Master Continuity Equation with Plainmath

Let f be continuous on [a,b] and differential on (a,b). Prove that if a >= 0 there are x1, x2, x3 ∈ (a,b) such that
$f^{\prime }(x_1)=(b+a)\frac{f^{\prime }(x_2)}{2x_2}=(b^2+ba+a^2)\frac{f^{\prime }(x_3)}{3x_3^2}$
I think this problem will use the generalized mean value theorem to solve. However, I don't know how to apply it. Can you suggest a solution? Thank you very much!

In real functions, do we have a notion of one-sided measure theoretic limits? I want to define them with the following:
${\displaystyle \underset{x\to <!--\; \to \; -->c^{+}}{lim}f(x)=L}$
iff
$\mathrm{\forall <!--\; \forall \; -->}\mathrm{ϵ<!--\; ϵ\; -->},\mathrm{\exists <!--\; \exists \; -->}\mathrm{\delta <!--\; \delta \; -->}>0,J:=f((c,c+\mathrm{\delta <!--\; \delta \; -->})),\mathrm{\mu <!--\; \mu \; -->}(J\cap <!--\; \cap \; -->B_{\mathrm{ϵ<!--\; ϵ\; -->}}(L))=\mathrm{\mu <!--\; \mu \; -->}(J)$

Let $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ with intermediate value property and increasing over $\mathbb{R}$\$\mathbb{Q}$. Then $f$ is continous on $\mathbb{R}$. How to try?

Given metric spaces $(X,d)$ and $(Y,d^{\prime })$ and continuous mapping $S$ and $T$ from $X$ into $Y$, prove that the set $\{x\in <!--\; \in \; -->X:Sx=Tx\}$ is closed in $(X,d)$.
I've run out of any ideas where I should start

Let $\mathrm{\Omega <!--\; \Omega \; -->}=\mathbb{C}\mathrm{\setminus <!--\; \setminus \; -->}[-<!--\; -\; -->1,1]$, i.e. deleting ``the line'' only, is there a function $f:\mathrm{\Omega <!--\; \Omega \; -->}\to <!--\; \to \; -->\mathbb{C}$ such that $f$ satisfies $f(z{)}^2=1-<!--\; -\; -->z^2$ and is continuous on this region?

My guess is that such a function would exist, but requires a piecewise definition. A candidate solution I have been working on is $f(z)=e^{\frac{1}{2}\log <!--\; \; -->(1-<!--\; -\; -->z^2)}$. The problem with this solution is having the domain, as I realize it is possible to have $1-<!--\; -\; -->z^2>1$, where my solution is well-defined. Is there a way to work around this or should I try something else?

A function is said to be continuous on an interval when the function is defined at every point on that interval and undergoes no interruptions, jumps, or breaks. If some function f(x) satisfies these criteria from x=a to x=b, for example, we say that f(x) is continuous on the interval [a, b]. The brackets mean that the interval is closed -- that it includes the endpoints a and b. In other words, that the interval is defined as a ≤ x ≤ b. An open interval (a, b), on the other hand, would not include endpoints a and b, and would be defined as a < x < b.

So, if a function is continuous on an open interval $(a,b)$ or even semi-open such as $[a,b]$, $(a,b]$ is it bounded?

Let $H_1$ and $H_2$ be Hilbert spaces, let $T\in <!--\; \in \; -->B(H_1,H_2)$. Suppose that $\mathrm{Ker}<!--\; \; -->T$ is finite-dimensional and that $\mathrm{Im}<!--\; \; -->T$ is closed in $H_2$. Prove that $\mathrm{Ker}<!--\; \; -->(T+K)$ is finite-dimensional for each $K\in <!--\; \in \; -->K(H_1,H_2)$.

