To show that  $\frac{d}{dx}:A \to L^2 (a,b)  $ is not continuous, where $$A=\left\{u \in L^2 (a,b): \frac{du}{dx} \in  L^2 (a,b) \right\} $$  I'm trying take a map's sussection that $|du_n|&gt;c_n|u_n|$ but I cant find.
23320
If we define $D$ as the set of functions $f \in L^2 (\mathbb{R})$ such that $f' \in L^2 (\mathbb{R})$ too, then it can be shown (I think) that the linear operator $\frac{d}{dt}: D \rightarrow L^2 (\mathbb{R})$ is not bounded with respect to the norm on $L^2 (\mathbb{R})$, which is $||f|| = \sqrt{\int_{-\infty}^{\infty}|f(t)|^2 dt}$.   But I'm having trouble proving this. What I'd like to do is find a sequence $(f_n)$ in $D$ with $||f_n||=1$ $\forall n \in \mathbb{N}$ but with $||f_n '||$ increasing and becoming arbitrarily large with $n$.   I understand that if we were working with the space of continuous functions on $\mathbb{R}$ instead of $L^2 (\mathbb{R})$, with the nice supremum norm, we could just use $f_n (t) = \sin(nt)$, but that won't work here. I tried the sequence $f_n(t) = \sqrt{\frac{\sin(nt)}{t}}$ because that's got a handy $L^2$-norm of $\sqrt{\pi}$ for all $n\in\mathbb{N}$, but its derivative isn't in $L^2 (\mathbb{R})$ so I feel stuck.   I think I'm approaching the problem correctly, but the norm in question here isn't easy to work with -- is there a simple sequence I could use?
87532
Let $f: X \times Y \rightarrow \mathbb{R}$ be a continuous map. Show that if $Y$ is compact then the function $g: X \rightarrow \mathbb{R}$ defined by $g(x) = \inf \{f(x,y): y \in Y\}$ is also continuous.  No clue here. Can you please help?
11185
Suppose that $f : [a, b] \times [c, d] \to\mathbb R$ is a continuous function. Let $$G(y)= \int_a^b f(x, y) \, dx$$ $$H(x)= \int_a^b f(x, y) \, dy$$  Prove that $G$  is continuous on $[c, d]$ and $H$ is continuous on $[a, b]$.  So I know that the function will be uniformly continuous on the interval but I don't know where to go from there
74232
Suppose $f(x)$ is continuous on $[a, b]$ except at a point $c$ in $(a, b)$ at which $f(x)$ has a jump discontinuity. For $x$ in $(a,b)$, set $F(x)=\int_a^xf(t) \, dt$. Show that $F(x)$ is continuous but not differentiable at $x =c$.  Hint: first explain why $f(x)$ is integrable on $[a,b]$, then consider one-sided derivatives of $F(x)$ at $x =c$.  I don't know how to step up because it seems conflict to me. $f(x)$ is not continuous because it has a jump but I have to prove it continuous, how? Can someone guide me, please.
44837
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function.  Show that the points at which $f$ is continuous is a $G_{\delta}$ set.  $$A_n = \{ x \in \mathbb{R} | x \in B(x,r) \text{ open }, f(x'')-f(x')&lt;\frac{1}{n}, \forall x',x'' \in B(x)\}$$  I saw that this proof was already on here, but I wanted to confirm and flesh out more details.  "$\Rightarrow$" If f is continuous at $x$, then $f(x'')-f(x')&lt;\frac{1}{n}$ for $x'',x' \in B(x, r_{n})$.  That is, there is a ball of radius $r$ where $r$ depends on $n$.  Then $x \in A_n$ and thus $x \in \cap A_n$.  "$\Leftarrow$" If $x \in \cap A_n$, then there is an $\epsilon &gt; 0$ and a $\delta &gt; 0$ such that $x' , x'' \in B(x, \delta_n)$ for all $n$ and $$|f(x'')-f(x')|&lt;\epsilon.$$ Take $\epsilon = \frac{1}{n}$.
18829
I have a question:     Suppose $f$ is continuous and even on $[-a,a]$, $a&gt;0$ then prove that   $$\int\limits_{-a}^a \frac{f(x)}{1+e^{x}} \mathrm dx = \int\limits_0^a f(x) \mathrm dx$$   How can I do this? Don't know how to start.
