# Tag Info

4

If it were $L^2$ then it would satisfy Cauchy-Schwarz, i.e. you would have $|f(0)| \leq C \| f \|_{L^2}$ for some $C$. Construct a sequence of functions $f_n$ such that $|f_n(0)|>n \| f_n \|_{L^2}$ to contradict this.

3

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 $L^2([-\pi,\pi])$

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 ...

2

First note that for $v_1,v_2\in S^{n-1}$ we have that $$|f(x+v_1t_1+v_2t_2)- f(x)| \\= |f(x+v_1t_1+v_2t_2)-f(x+v_1 t_1)+f(x+v_1 t_1) -f(x)| \\\leq |t_2|L(v_2) + |t_2|L(v_1)$$ Now take an orthonormal basis $(e_i)$ of $\mathbb{R}^n$ and define $$L_i = L(e_i)$$ to get $$|f(x+ \sum_i t_i e_i)-f(x)| \leq \sum_i |t_i| L_i \leq \sqrt{\sum_i ... 1 You have$$ win(c, 0) = \frac{1}{2 \pi i} \int_c \frac{dz}{z} = \frac{1}{2 \pi i} \int_0^1 \frac{c'(t)}{c(t)} \, dt $$and analogously for c_1 and c_2. From c(t) = \frac{c_1(t)}{c_2(t)} one can easily compute a relationship between the "logarithmic derivatives"$$ \frac{c'(t)}{c(t)} \, , \, \frac{c_1'(t)}{c_1(t)} \, , \, \frac{c_2'(t)}{c_2(t)}  ...

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