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2

If it's continuous, it's bounded (image of the compact set $[-\pi,\pi]$ is a compact hence bounded set), so $0\le |f(x)|^2\le M$. Then by Lebesgue's DCT or Fatou's lemma or whatever you want, $$\int_{-\pi}^\pi |f(x)|^2\,dx\le \int_{-\pi}^\pi M\,dx = 2\pi M < \infty$$ which is exactly what it means to be in $L^2([-\pi,\pi])$

0

The expression you want to keep small is $$\frac{1}{a+x}-\frac{1}{a+b} =\frac{b-x}{(a+b)(a+x)}$$ and you can assume $x>0$ by taking $|x-b|<b$. Then $a+x>a$ and so $1/(a+x)<1/a$. Therefore $$\left|\frac{1}{a+x}-\frac{1}{a+b}\right|= \left|\frac{b-x}{(a+b)(a+x)}\right|< |b-x|\frac{1}{a(a+b)}$$ So, if $$|b-x|<\varepsilon a(a+b)$$ you ...

2

The given ODE is a linear homogeneous equation of second order. This implies that the set of solutions is a two-dimensional complex vector space. If the equation $a\lambda^2+b\lambda +c=0$ has two different solutions $\lambda_1$, $\lambda_2\in{\mathbb C}$ then the general solution is given by $$y(t)=C_1e^{\lambda_1 t}+C_2e^{\lambda_2 t}\ .$$ If the ...

1

For a very trivial example, let $X$ be any set, and let $d$ be the discrete metric on $X$; $d(x,y)=0$ if $x=y$, and $d(x,y)=1$ otherwise. Clearly this $d$ is an ultrametric, and it generates the discrete topology. For a much more interesting example, let $D=\{0,1\}$ with the discrete topology, and let $X=D^{\Bbb N}$, the Cartesian product of countably ...

-1

Use Laplace transform: $$ay''(t)+by'(t)+cy(t)=0\Longleftrightarrow$$ $$\mathcal{L}_t\left[ay''(t)+by'(t)+cy(t)\right]_{(s)}=\mathcal{L}_t\left[0\right]_{(s)}\Longleftrightarrow$$ $$a\left[s^2y(s)-sy(0)-y'(0)\right]+b\left[sy(s)-y(0)\right]+cy(s)=0\Longleftrightarrow$$ $$as^2y(s)-asy(0)-ay'(0)+bsy(s)-by(0)+cy(s)=0\Longleftrightarrow$$ ...

1


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