Show that if $f:\omega_1 \to \mathbb{R}$ is continuous, then there is $a \in \omega_1$ such that if $x\geq a$, then $f(x) = f(a)$.  I'm stuck on this question for a while, although it seems to me that the proof will be related to the fact that $f$ is bounded (I proved this as folows: $f$ is continuous and $w_1$ has the least element of the set of its limit point, say $a$, then $f(a)$ should be the upper bound of $f:\omega_1 \rightarrow \mathbb{R}$. Is this correct?). Can anyone please help on this problem?
26050
show that every continuous real-valued function defined on $S_{\mathbb{\Omega}}$ is eventually constant.Where $S_{\mathbb{\Omega}}$ denote the first uncountable ordinal.  There is a hint that for each $\epsilon$,there is an element $\alpha$ of $S_{\mathbb{\Omega}}$  such that $|f(\alpha)-f(\beta)|&lt;\epsilon$ for all $\beta&gt;\alpha$. However, I couldn't figure out how to prove this statement.
33427
Using the definition of continuity and the $\epsilon$-$\delta$ definition of a limit, how would you proceed to show that the following statement holds true?  Let $f : \Bbb{R} \rightarrow \Bbb{R}$ be a continuos function at $x=0$, with the property that if $x\neq 0$ then $f(x)\geq 0$. Then $f(0) \geq 0$.  My attempt so far has been to assume that $f(0) \lt 0$, which means that $$f(0)=\lim_{x\to 0} f(x)\lt 0,$$ because of the continuity at the given point. Furthermore, I think this means that when $x$ approaches $0$ from both sides, at some point the value $f(x)\lt 0$, thus forming a contradiction.   I can't seem to see what to do afterwards. Is this a way of approaching the problem? And how will the $\epsilon-\delta$ definition come into play if my approach is usable?
27268
The following is the well known  for $H_0^1(\Omega)$:       Suppose that $\Omega$ is an open set in $\mathbb{R}^n$ that is bounded. Then there is a constant $C$ such that   $$ \int_\Omega u^2\ dx\leq C\int_\Omega|Du|^2\ dx\quad \textrm{ for all   }\ \ \color{red}{u\in H_0^1(\Omega)}. $$   Here is my questions:    Can the condition $u\in H_0^1(\Omega)$ be relaxed so that the inequality is still true? More precisely, suppose $H$ is a subspace of $H^1(\Omega)$ such that for any $u\in H$, $u|_\Gamma=0$, where $\Gamma$ is a ([Added:]nonempty) closed proper subset of the boundary $\partial\Omega$ and the idenity is in the sense of trace. In other words, $u$ is only zero on a part of the boundary. Do we still have the Poincaré ineqaulity in $H$?    [Added:] This question is motivated by the following exercise
103954
If $f$ Continuous bounded on $\mathbb{R}$，and $$\lim_{h\to0}\sup_{x\in\mathbb{R}}\left|f(x+h)-2f(x)+f(x-h)\right|=0.$$ prove that $f$ uniformly continuous on $\mathbb{R}$. I try to find a $\delta&gt;0$ to make it arbitrary $x_1,x_2\in \mathbb{R}$and$\left|x_1-x_2\right|&lt;\delta$have$$\left|f(x_1)-f(x_2)\right|&lt;\varepsilon.$$but I don't know how to use this complex limit expression.I hope you can give me some tips, thank you!
49493
Suppose $f\colon [a, b] \to \Bbb R$ is continuous. Define $F(x):= \sup \{f(y)\mid y \in [a, x]\}$. Prove $F(x)$ is continuous.   What I have so far:  For $x \le b$, the set $\sup\{f(x)\mid y \in [a, x]\} \not= \emptyset$. By Completeness then, $\sup\{f(y)\mid y \in [a, x]\}$ exists and is finite. So by the Intermediate Value Theorem, there exists a $x_0\in [a, x]$ such that $f(x_0) = s_{0_k}$, where $s_{0_k} = \sup\{f(x)\mid y \in [a, x]\}$.  Then choose a sequence $\{s_n\}$ in $[a, x]$ such that $s_n\to s_{0_k}$. Since there exists a sequence in a closed, bounded set that converges to the supremum of that set.  Also then $s_{0_k}\in [a, x]$ by the Comparison Theorem. And since $f$ is continuous, $\lim f(s_n) = s_{0_k}$.   Observe that as $x_n\to b$ as $n \to \infty$ and that $s_{0_k}$ is monotone increasing. Thus each $s_{0_k}\to s_0$ where $s_0=\sup\{f(y)\mid y \in [a, x]\}$. Thus each $s_{0_k}\to s_0$, hence $F(x)$ is continuous at each $s_{0_k} \in [a, b]$.  First off, is this correct? Secondly, It seems to me there is a simple answer, but I can't see it any ideas?