4998
I'm asked to show that  $$\frac{d(\hat{A}\hat{B})}{d\lambda} ~=~ \frac{d\hat{A}}{d\lambda}\hat{B} + \hat{A}\frac{d\hat{b}}{d\lambda}$$ With $\lambda$ a continuous parameter. Should I use the definition $$\frac{d\hat{A}}{d\lambda} ~=~ \lim_{\epsilon \to 0} \frac{\hat{A}(\lambda + \epsilon) - \hat{A}(\lambda)}{\epsilon}$$ applied to $\hat{A}\hat{B}$ like $$\frac{d(\hat{A}\hat{B})}{d\lambda} ~=~ \lim_{\epsilon \to 0} \frac{\hat{A}(\lambda + \epsilon)\hat{B}(\lambda + \epsilon) - \hat{A}(\lambda)\hat{B}(\lambda)}{\epsilon}$$ and do some algebra to get the RHS of the first equation, or I'm missing something? Another interesting derivative to pay attention to is: $$\frac{d}{d\lambda}\exp(\hat{A}(\lambda) )~?$$
56323
Let $f,g\in PC(2\pi)$ continuous. Show that $$\lim_{n\to \infty}\int_{-\pi}^{\pi}f(x)g(nx)\,dx=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)\,dx\int_{-\pi}^{\pi}g(x)\,dx$$ Where if $f\in PC(2\pi)$ then $\int_{-\pi}^{\pi}f(x)\,dx=\int_c^{c+2\pi}f(x)\,dx \quad\forall c\in\mathbb{R}.$ I've been struggling with this proof. Any suggestions would be great!
11555
Let $f(x,y)=     \begin{cases}       \frac{2x^2y}{x^2+y^2} &amp; if &amp; (x,y) \neq (0,0) \\       0       &amp; if &amp; (x,y)=(0,0) .  \end{cases}$  (i) Prove the directional derivatives of $f$ exist in any direction at the point $(0,0)$.  (ii)¿Is $f$ continuous on (0,0)?  For (i) I took $u=(u_{1},u_{2}) \in \mathbb{R^{2}}$ such $\|u\|=1$ and $0=(0,0)$. So $$\lim_{t \to 0} \frac{f(0+ t u)-f(0)}{t}=\lim_{t \to 0}\frac{2(tu_{1})^2 (tu_{2})}{(tu_{1})^4+(tu_{2})^2}=\lim_{t \to 0}\frac{(t^2 u_{1}^2)(t^2 u_{1}^{2}) tu_{2}}{t^{4}u_{1}^{4}+t^{2} u_{2}^{2}}.$$ But I cannot find the limit when $t$ aproaches to $0$. Which by finding them I would prove the directional derivative exist at every direction right?  For (ii) I got the intuition $f$ is not continuous at $(0,0)$ so I took $\lbrace (\frac{1}{k},\frac{1}{k}) \rbrace_{k \in \mathbb{N}}$ and $\lbrace (\frac{1}{k},0) \rbrace_{k \in \mathbb{N}}$ which are two different sequences in $\mathbb{R}^{2}$ converging to $(0,0)$. However, $\lbrace f(\frac{1}{k},\frac{1}{k}) \rbrace_{k \in \mathbb{N}}=\lbrace \frac{1}{k^{2}} \rbrace_{n \in \mathbb{N}} \to 0$ and $\lbrace f(\frac{1}{k},0) \rbrace_{k \in \mathbb{N}} \to 0$. So maybe my intuition was not right? Can anyone help me end the proof of continuity or not continuity , please?
2881
Let $S = [0,1) \cup [2,3]$ and $f\colon S \rightarrow \mathbb R$ be such that $f(S)$ is connected . Which of the following are true:  a) $f$ is discontinuous exactly at one point.  b) $f$ is discontinuous exactly at two points.  c) $f$ is discontinuous at infinite many points.  d) $f$ is continous.  I think $f$ is continuous . We need to show that $\lim_{x \to a^+}f(x) = f(a)$ for any $a \in S$. Since $f(x)$ is increasing, $\lim_{x \to a^+}f(x) \leq f(a)$. Suppose $\lim_{x \to a^+}f(x) &lt; f(a)$. Then there is $l$ such that $\lim_{x \to a^+}f(x) &lt; l &lt;f(a)$, which contradicts connectedness of $f(S)$. Similarly we can show that $\lim_{x \to a^-}f(x) = f(a)$.  I would be thankful if anyone would check my solution and correct me if there are any errors.