1. Define Hilbert space as direct sum of two complemented subspaces
2. For compact operator use Hilbert Schmidt decomposition

Main idea is to prove that intersection of Spectrum of such $T and compact K is finite

Given a continuous function $f:\mathcal{R}\to <!--\; \to \; -->\mathcal{R}$ with:
$\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}f(x)=0$
$\underset{x\to <!--\; \to \; -->-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}f(x)=0$
$f(x)>0\mathrm{\forall <!--\; \forall \; -->}x\in <!--\; \in \; -->\mathcal{R}$
Show that there exists $x,y\in <!--\; \in \; -->\mathcal{R}$ with $x\ne <!--\; \ne \; -->y$ so that $f(x)=f(y)$.
In this question no interval is given. What I want to do is to show that there is a maximum and then use the Intermediate value theorem. But I don't know how to do that without any interval given. Is it sufficient enough to say that because the limit for x to infinity and -infinity and f(x)>0 there must exist a maximum value? Since f(x) cannot be zero.

Let $X$ be a non-empty set and let $x_0\in <!--\; \in \; -->X$. The topology $T$ is defined by the collection of subsets $U\subset <!--\; \subset \; -->X$ such that either $U=\mathrm{\varnothing <!--\; \varnothing \; -->}$ or $U\ni <!--\; \ni \; -->x_0$.

Is it true that if $f:X\to <!--\; \to \; -->X$ is continuous, does it follow that $f(x_0)=x_0$ with respect to the topology $T$?

I under stand that the converse is true, that is for any function $f:X\to <!--\; \to \; -->X$ with $f(x_0)=x_0$ is continuous with respect to the topology T but i don't know how to prove if this case is true.

Let $h(t):[0,1]\to <!--\; \to \; -->\mathbb{C}$ be a continuous function. Prove that $f:\mathbb{C}\setminus <!--\; \setminus \; -->[0,1]\to <!--\; \to \; -->\mathbb{C}$ defined by $f(z)={\int <!--\; \int \; -->}_0^1\frac{h(t)}{z-<!--\; -\; -->t}$ is holomorphic.

My attempt:
$$\begin{gathered} \underset{z\to <!--\; \to \; -->0}{lim}\frac{f(z+z_0)-<!--\; -\; -->f(z_0)}{z}=\underset{z\to <!--\; \to \; -->0}{lim}\frac{{\int <!--\; \int \; -->}_0^1\frac{h(t)}{z+z_0-<!--\; -\; -->t}-<!--\; -\; -->{\int <!--\; \int \; -->}_0^1\frac{h(t)}{z_0-<!--\; -\; -->t}}{}= \\ \underset{z\to <!--\; \to \; -->0}{lim}\frac{{\int <!--\; \int \; -->}_0^1(\frac{(z_0-<!--\; -\; -->t)h(t)}{(z+z_0-<!--\; -\; -->t)(z_0-<!--\; -\; -->t)}-<!--\; -\; -->\frac{(z+z_0-<!--\; -\; -->t)h(t)}{(z_0-<!--\; -\; -->t)(z+z_0-<!--\; -\; -->t)})dt}{z}=\underset{z\to <!--\; \to \; -->0}{lim}\frac{{\int <!--\; \int \; -->}_0^1\frac{-<!--\; -\; -->zh(t)}{(z+z_0-<!--\; -\; -->t)(z_0-<!--\; -\; -->t)}dt}{z}= \\ -<!--\; -\; -->\underset{z\to <!--\; \to \; -->0}{lim}{\int <!--\; \int \; -->}_0^1\frac{h(t)}{(z+z_0-<!--\; -\; -->t)(z_0-<!--\; -\; -->t)}dt=-<!--\; -\; -->{\int <!--\; \int \; -->}_0^1\frac{h(t)}{(z_0-<!--\; -\; -->t{)}^2}dt \end{gathered}$$
Where I'm not sure about the last equality.
Is my solution correct? Any help would be appreciated.