32367
Show that für $f\colon (\Omega,\mathcal{A})\to\mathbb{R}_{\geq 0}$ measurable, it is     $$ \sum_{k=1}^{\infty}\mu(\left\{f\geq k\right\})\leq\int f\, d\mu\leq\sum_{k=0}^{\infty}\mu(\left\{f&gt;k\right\}). $$      I do not know if one needs it here, but we defined the integral of a non-negative measurable function $f\colon (\Omega,\mathcal{A})\to (\overline{\mathbb{R}},\overline{\mathcal{B}})$ by $$ \int_{\Omega}f\, d\mu=\sup\left\{\sum_{k=1}^{n}(\inf\left\{f(x): x\in A_k\right\})\mu(A_k): n\in\mathbb{N}, A_1,\ldots,A_n\in\mathcal{A}\mbox{ disjoint },\Omega=\bigcup_{k=1}^{n}A_k\right\}. $$    My first idea was to use this definition, but I did not come along with it...  Maybe you can help me to show that?  (I think, maybe it has something to do with monotone convergence theorem or majorated convergence?)
35614
I want to prove the following fact :  Consider $\Omega \subset R^n$ a bounded and open set. Let $v \in H^{1}_{0}(\Omega)$ a nonnegative function. Then exists a sequence $v_m$ in $C^{\infty}_{0}(\Omega)$ of nongative functions converging to $v$ in $H^{1}( \Omega)$.  I know how to prove this fact (I think this fact can help): If $u_m$ is a sequence converging to $u$ in $H^1(\Omega)$ , then ${u^{+}_m} \rightarrow  u^+$.  Someone can give me a hint ?  Thanks in advance
79979
A problem that I have been thinking: For what kind of functions $f:[a,b]\to \mathbb R$ is  $x\mapsto\sup_{t\in [a,x]}f(t)$ continuous? This is not true for bounded functions, take for example $f(x)=0$ in $[0,1]$ and $f(x)=1$ in $(1,2]$. It is however trivially true for continuous monotone functions. Assuming mere continuity for $f$ I can show right continuity of $\max_{t\in[a,x]}f(t)$:   For $\epsilon$ and some $\delta&gt;0$,  $$x\in [x_0-\delta,x_0+\delta]\implies f(x_0)-\epsilon&lt;f(x)&lt;f(x_0)+\epsilon$$ If $x\in [x_0,x_0+\delta]$ then $$\max_{t\in [a,x]}f(t)=\max\left\{\max_{t\in [a,x_0]}f(t),\max_{t\in [x_0,x]}f(t)\right\}$$ If $max_{t\in [a,x_0]}f(t)\ge \max_{t\in [x_0,x]}f(t)$ then $$\left|\max_{t\in [a,x]}f(t)-\max_{t\in [a,x_0]}f(t)\right|=0&lt; \epsilon$$ Otherwise, $$\left|\max_{t\in [a,x]}f(t)-\max_{t\in [a,x_0]}f(t)\right|\le \max_{t\in [x_0,x]}f(t)-\max_{t\in [a,x_0]}f(t)&lt;f(x_0)+\epsilon-\max_{t\in [a,x_0]}f(t)\le \epsilon$$ which proves right continuity. When I tried to prove left continuity I was stuck on showing $$\max_{t\in [x,x_0]}f(t)-\max_{t\in [a,x]}f(t)&lt;\epsilon$$ Can this be circumvented?
89560
I am having difficulties showing that this is true. How would you prove this statement?  Given that the function $h: \mathbb{R}\rightarrow\mathbb{R}$ is increasing. Show that $h$ is continuous if and only if  $h(\mathbb{R})$ is connected.
14432
Suppose that $f: \mathbb R \to\mathbb  R$ satisfies $f(x + y) = f(x) + f(y)$ for each real $x,y$. Prove $f$ is continuous at $0$ if and only if $f$ is continuous on $\mathbb R$.  Proof: suppose $f$ is continuous at zero. Then let $R$ be an open interval containing zero. Then $f$ is continuous at zero if and only if $f(x) \to f(0)$ as $x \to 0$. Then $|f(x) - f(0)| &lt; \epsilon$. Can anyone please help me? I don't really know how to continue. Thank you.
