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13

Yes, it does. Suppose not. Then $a-b > 0$ and in particular we can take $\epsilon = a-b$. The statement then says that $a-b < a-b$, which is absurd.


6

The argument is pretty much fine, but what you're really relying on (without mentioning explicitly) is the fact that an uncountable set in $\mathbb{R}$ must contain a convergent sequence. This property is essentially just the fact that $\mathbb{R}$ cannot contain uncountably many disjoint intervals, which gives you a simpler proof. Your set $E_{n_0}$ would ...


3

You've already been given a counterexample, very similar to the example I was going to give. The point to this answer is to motivate one way one might find such an example: Observe that $f/f'\to 0$ is the same as $g'\to\infty$, if $g=\log f$.


3

If $\{f_n\}_n$ is Cauchy, for each $k\in\Bbb N$ we can find an $N_k$ (depending on the $k$ we just chose) such that $$ \|f_n - f_m\| < 2^{-k} \quad\text{ whenever } n,m\geq N_k.$$ For $k+1$, we likewise get $$ \|f_n - f_m\| < 2^{-k-1} \quad\text{ whenever } n,m\geq N_{k+1}.$$ Now, both inequalities will hold for $n,m \geq \max\{N_k,N_{k+1}\}$, so ...


3

Split the integration range into positive and negative regions, drop the absolute value in the positive regions, negate the integrand in the negative regions, evaluate each partial integral independently, and then sum.


3

The easiest way to do this this particular function is to begin by graphing it, recalling that the integral of a function over an interval is just the signed area under the function on said interval. If we graph $f(x)=|x-1|$ and focus on the interval $[0,2]$, we see something like the following: To make things even clearer, what we want to find is the area ...


3

I try to prove this without invoking anything about Lebesgue integral. Moreover, I do not assume the fact that $fg$ is Riemann integrable nor the inequality $| \int_a^b f(x)g(x) dx | \leq \int_a^b |f(x)g(x)|dx$. $f$ is Riemann integrable $\Rightarrow$ $f$ is bounded. Choose $M>0$ such that $|f(x)|\leq M$ for all $x\in[a,b]$. Let $\varepsilon>0$ be ...


3

Yes. For example if $S=[1,2]$, then $T$ contains $0$, because $0$ exceeds no element of $S$. In fact it will be true that $(-\infty,\inf S]\subset T$ for the same reason.


2

The point is that the inequality $$ | u ( x , t ) - f ( x ) | \leq \int _ { \mathbb { R } } | \widehat { f } ( \xi ) | \left| e ^ { - 4 \pi ^ { 2 } a ^ { 2 } \xi ^ { 2 } t } - 1 \right| $$ holds independently of $x$ and that the term on the RHS converges to $0$ as $t\to 0$. This means that the convergence is uniform in $x$. The Dominated Convergence ...


2

The answer to your question is $$ z_1=1\\ z_2=-\frac{1}{2}+\frac{\sqrt{3}}{2}\cdot i\\ z_3=-\frac{1}{2}-\frac{\sqrt{3}}{2}\cdot i $$ If $\vec{z}_1=(x,y)$, $\vec{z}_2=(z,w)$ and $\vec{z}_3=(u,v)$ then by figure \begin{align} (z-u)^2+(w-v)^2=&(z-x)^2+(w-y)^2\\ (z-x)^2+(w-y)^2=&(u-x)^2+(v-y)^2\\ (z-u)^2+(w-v)^2=&(u-x)^2+(v-y)^2\\ \end{align} By $\|...


2

For the modulus, between $0$ and $1$, $|x-1|$ is negative so this will go to $-(x-1)$ which removes the modulus turning it into $(1-x)$. Then between $1$ and $2$, $|x-1|$ is always positive so it can remain as $(x-1)$. So: $$\int_0^1 (1-x) \,dx+ \int_1^2 (x-1) = \frac{1}{2} + \frac{1}{2} =1$$


2

In the special case of a first degree polynomial function $f(x)=Ax+B$ one may make use of the identity $$ \int |u|\,du=\frac{1}{2}u|u|+c$$ and the substitution $u=Ax+B,\,dx=\frac{1}{A}du$ to obtain $$ \int |Ax+B|\,dx=\frac{1}{A}\int|u|\,du=\frac{1}{2A}(Ax+B)|Ax+B|+c $$ For a definite integral, you may actually solve the integral at the $u$ stage, of ...


2

Hint: For every $x \in X$, the function $$y\in A\to d(x,y)\in \mathbb R$$ is continuous because $d(y,y')\le\delta$ implies $d(y,x)\le d(y,y')+d(y',x) \implies |d(y,x)-d(y,'x)|\le d(y,y')\le\delta$. Since $A$ is compact and $d(-,x)$ is a real valued continuous function, what can you say about the image?


2

This might be a counterexample, but take it with some care. It builds on the fact that every $L^p(0,1)$ contains a Hilbert subspace which is complemented, see Remark on p. 143 in Johnson&Lindenstrauss Let us call this subspace $X_p$. We get an isometric isomorphism $T_p : X_p \to \ell^2$ and a bounded projection $P_p : L^p(0,1) \to X_p$. Now, we ...


