Let $(X,S,\mu)$ is a measure space and $\mu(X)&lt;\infty$. Define $d(f,g)=\int\frac{|f-g|}{1+|f-g|}d\mu$ is a metric on the space of measurable functions. My work- symmetry, let's show $d(f,g)=d(g,f)$ $d(f,g)=\int\frac{|f-g|}{1+|f-g|}d\mu$ and $d(g,f)=\int\frac{|g-f|}{1+|g-f|}d\mu$ It's obvious that, $d(f,g)=d(g,f)$ My concern is triangular inequality, What want to show is, $d(f,h)\leq d(f,g)+d(g,h)$ so the right hand side becomes, $\int\frac{|f-g|}{1+|f-g|}d\mu+\int\frac{|g-h|}{1+|g-h|}d\mu= \int\frac{|f-g|}{1+|f-g|}+\frac{|g-h|}{1+|g-h|}d\mu$ I don't think simplification of right side will give the result easily. Can someone give me a hints. Thank you in advance
104930
I'm working in the following exercise:  Suppose $(X, \mathcal A, \mu)$ is a finite measure space and suppose $\mathcal F$ is the set of all $\mathcal A$-measurable functions $f: X \rightarrow \mathbb R$. For $f, g \in \mathcal F$, let $$d(f, g)=\int_X\frac{|f-g|}{1+|f-g|}d\mu.$$  Show that:   $d(f, g)=0$ if and only if $f=g$ almost everywhere. $d(f, g)=d(g, f)$ $d(f, g)\leq d(f, h)+d(h, g)$ If $f_n$ is a sequence in $\mathcal F$ and if $f \in \mathcal F$ then $d(f_n, f)\rightarrow 0$ if and only if for every $\delta&gt;0$ the following holds: $$\lim_{n\rightarrow\infty}\mu(\{x \in X: |f_n(x)-f(x)|\geq\delta\})=0.$$ If $f_n$ is a Cauchy sequence on $\mathcal F$ then there exists $f \in \mathcal F$ such that $d(f_n, f)\rightarrow 0$.   i.e., I must show that this thing is a complete pseudometric space.  I have already done 1. 2. 3. 4., but I'm stuck on 5. I don't know which $f$ should the $f_n$ converge to.
48616
Can anyone please tell me whether the following function $d:\mathbb{R}^2 \times \mathbb{R}^2 \to  \mathbb{R}$ is really a metric or not on $\mathbb{R}^2$.  $$ d((x_1,x_2),(y_1,y_2))= \begin{cases} |x_1-y_1|, &amp;\text{ if } x_2=y_2\\ |x_1|+|y_1|+|x_2-y_2|, &amp;\text{ if } x_2 \neq y_2. \end{cases} $$  If it really is indeed a metric, please give the proof of triangle inequality.(I'm OK with the other two conditions).
69524
Let $X=C[0,1]$ be the set of all continuous functions on $[0,1]$. For any two functions $f,g\in X$, set $$d(f,g)=\|f-g\| = \sup \limits_{0\leq x \leq 1} |f(x)-g(x)|.$$  Prove $(X,d)$ is a metric space.    My attempt:  This question comes down to showing that $d$ is a metric, which means that:   $d(f,g)\geq 0$ $d(f,g)=0\iff f=g$ $d(f,g)=d(g,f)$ Triangular sadness.     1) Nonnegativity is guaranteed by taking the supremum of the absolute value of a sum, since the absolute value of a sum is non-negative, the supremum of non-negative numbers is non-negative.  $\square$    2) For the supremum of $|f(x)-g(x)|$ to be equal to $0$, the greatest number for $|f(x)-g(x)|$ from $0\leq x\leq 1$ must be $0$, hence $|f(x)-g(x)|=0$ for all values on this range, and thus $f(x)=g(x)$ is the constant zero map.  $\square$    3) $|f(x)-g(x)|=|g(x)-f(x)|$ implies that $\sup \limits_{0\leq x \leq 1} |f(x)-g(x)|=\sup \limits_{0\leq x \leq 1} |g(x)-f(x)|$ $\square$    4) Triangular sadness:  We need to prove that $d(f,g)\leq d(f,h) + d(g,h)$  I.e.  $\sup \limits_{0\leq x \leq 1} |f(x)-g(x)|\leq \sup \limits_{0\leq x \leq 1} |f(x)-h(x)|+\sup \limits_{0\leq x \leq 1} |g(x)-h(x)|$  Now I am not sure how I can 'behave' with the sup 'function'. Can I do the following?  $$\sup|f(x)-g(x)| = \sup|f(x)-g(x)+h(x)-h(x)|=\sup|f(x)-h(x)+(h(x)-g(x))|$$ Then use normal triangular inequality:  $$\sup|f(x)-h(x)+(h(x)-g(x))|\leq \sup(|f(x)-h(x)|+|h(x)-g(x)|)=\sup(|f(x)-h(x)|+|g(x)-h(x)|)$$   This is obviously less than $\sup \limits_{0\leq x \leq 1} |f(x)-h(x)|+\sup \limits_{0\leq x \leq 1} |g(x)-h(x)|$, but I don't know how to prove it. How do I finish that one off.