The function,
$f(x)=\{\begin{array}{ll}(e^x-<!--\; -\; -->1)/x& x\ne <!--\; \ne \; -->0\\ 1& x=0\end{array}$
is continuous and differentiable at $x=0$. By composition, the $x\ne <!--\; \ne \; -->0$ case is analytic everywhere it is defined, and integrating $f$ around 0 in the complex plane yields 0 and suggests that there is no pole there. Furthermore, working from the Taylor series of $e^x$, the $x\ne <!--\; \ne \; -->0$ case can be seen to have a series that is defined at 0 and equal to 1:
$\frac{e^x-<!--\; -\; -->1}{x}=\frac{-<!--\; -\; -->1+\underset{k=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{x}^k/k!}{x}=\frac{\underset{k=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{x}^k/k!}{x}=\underset{k=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{1}{k+1}\frac{x^k}{k!}$
From this we can also see that $f$ is analytic and infinitely differentiable.

All taken together, it seems strange that such a nicely behaved function would have an irremovable "patch" necessary at $x=0$. Nonetheless, I have not been able to come up with another way to write the function that allows it to be defined at zero without a special case there. Is it impossible to write this analytic function without a case block?

Suppose $y(x)=x^2$ is a solution of $y"+P(x)y^{\prime }+Q(x)y=0$ on (0,1) where $P$ and $Q$ are continuous functions on (0,1). Can both $P$ and $Q$ be bounded functions? Justify your answer.
I have tried differentiating $y=x^2$, $y^{\prime }=2x$ and $y^{\prime }/x=2$. Differentiating this once again yielded $y"-<!--\; -\; -->y^{\prime }/x=0$. So I don't think $P(x)$ can be bounded in (0,1). But Iam not so sure about whether this is correct or not. So it would be great if anyone has some ideas to share.

I have seen it stated that for an open subset $Y\subseteq <!--\; \subseteq \; -->X$ such that $X$ is a compact Hausdorff space we get an identification of the $C^{\ast <!--\; \ast \; -->}$-algebras : $C(X\setminus <!--\; \setminus \; -->Y)\cong <!--\; \cong \; -->C(X)/C_0(Y)$.
I suppose that this relies on the Tietze extension theorem but I fail to connect the dots.
How do I realize this?