92142
We have the function $f: \mathbb{R} \to \mathbb{R}$.     Question: Show that if the sets $\{x:f(x) &gt; a\}$ and $\{x:f(x) &lt; a\}$ are open for every $a\in\mathbb{R}$, then $f$ is continuous.   My attempt:$f$ is continuous if $A \subset \mathbb{R}$ is open, then $f^{-1}(A) \subset \mathbb{R}$ is open. I can show that the image of $\{x:f(x) &gt;a\}$ is open. Namely $\{f(x) &gt; a\} = (a,\infty)$ is open in $\mathbb{R}$. But this doesn't prove that for any open set $A \subset \mathbb{R}$, we have that $f^{-1}(A)\subset\mathbb{R}$ open as well.    Thanks in advance!
11660
I am given $f,f' \in L^1(\mathbb{R})$, and f is absolute continuous, I want to show that:  $$\lim_{|x|\rightarrow \infty} f(x)= 0$$  Not sure how to show this, I know that $f(x)=\int_0^x f'(t) \, dt+f(0)$, and I can assume without loss of generality that $f(0)=0$, any help?  Thanks in advance.
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.
1759
Suppose that $\|f\|_p &lt \infty$ for all $1\leq p &lt p'$, I want to know if the the following is true and in that case how to show it     $p \mapsto \|f\|_p$ is continuous on $[1,p')$   Or maybe we need to impose some more constraints such as finite measure space. In case of finite measure space, I tried to use .
60477
exercise: Suppose that $f:(0,∞) → R$  satisfies $f(x) - f(y) = f(x/y)$  for all $x,y$ in $(0, ∞)$ and $f(1) = 0$.   Prove f is continuous on $(0, ∞)$ if and only if f is continuous at $1$.    (proof→:) Suppose that $f:(0,∞) → R$ satisfies $f(x) - f(y) = f(x/y)$ for all $x,y$ in $(0, ∞)$ and $f(1) = 0$. Suppose f is continuous on (0,∞), then by definition f is said to be continuous at a point $a$ if for every $ε &gt; 0$  there is a $δ &gt; 0$ such that $|x-a| &lt; δ$ implies $|f(x) - f(a)| &lt; ε$.   Then suppose f is continuous at $1$, then by our defined function  f(x) - f(y) = f(x/y) , we get $|x-1| &lt; δ$ implies $|f(x) - f(1)| = |f(x) - 0| = |f(x)| =  &lt; ε$.   (proof ←) :Suppose that $f$ is continuous at $1$, then we need to show $f$ is continuous on every point $a$ on $(0, ∞)$. Thus, by definition, f is said to be continuous at a point $a$ if for every $ε &gt; 0$ there is a $δ &gt; 0$ such that $|x-a| &lt; δ$ implies $|f(x) - f(a)| &lt; ε$. Thus, $|x-a| &lt; δ → |f(x) - f(a)| = |f(x/a)| &lt; ε$. Then $f$ is continuous on on $(0, ∞)$.  Please can anyone please help me verify this makes sense. If not, can someone please give any feedback/hints. Thank you in advance.
49805
As the title indicates, I'd like to prove the following:     If $f:\mathbb R\to\mathbb R$ is a continuous function on $[a,b]$, then $f$ attains its maximum.   Now, I do have a working proof: $[a,b]$ is a connected, compact space, which means that because $f$ is continuous, $f([a,b])$ is compact and connected as well.  Therefore, $f([a,b])$ is a closed interval, which means it has both a minimum and, as desired, a maximum.  What I would like, however, is a proof that doesn't require such general or sophisticated framework.  In particular, I'd like to know if there's a proof that is understandable to somebody beginning calculus, one that (at the very least) doesn't invoke compactness.  Any comments, hints, or solutions are welcome and apreciated.
105695
Let $\Omega$ be a bounded open subset of $\mathbb{C}$ and $f:\Omega\rightarrow\Omega$ be holomorphic in $\Omega$. Prove that if there exists a point $z_0$ in $\Omega$ such that $$f(z_0)=z_0~~~~\text{and}~~~~f'(z_0)=1$$ then $f$ is linear.  Please give some hints. Thanks in advance!