75574
I have some problems understanding  of this problem:  Given $f,g : \mathbb{R} \rightarrow \mathbb{R}$ - continuous functions with period $1$, prove that  $$\lim _{n \rightarrow \infty} \int_0^1 f(x)g(nx) \, dx = \int_0^1 f(x) \, dx \cdot \int_0^1g(x) \, dx$$  It says there that if we split this integral into $n$ parts to get  $$\int_0^1 f(x)g(nx) \, dx = \sum_{r=0}^{n-1} \int_{\frac{r}{n}}^{\frac{r+1}{n}} f(x)g(nx) \, dx$$  we will have   $$\int_0^1 f(x)g(nx) \, dx = \sum_{r=0}^{n-1} f\left(\frac{r}{n}\right) \int_{\frac{r}{n}}^{\frac{r+1}{n}}g(nx) \, dx$$  because for large $n$, $f$ is roughly constant over the range.  Could you tell me why $f$ is roughly constant over the range for $n$ large enough?  I would really appreciate all your insight.
61645
Given that $(X,d)$ is a metric space, with $d$ the French railway metric. Show that $(X,d)$ is a complete metric space.  Suppose $\{x_n\}$ is a Cauchy sequence in $(X,d)$. Choose $N$ such that for all $m,n$ $\geq$ N: d($x_{n},x_{m}$) &lt; $\epsilon$. (Note that: d($x_{n},x_{m}$) = d($x_{n},p)+d(p,x_{m}$) , $p \in X$).  Then for all $n \geq N: d(x_{n},x) \leq$ .....  Thats where im stuck. Help would be much appreciated.
11188
Let $f$ be a continuous mapping of a compact metric space $(X, d)$ onto a Hausdorff space $(Y, \tau_1)$. Then $(Y, \tau_1)$ is compact and metrizable.   In one proof the following metric is constructed:  $d_1(y_1, y_2) = inf\{d(a, b) : a \in f^{-1}\{y_1\},\ b \in f^{-1}\{y_2\}\}$, for all $y_1$ and $y_2$ in $Y$.  I'm thinking about how to prove that triangle inequality holds for $d_1$, s.t. $d_1(x, z) \leq d_1(x, y) + d_1(y, z)$  And how about the case when triangle inequality does not hold?    Update: @daniel-schepler suggested to use metric $d_1(x, y) = inf\{d(x^*, x_1) + d(y_1, x_2) + \cdots + d(y_{n-1}, x_n) + d(y_n, y^*)\}$ where $f(x^*) = x$, $f(y^*) = y$, and $f(x_i) = f(y_i)$ for each $i$.  Now we need to prove 3 property of metric: positivity, symmetry and triangle inequality. This is my sketch:   Positivity. Chose two points $x,y \in Y$ s.t. $x \neq y$. Singleton sets $\{x\}$ and $\{y\}$ are disjoint closed sets by Hausdorfness of $Y$ and their preimages are also disjoint closed in $X$ by continuity of $f$. $X$ is compact metric space by hypothesis and therefore a normal space. In normal space every two disjoint closed sets have disjoint open neighborhoods. Let $f^{-1}(x) \subseteq U$ and $f^{-1}(y) \subseteq V$ s.t. $U,V$ are open and $U \cap V = \varnothing$. Therefore there exists an $\varepsilon &gt; 0$, s.t. $x^* \in B_\varepsilon(x^*) \subseteq U$ and $y^* \in B_\varepsilon(y^*) \subseteq U$ and distance between any two closed sets is strictly positive and $d_1$ is positive and zero iff $x = y$. Symmetry. Clearly $d_1$ is symmetric. Triangle inequality. For triangle inequality we can show that in the "worst" case $d_1(x, z) = d_1(x,y) + d_1(y, z)$.
