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No. Arrange the intervals $[\frac {i-1} {2^{k}},\frac i {2^{k}})$ in a sequence $A_1,A_2,...$. Let $f_n$ be the characteristic function of $A_n$. Then $nf_n(x)$ is not bounded for any $x \in (0,1)$ but $nf_n \to 0$ in every $L^{p}$ ($1\leq p<\infty$).


Yes, it does. Let $$\eqalign{f_N(x) &= \cases{f(x) & if $|f(x)| \le N$\cr 0 & otherwise}\cr \phi_N(g) &= \int_\Omega f_N(x) g(x)\; dx\cr \phi(g)&= \int_\Omega f(x) g(x)\; dx\cr}$$ By Dominated Convergence, $\phi_N(g) \to \phi(g)$ as $N \to \infty$ for all $g \in L^q$. By the Uniform Boundedness Principle, there is $M$ such that all $\|\...


\begin{align} e^{x^2}\operatorname{erf}(x) &=\frac{1}{\sqrt{\pi}}\int_{-x}^{x}e^{x^2-t^2}\,dt &&\text{[definition of $\operatorname{erf}(x)$]} \\&=\frac{2x}{\sqrt{\pi}}\int_{0}^{1}e^{4x^2y(1-y)}\,dy &&\text{[substitution $t=x(1-2y)$]} \\&=\frac{2x}{\sqrt{\pi}}\sum_{n=0}^{\infty}\frac{(4x^2)^n}{n!}\int_{0}^{1}y^n(1-y)^n\,dy &&...


DISCLAIMER: I'm not terribly familiar with topology so I don't know whether I've been rigorous enough with this answer. I think it brings the idea across if nothing else. The set $\mathcal{N}$ can be characterised as the set: $$ \mathcal{N}=\left\{\begin{bmatrix}a&b\\c&d\end{bmatrix}:a^2+b^2+c^2+d^2=1\quad\land\quad ad-bc=0\right\} $$ If we rewrite ...


Hint: $$\|f_n-f\|^p_p=\int_{A_n}|f_n-f|^p d \mu+\int_{A_n^c}|f_n-f|^p d \mu=0+\int_{A_n^c}|f|^p d \mu.$$ Now split the last integral into two cases $f>n$ and $0<f<\frac{1}{n}$. Think, one of the integrals is over a set of measure zero (which one and why?)


Let $(X,d)$ be a metric space with $d$ unbounded and consider the metric $d'(x,y)=\min\{d(x,y),1\}$. This induces the same topology, but is not equivalent to $d$, since it is bounded

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