2

Yes, it does. As $0\leqslant x-\ln(1+x)\leqslant x^2/2$ for $x\geqslant 0$, we have $$0\leqslant\sum_{k=1}^{n}\big(x_{n,k}-\ln(1+x_{n,k})\big)\leqslant\frac{1}{2}\sum_{k=1}^{n}x_{n,k}^2\leqslant\frac{1}{2}(\max_{1\leqslant k\leqslant n}x_{n,k})\sum_{k=1}^{n}x_{n,k}\underset{n\to\infty}{\longrightarrow}0$$ which implies $\displaystyle\lim_{n\to\infty}\sum_{k=...


2

Just a little note $x \in (0, \pi)$ means that you have $-1 < \cos(x) < 1$, it's important that $\cos(x)$ is neither 1 nor -1. Now given that $-1 < \cos(x) < 1$ you have $|\cos(x)| < 1$. Remember the geometric series formula $$\sum_{n = 0}^{\infty} r^n = \frac{1}{1 - r}$$ for $|r| < 1$. Also note that your sum starts from $n = 1$ while ...


2

For a given real number $x$ the quantity $\cos x$ is a constant. Therefore you have a convergent geometric series defined on the interval $(0,\pi)$ The formula for geometric series gives your answer as $$\frac {\cos x}{1-\cos x}$$


1

From About generalized Minkowski inequality on MathOverflow: The question that you are asking was asked in "On Generalizations of Minkowski's Inequality in the Form of a Triangle Inequality" by F. Mulholland (1949). In that paper, Mulholland established a sufficient condition on $f$, namely that it should satisfy $f(0)=0$, be increasing on $x \ge ...


1

Assuming you take only real $x$ and $a_i$ and $b_i$ are real (inequality isn't defined on $\mathbb{C}$), it doesn't. Take, for example $g(x) = \frac{1}{2 + x^2}$ and $h(x) = \frac{1}{1 + x^2}$. Then $g(x) < h(x)$, but $R_1 = \sqrt{2}$ and $R_2 = 1$. To get example for the other side (when smaller function has smaller radius of convergence), take $\frac{...


1

Hint: Draw the area of integration and observe that $$\int_{0}^{1} \int_{x}^{1} h(t,x) dt dx = \int_{0}^{1} \int_{0}^{t} h(t,x) dx dt.$$


1

At least pointwise convergence isn't guaranteed. Let $f_n(x) = \begin{cases} \frac{1}{2} - \frac{1}{n}, && x < 1\\ 1,&& x \geqslant 1 \end{cases}$. Then $f_n^-(\frac{1}{2}) = 1$, but $f^-(\frac{1}{2}) = 0$. If $f^-$ is continuous then $f^-_n$ converges to $f^-$ at least pointwise. If $f^-$ is continuous then $f$ is strictly increasing (...


1

Hints: recall that an infinite product of positive numbers is called convergent if the finite products converge to a (strictly) positive number. For $0<a_i<1$, $\prod a_i$ converges iff $\sum \log(a_i)$ converges. Use the inequality $\log (1+x) \leq x$ and the fact that $\sum_i (1-\epsilon)^{i}<\infty$ to prove that the the infinite product is ...


1

It's just a prefabricated definition, it was written that way so that (i) and (ii) hold, it has no other use than to evaluate the skills of the participants.


1

Hint: You need to show that every point on y-axis between -1 and +1 is a limit point of your set $A$.


1

Your $d(y, A)$ is defined to be $d(y, A)=\inf\{ d(y,x):x\in A \}$. It follows that there is a sequence $\{x_n\}\subset A$, such that $$d(y,x_n)\xrightarrow{n\to\infty}d(y,A).$$ Since $A$ is compact, there exists a convergent subsequence $\{x_{n_k}\}$ of $\{x_n\}$ such that $$x_{n_k}\xrightarrow{n_k\to\infty}x_0\in A.$$ Now we claim that $d(y,A)=d(y,x_0).$ ...


1

As you have observed the function is not continous at $x$ if $x \neq 0$ so it is not differentiable either. For differentiability at $x=0$ consider $\lim_{x \to 0} \frac {|x|^{p}} x$. If $p>1$ this limit is $0$ because $|\frac {|x|^{p}} x|=|x|^{p-1} \to 0$. For $p<1$ it is clear that $|\frac {|x|^{p}} x|=|x|^{p-1} \to \infty$. For $p=1$ consider right ...


1

Take $x\in X$ and suppose that $f$ is continuous at $x$ with repect to the metric $d_1$. Take $\varepsilon>0$. You know that there is a $\delta>0$ such that$$d_1(x,y)<\frac\delta m\implies\bigl\lvert f(y)-f(x)\bigr\rvert<\varepsilon.$$But then$$d_2(x,y)<\delta\implies md_1(x,y)<\delta\iff d_1(x,y)<\frac\delta m\implies\bigl\lvert f(y)-f(...


1

Let $g(x) = f(x-1)$. Then we obtain $g(x) = f(x-1) = 8f(2x-1) = 8g(2x)$, and $g(-1)=1$. Note that values of $g$ for positive $x$ cannot be related to negative $x$. Clearly, $g(0)$ is arbitrary. To find solutions for positive $x$, we let $h(x) = g(2^x)$, giving $h(x) = 8h(x+1)$. This has solutions $8^{-x} h_1(x)$ for any function $h_1$ with period $1$. ...


1

Yes, it is true. Let $L^\infty _0 (\Omega)$ be the closure in $L^\infty$ of $C_0 ^\infty (\Omega)$ (the space of smooth functions with compact support). This will encode the fact that we are looking for essentially bounded solutions that vanish at the boundary. Consider the two auxiliary problems $$\left\{ \begin{eqnarray} &v_t - \Delta v &= 0 \\ &...


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