I have been assigned this question but the solution given uses a different method so i don't know if my solution is correct
$\mathrm{\exists <!--\; \exists \; -->}\stackrel{¯<!--\; ¯\; -->}{\mathrm{ϵ<!--\; ϵ\; -->}}><!--\; >\; -->0\text{s.t.}|\stackrel{¯<!--\; ¯\; -->}{x}-<!--\; -\; -->\stackrel{¯<!--\; ¯\; -->}{y}|<<!--\; <\; -->\mathrm{\delta <!--\; \delta \; -->}$
$\mathrm{\exists <!--\; \exists \; -->}\stackrel{¯<!--\; ¯\; -->}{x},\stackrel{¯<!--\; ¯\; -->}{y}\in <!--\; \in \; -->D\text{s.t.}|f(\stackrel{¯<!--\; ¯\; -->}{x})-<!--\; -\; -->f(\stackrel{¯<!--\; ¯\; -->}{y})|\ge <!--\; \ge \; -->\mathrm{ϵ<!--\; ϵ\; -->}$
$\mathrm{\forall <!--\; \forall \; -->}\mathrm{\delta <!--\; \delta \; -->}><!--\; >\; -->0\leftrightsquigarrow <!--\; \leftrightsquigarrow \; -->\mathrm{\forall <!--\; \forall \; -->}n\in <!--\; \in \; -->\mathbb{N}$
$\mathrm{\forall <!--\; \forall \; -->}\mathrm{\delta <!--\; \delta \; -->}><!--\; >\; -->0\mathrm{\exists <!--\; \exists \; -->}n\in <!--\; \in \; -->\mathbb{N}\text{s.t.}\frac{1}{n}<<!--\; <\; -->\mathrm{\delta <!--\; \delta \; -->}$
$|\stackrel{¯<!--\; ¯\; -->}{x}-<!--\; -\; -->\stackrel{¯<!--\; ¯\; -->}{y}|<<!--\; <\; -->\frac{1}{n}<<!--\; <\; -->\mathrm{\delta <!--\; \delta \; -->}$
$\stackrel{¯<!--\; ¯\; -->}{x}={\stackrel{¯<!--\; ¯\; -->}{x}}_n\stackrel{¯<!--\; ¯\; -->}{y}={\stackrel{¯<!--\; ¯\; -->}{y}}_n$
$|{\stackrel{¯<!--\; ¯\; -->}{x}}_n-<!--\; -\; -->{\stackrel{¯<!--\; ¯\; -->}{y}}_n|<<!--\; <\; -->\frac{1}{n}{\stackrel{¯<!--\; ¯\; -->}{y}}_n={\stackrel{¯<!--\; ¯\; -->}{x}}_n-<!--\; -\; -->\frac{1}{2n}$
$|{\stackrel{¯<!--\; ¯\; -->}{x}}_n-<!--\; -\; -->{\stackrel{¯<!--\; ¯\; -->}{y}}_n|=|{\stackrel{¯<!--\; ¯\; -->}{x}}_n-<!--\; -\; -->{\stackrel{¯<!--\; ¯\; -->}{x}}_n+\frac{1}{2n}|=\frac{1}{2n}<<!--\; <\; -->\frac{1}{n}$
${\stackrel{¯<!--\; ¯\; -->}{y}}_n={\stackrel{¯<!--\; ¯\; -->}{x}}_n-<!--\; -\; -->\frac{1}{2n}$
$|{\stackrel{¯<!--\; ¯\; -->}{x}}_n-<!--\; -\; -->{\stackrel{¯<!--\; ¯\; -->}{y}}_n|=\frac{1}{2n}<<!--\; <\; -->\frac{1}{n}$
$\text{choose }\stackrel{¯<!--\; ¯\; -->}{\mathrm{ϵ<!--\; ϵ\; -->}}=1$
$|f(\stackrel{¯<!--\; ¯\; -->}{x})-<!--\; -\; -->f(\stackrel{¯<!--\; ¯\; -->}{y})|\ge <!--\; \ge \; -->\stackrel{¯<!--\; ¯\; -->}{\mathrm{ϵ<!--\; ϵ\; -->}}=1$
$|{\stackrel{¯<!--\; ¯\; -->}{x}}_n^3-<!--\; -\; -->({\stackrel{¯<!--\; ¯\; -->}{x}}_n^3-<!--\; -\; -->\frac{3{\stackrel{¯<!--\; ¯\; -->}{x}}_n^2}{2n}+\frac{3{\stackrel{¯<!--\; ¯\; -->}{x}}_n}{4n^2}-<!--\; -\; -->\frac{1}{8n^3})|$
$=|\frac{3{\stackrel{¯<!--\; ¯\; -->}{x}}_n^2}{2n}-<!--\; -\; -->\frac{3{\stackrel{¯<!--\; ¯\; -->}{x}}_n}{4n^2}+\frac{1}{8n^3}|$
$\text{choose }{\stackrel{¯<!--\; ¯\; -->}{x}}_n=2n^2$
$=|\frac{3(2n^2{)}^2}{2n}-<!--\; -\; -->\frac{3(2n^2)}{4n^2}+\frac{1}{8n^3}|$
$=|6n^3-<!--\; -\; -->\frac{3}{2}+\frac{1}{8n^3}|\ge <!--\; \ge \; -->1$
$\text{thus }f(x)=x^3,(x\in <!--\; \in \; -->\mathbb{R})\text{ is not uniformly continuous, as required}$

$f$ is continuous at $x_0$ and $f(x_0)<0$ then there exists a neighborhood of $x_0$ in which $f(x)<0$ .
Is it possible to prove this statement, and if so what is the best way to go about it? Contradiction, Induction, Deduction, etc... ?

I think it's obvious but I cannot show it explicitly.
If a function $f$ is continuous and positive definite, that is, $f(0)=0$ and $f(x)>0$ for all nonzero $x$.
Then, how to show that $\underset{t\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}f(x_t)=0⟹<!--\; ⟹\; -->\underset{t\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{x}_t=0$?

