption. If no such $x_n$ existed for some $n \in \Bbb{N}$, then this would imply that $f(x) < n$ for all $x \in [a, b]$, which is to say $f$ is bounded above by $n$. In answer to your second question, this follows from the first highlight. You know that $f(x_n) \ge n$ for all ...


2

You don't need to use the $\varepsilon$-$\delta$ definition. First you should know that that the metric as a map $d: M \times M \to \Bbb R$ is continuous, where $M \times M$ has the product topology (or product metric, if you prefer), and $\Bbb R$ the standard topology. This is a standard fact proven many times over on this site. Then $(1,f): M \to (x, f(...


1

For an explicit construction of the type of function suggested in the comment above take $n=1$ and $f(x)=\sum_k I_{(k-\frac 1 {k^{3}},k+\frac 1 {k^{3}}) }(x) (k^{4}|x-k|-k)$. This function is continuous and integrable but $|f(k)|=k \to \infty$.


1

Counterexample that's easy to write down: $$f(x) = x\left ( \dfrac{2+\sin x}{3} \right )^{x^5},$$ although it requires some work to verify.


1

This is clearly continuous for $ab\ne 0$ so the question is how to define it on (0, b, c) and (a, 0, c). I would rewrite it as $c\frac{ sin(ac)}{ac}\frac{1- cos(ab)}{(ab)^2}$.


1

No, you must show that there is an open set $O$ in the box topology so that $f^{-1}[O]$ is not open (not the same as closed!). Try $$O= \prod_{n =1}^\infty (-\frac1n,\frac1n)$$ and show that $f^{-1}[O]=\{0\}$, indeed not open.


1

A proof not using contradiction: Let $A_n := \{x \in X \mid f(x) < n\}.$ As $f$ is upper-semicontinuous, $A_n$ is an open set. We have: $X = \cup_{n \in \mathbb N} A_n.$ $X$ is compact, therefore it has a finite subcover $F \subset \mathbb N$, i.e., $X = \cup_{n \in F} A_n$, which shows that $f$ is bounded from above by $M = \max_{n \in F} n.$


1

Let $\{q_1,q_2,....\} $ an enumeration of the rationals.Take the function $$f(x)=\sum_{q_n <x}\frac{1}{n^2}$$ The functions is clearly bounded and increasing by density of rationals. Now let $q_N \in \Bbb{Q}$ Then, for all $x > q_N$, note that $$f(x) = \sum_{ x > q_n} \frac 1{n^2} = \sum_{ q_N > q_n} \frac 1{n^2} + \sum_{ x > q_n \geq ...


1

We have $f(x)$ continuous if for all $a\in [0,1]$ hold $\lim_{x\to a}f(x)=f(a)$. By limit proprerties we have \begin{align} \lim_{x\to a}f(x) =& \lim_{x\to a} \frac{e^{x^2}\cdot\sqrt{\sin(x)}}{\cos(x)} \\ =& \frac{\lim_{x\to a} e^{x^2}\cdot \lim_{x\to a} \sqrt{\sin(x)}}{\lim_{x\to a} \cos(x)} \\ =& \frac{ e^{\lim_{x\to a} x^2}\cdot \sqrt{\...


1

For any given $\alpha>0$, choose an integer $n$ so large that $$n>\frac{1}{2\pi\alpha}.$$ Define \begin{align*} x_1&\equiv\frac{1}{2n\pi},\\x_2&\equiv\frac{1}{(2n+1)\pi}. \end{align*} Clearly, $x_1$ and $x_2$ are both in $(0,\alpha)$. Moreover, \begin{align*} f'(x_1)&=2x_1\sin\left(\frac{1}{x_1}\right)-\cos\left(\frac{1}{x_1}\right)+c\\ &...


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