54280
Let $\mu$ be a general measure, suppose $f,g$ has compact support on $\mathbb{R}$, when does the integration by parts formula hold $$\int f'g d\mu = - \int g'fd\mu?$$ I know in general this is false, we can take $\mu$ to be supported on a point, say $0$, then it is not necessarily true that  $$f'(0)g(0) = -g'(0)f(0).$$  If $\mu$ is absolute continuous w.r.t. Lebesgue measure, we have $\frac{d\mu}{dx} = h$ $$\int f'gd\mu = \int f'gh dx = -\int f(gh)'dx$$ where $(gh)'dx$ might be a measure. but we can not recover the form $\int g'fh dx$.  Thank you very much!
79930
Why do we have the following equality$$\int_{-\infty}^\infty |g(x)|\,dx = \int_0^\infty \mu(\{x : |g(x)| \ge t\})\,dt,$$where $\mu$ is Lebesgue measure?
71298
I am solving the following problem and its parts. Let (C[0,1],$d_\infty$) be the metric space of continuous functions on [0,1] where the distance function is defined by $d_\infty(f,g)=\sup_{x∈[0,1]}|f(x)−g(x)|. $ Let $T : (C[0, 1], d_\infty)\to (C[0, 1],d_\infty$) be defined by $(Tf)(x)=\int_0^xf(t)dt$ Prove that: T is not a contraction, i.e. there does not exist 0 &lt; K &lt;    1 such that $d_\infty(T f, T g)\leq K · d_\infty(f, g)$ holds for any $f,g ∈ C[0,1]$. Ive tried an example but I haven't really got anywhere. My work is as follows: $$f'(t)=\int_0^xtdt=\frac{x^2}{2}=F(x)-F(0)$$ so we pick $t$ &amp; $t^2$ such that $d$($t$,$t^2$)=sup[$t$-$t^2$] with max value = $\frac{1}{4}$ and $d$(T$t$,T$t^2$)=sup[$\frac{t^2}{2}-\frac{t^3}{3}$] with max value =.167
47817
Let $(M,d)$ be a compact metric space and $f:M \to M$ such that $d(f(x),f(y)) \ge d(x,y) , \forall x,y \in M$ ; then how to prove that $d(f(x),f(y))=d(x,y) , \forall x,y \in M$ i.e. that $f$ is an isometry  ? I wanted to prove by contradiction , that suppose $\exists a,b \in M$ such that $d(f(a),f(b)) &gt; d(a,b)$ ; I wanted to use the sequences $\{f^n(a)\}$ and $\{f^n(b)\}$ , these have convergent subsequences $\{f^{r_n}(a) \}$ and $\{f^{r_n}(b)\}$ ( without loss of generality we have extracted a common subsequence ) converging to say $x,y \in M $ respectively . Then $d (f^{r_n}(a) , f^{r_n}(b))$ converges to $d(x,y) &gt;d(a,b)$ ;  but I don't know what to do next , please help . Thanks in advance
20957
I tried $|d(x, z) - d(y, z)| \leq d(x, z) + d(y, z)$, but the the triangle equation goes the wrong side. Is there an analogue to the reverse triangle inequality with metrics? The reason I want to know this is to proof that $d(x, A)$ is continious where $A$ is a set. (Let $|x - y| &lt; \delta$, then after some algebra $|d(x, A) - d(y, A)| = \inf(\{|d(x, a) - d(y, a)| \ | \ a \in A \})$.