While proving that the quotient space ${\stackrel{¯<!--\; ¯\; -->}{B}}^n/S^{n-<!--\; -\; -->1}$ is homeomorphic to $S^n$, I needed to construct a continuous function $p:{\stackrel{¯<!--\; ¯\; -->}{B}}^n\to <!--\; \to \; -->S^n$. I figured that by fixing two points $R=(0,0,\dots <!--\; \dots \; -->,r)$ and $L=(0,0,\dots <!--\; \dots \; -->,-<!--\; -\; -->r)$ in the closed $n$-ball ${\stackrel{¯<!--\; ¯\; -->}{B}}^n$ of radius $r$, I could then use the function
$p(x)=\{\begin{array}{l}(x_1,\dots <!--\; \dots \; -->,x_n,\sqrt{r^2-<!--\; -\; -->||x|{|}^2}):(x_n-<!--\; -\; -->r{)}^2\le <!--\; \le \; -->(x_n+r{)}^2\\ (x_1,\dots <!--\; \dots \; -->,x_n,-<!--\; -\; -->\sqrt{r^2-<!--\; -\; -->||x|{|}^2}):(x_n+r{)}^2<(x_n-<!--\; -\; -->r{)}^2\end{array}$
(where $||x||$ is the standard norm in ${\mathbb{R}}^n$) to determine the sign of the $(n+1)$th component of the image of $x\in <!--\; \in \; -->{\stackrel{¯<!--\; ¯\; -->}{B}}^n$. However, knowing that the closed unit interval can be mapped to the circle by the parametrization $x\mapsto <!--\; \mapsto \; -->(\cos <!--\; \; -->(2\mathrm{\pi <!--\; \pi \; -->}x),\sin <!--\; \; -->(2\mathrm{\pi <!--\; \pi \; -->}x))$, I can't help but to wonder whether some similar construction could be made from ${\stackrel{¯<!--\; ¯\; -->}{B}}^n$ to $S^n$?

That is, do you happen to know some nice/comfortable continuous mappings from a general closed $n$-ball to $n$-sphere, similar to the parametrization of a unit circle by a unit interval?

Theorem: Let $(X,d_X)$ and $Y(d_Y)$ be metric spaces. A function $f:X\to <!--\; \to \; -->Y$ is continuous if and only if for every open $U\subset <!--\; \subset \; -->Y$, then
$f^{-<!--\; -\; -->1}(U)=\{x\in <!--\; \in \; -->X:f(x)\in <!--\; \in \; -->U\}$
is open in $X$.


If I want to show that the function $f:[0,1]\cup <!--\; \cup \; -->[2,4]\subset <!--\; \subset \; -->\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ defined by
$f(x)=\{\begin{array}{ll}1,& x\in <!--\; \in \; -->[0,1];\\ 2,& x\in <!--\; \in \; -->[2,4].\end{array}$
is continuous, then I need to consider any open set $U\subset <!--\; \subset \; -->\mathbb{R}$ and show that
$f^{-<!--\; -\; -->1}(U)=\{x\in <!--\; \in \; -->[0,1]\cup <!--\; \cup \; -->[2,4]:f(x)\in <!--\; \in \; -->U\}$
is open.

I understand that if $1\notin <!--\; \notin \; -->U$ and $2\notin <!--\; \notin \; -->U$, then $f^{-<!--\; -\; -->1}(U)=\mathrm{\varnothing <!--\; \varnothing \; -->}$. But I don't understand what happens when only $1\in <!--\; \in \; -->U$, or only $2\in <!--\; \in \; -->U$, or $1,2\in <!--\; \in \; -->U$.

If only $1\in <!--\; \in \; -->U$, then is it correct to say that $f^{-<!--\; -\; -->1}(U)$=[0,1]? If only $2\in <!--\; \in \; -->U$, then is it correct to say that $f^{-<!--\; -\; -->1}(U)$=[2,4]? But I am really confused since these are closed intervals. I am lost.

Can someone help me to find $f^{-<!--\; -\; -->1}(U)$ for those cases? Any clues or hints will be appreciated.