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2

For (1), depending on whether you are a geometer or an analyst you may have a different sign on your Laplacian. However, if you locally write $$ \Delta f = \sum \frac{\partial^{2} f}{\partial x_{j}^{2}} $$ then this is negative definite. This being the classic example of the sort of operator $L$ you are interested in, the author might be assuming $L$ to ...


2

Your operator is $$Tf(x) = \int^\infty_{-\infty}1_{y\leq x}e^{-(x-y)}f(y)\,dy $$ which by change of variables $z=x-y$ becomes $$Tf(x) = \int^\infty_{-\infty}1_{z\geq 0}e^{-z} f(x-z)\,dz. $$ This is just the convolution operator $$f\mapsto \phi*f $$ with $$\phi(z) = 1_{z\geq 0}e^{-z}.$$ Now one of the fundamental inequalities (not hard to prove) about ...


2

$Tf=\phi*f$, where $\phi(t)=e^{-t}\chi_{(0,\infty)}(t)$. (I had $\phi$ backwards in the first version; noticed that guest's $\phi$ was different, then noticed his was right.) So $$\widehat{Tf}=\hat\phi\hat f.$$You can easily calculate $\hat\phi$; now the norm of $T$ is $||\hat\phi||_\infty$ and the spectrum of $T$ is the essential range of $\hat\phi$. (In ...


2

You proved that $T(t)x$ is a solution. So, it remains to show that the problem has only one solution. Let $u$ and $v$ be two solutions. Notice that the function $w=u-v$ satisfies $$\left\{\begin{align} &\frac{dw}{dt}=Aw, & 0\le t \le t_{e} \\ &w(0)=0. \end{align}\right.$$ Pick $s\in[0,t_e]$ and define $\phi:[0,s]\to X$ be setting ...


2

Notice that, the way you have defined $T=MS_R$, we have \begin{equation}T^1(a_1,a_2,a_3,\cdots)=(0,c_2a_1,c_3a_2,c_4a_3,\cdots),\end{equation} \begin{equation}T^2(a_1,a_2,a_3,\cdots)=(0,0,c_3c_2a_1,c_4c_3a_2,c_5c_4a_3,\cdots),\end{equation} and so on. In particular, if $(e_i)_{i=1}^\infty$ is the canonical basis for $\ell_p$, $p=2$, then we have ...


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As the comment above indicates, there is no such inner product. One can deduce that this is the case by noting that the parallelogram identity fails to hold.


1

For the other direction, if $v \neq 0$, then $||Tv|| = ||v|| \cdot||T( \frac{v}{||v||})|| \le ||v|| \cdot \sup_{||v|| = 1} ||Tv||$. So, $||T|| \le \sup_{||v|| = 1} ||Tv||$


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The polynomial $p(\lambda)=\lambda(\lambda^{k-1}-1)$ has distinct roots $$ 0,\alpha,\alpha^{2},\cdots,\alpha^{k-1},\;\;\;\; \alpha=e^{2\pi i/(k-1)}. $$ Therefore, $X$ decomposes into the direct (not necessarily orthogonal) sum $$ X = M_0\oplus M_1\oplus M_2\oplus \cdots \oplus M_{k-1},\\ ...


1

If $M$ is a vector space, then we refer $A$ as an endomorphism. There are differences when a function takes an element from one space to an element in the same space. Take for example: $f:C[0,1]_{\|\cdot\|_1}\to C[0,1]_{\|\cdot\|_2}$, where $\|\cdot\|_1 \to$ integral norm $\|\cdot\|_2 \to$ maximum norm Note that $C[0,1]$ is complete under maximum ...



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