40132
Let $f : \mathbb{R} \to [ 0, \infty)$ be a measurable function. If $\int_{- \infty}^{\infty} f(x) dx =1$. Then I want show that $ \int_{- \infty}^{\infty}  \frac{1}{1+ f(x)} dx = \infty.$  Any help will be appreciated.  Thank you in advance.
22019
Let $(E,d)$ be a metric space and $\rho(x,y)= \frac{d(x,y)}{1+d(x,y)}$. Show that $(E,\rho)$ is a metric and $\rho$ and $d$ are equivalent.   I have proved the part about the metric space. However, I cant remember how to prove the equivalence. Also I'm stuck with a follow up which says:     Show that there are no constants $c,C &gt;0$ such that for all $x,y \in E:$ $$cd(x,y) \leq \rho(x,y) \leq Cd(x,y)$$
100102
Let $(X, \mathcal{B}, \mu)$ be a measure space and $f$ a measurable function on $X$ and suppose that $\forall E \in \mathcal{B}$ we have that $\int_E f = 0$.  Then I want to show that $f = 0$ almost everywhere (a.e.).   Suppose for sake of contradiction that $f \ne 0$ a.e. Then $\not\exists E \in \mathcal{B}$ s.t. $\mu(E) = 0$ and $f(x) = 0$, $\forall x \in X - E$  Then $\{x : f(x) \ne 0\} = A$ is not measure zero so that either $\mu(A) &gt; 0$ or $A \notin \mathcal{B}$. Now if $\mu(A) &gt; 0$ then it is easy to see that $\int_A f \ne 0$ so that we have a contradiction of our original hypothesis that $\forall E \in \mathcal{B}, \int_E f = 0$. But if on the other hand $A \notin \mathcal{B}$, I cannot no longer appeal to $\int_A f \ne 0$ since $\int_A f$ is non-sense.  So I'm having trouble with this part of the argument.
6281
Let $f,g,g´$ be continous on $[a,b]$ and $g$ monotone on $[a,b]$; then there exist $c\in (a,b)$ so that $$\int_{a}^{b}f(x)g(x)dx=g(a)\int_{a}^{c}f(x)dx+g(b)\int_{c}^{b}f(x)dx$$  Ineed to apply the first generalized mean value theorem for integrals i.e. $$\int_{a}^{b}f(x)g(x)dx=f(c)\int_{a}^{b}g(x)dx$$ to $g´$and $\int_{a}^{x}f(x)dx$ but I don´t know how please I would appreciate your help
8003
Let $(X,d)$ be a metric space and for $x,y \in X$ define $$d_b(x,y) = \dfrac{d(x,y)}{1 + d(x,y)}$$ a) show that $d_b$ is a metric on $X$  Hint: consider the derivative of $f(t)$ = $\dfrac{t}{1+t}$  b) show that $ d$ and $ d_b $ are equivalent metrics. c) let $(X,d) $ be $(\mathbb{R}, |\cdot|)$ Show that there exists no $ M&gt;0$ such that $ |x-y| $ $\leq$ $Md_b$ $(x,y)$ for all $ x,y$ $\in$ $\mathbb{R}$  I have calculated $f(t)$ and $f'(t)$ and from this I know $f(t)$ is an increasing function as $f'(t)$ is strictly positive. But I don't know where to go from here or how to do parts b) and c)
58969
For a variation of the metric $g^{\mu\nu}$ with respect to $g^{\alpha\beta}$ you might expect the result (at least I did):  \begin{equation} \frac{\delta g^{\mu\nu}}{\delta g^{\alpha\beta}}= \delta^\mu_\alpha\delta^\nu_\beta. \end{equation}  but then to preserve the fact that $g^{\mu\nu}$ is symmetric under interchange of $\mu$ and $\nu$ we should probably symmetrise the right hand side like this:  \begin{equation}\frac{\delta g^{\mu\nu}}{\delta g^{\alpha\beta}}= \delta^\mu_\alpha\delta^\nu_\beta + \delta^\mu_\beta\delta^\nu_\alpha.\end{equation}  Is this reasonable/correct? If not, why not?  It seems that I can derive some weird results if this is right (or maybe I'm just making other mistakes).