64888
If $f,g$ are continuous at $a$, show that $h(x)=\max\{f(x),g(x)\}$ and $k(x)=\min\{f(x),g(x)\}$ are also continuous at $a$. Here is my attempt at a proof. It feels very elaborate and I am not sure if it is correct. Can someone please point out any mistakes or places where I may improve. Thanks!  By the definition of continuity at $a$ of $f,g$ we have that $\lim\limits_{x\to a}f(x)=f(a)$ and $\lim\limits_{x\to a}g(x)=g(a)$. Suppose that $f(a)&gt;g(a)$. Then there exists a $\delta&gt;0$ such that $f&gt;g$ for all $x$ satisfying $|x-a|&lt;\delta$. Then for $x$ satisfying $|x-a|&lt;\delta$ we have $h(x)=f(x)$ and $k(x)=g(x)$ and thus $h$ and $k$ are continuous at $a$ because $f$ and $g$ are continuous at $a$. Now if $f(a)&lt;g(a)$ we simply relabel $f=\tilde{g}$ and $g=\tilde{f}$, so that $\tilde{f}(a)&gt;\tilde{g}(a)$ and we apply the previous result to show that again $h$ and $k$ are continuous at $a$.\  Now if $f(a)=g(a)$ then we can distinguish three cases. $f\geq g$ in a small neighborhood about $a$, $f\leq g$  in a small neighborhood about $a$ or $f\geq g$ on one side of $a$ and $f\leq g$ on the other side. In the first case we assume that $f\geq g$ in some small neighborhood about $a$. Then, since $f(a)=g(a)$ we have $h(x) = f(x)$ in this neighborhood and $k(x) = g(x)$ and again we see that $h$ and $k$ are continuous at $a$ due to the continuity of $f$ and $g$ at $a$. Similarly if $f\leq g$ in a neighborhood around $a$ then $h(x)=g(x)$ and $k(x)=f(x)$ in this neighborhood and $h$ and $k$ are continuous at $a$.   Suppose that just to the left of $a$ we have $f\geq g$ and just to the right of $a$ we have $f\leq g$. Then $\lim\limits_{x\to a^-}h(x)=\lim\limits_{x\to a^-}f(x)=f(a)=h(a)=g(a)=\lim\limits_{x\to a^+}g(x)=\lim\limits_{x\to a^+}h(x)$. Similarly $\lim\limits_{x\to a^-}k(x)=\lim\limits_{x\to a^-}g(x)=g(a)=k(a)=f(a)=\lim\limits_{x\to a^+}f(x)=\lim\limits_{x\to a^+}k(x)$. So $\lim\limits_{x\to a}h(x)=h(a)$ and $\lim\limits_{x\to a}k(x)=k(a)$ which means that $h$ and $k$ are continuous at $a$. Lastly if $f\leq g$ just to the left of $a$ and $f\geq g$ just to the right of $a$ we can relabel $f=\tilde{g}$ and $g=\tilde{f}$ and apply the last result to show $h$ and $k$ are continuous at $a$.   Again, please point out any mistakes I may have made. Thanks!!
108519
How does one go about proving that the function $x \sin x$ is not uniformly continuous on the set of real numbers ?  Any method that uses the sequential criterion for discontinuity would be preferred.  Thanks in advance !
74398
let $f(x,y)$ continuous in $[a,b] \times [c,d]$ and $\frac{\partial f}{\partial y} $ continuous in $[a,b] \times [c,d]$, define $$F(y) = \int_a^b f(x,y)dx$$   then $$F'(y) = \int_a^b \frac{\partial f}{\partial y}(x,y) dx$$   Can you explain the intuition behind this rule, in terms of geometry for example, or how did Leibniz  came up with this rule?
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.
39699
Greets   I came up the other day with the following question: Is it true that $f:\mathbb{R}^n\longrightarrow{\mathbb{R}^m}$ is continuous if and only if $f$ maps compact sets onto compact sets and maps connected sets onto connected sets?  I'm having trouble showing the "if", and I haven't found a counterexample to this part, so I would appreciate ideas to prove or disprove this.  Thanks
80328
Is function $f:\mathbb C-\{0\}\rightarrow\mathbb C$ prescribed by  $z\rightarrow \large{\frac{1}{z}}$ by definition discontinuous at $0$?  Personally I would say: "no". In my view a function can only be (dis)continuous at $z$ if $z$ belongs to its domain.  But I have heard other sounds, that made me curious.   This question was inspired by comments/answers on  question.
16143
I have to prove that $\frac1x $ with $x \in (0,\infty)$ is not uniformly continuous. I understand that the further you approach the $0$ the greater your $e$ and the smaller your $\delta$ becomes (in the limit even $\delta = 0 $ ?).  But I need help with the "technical" proof, can someone give me an instruction to do this?