How to show that $T(t), t\geq 0$ is strongly continuous? Is $\{T(t):t\geq 0\}$ a contraction semi-group?

Question

Let $\mathscr{X}=L^{2}(-\pi,\pi)$ and define operator: $$(T(t)f)(\theta)=\frac{1}{2\pi}\int_{-\pi}^{\pi}G(\theta-\xi,t)f(\xi)d\xi, t>0,$$ Where $G(\theta,t)=1+2\sum_{n=1}^{\infty}e^{-n^2t}\cos n\theta$, and $T(0)f=f$. How to show that $T(t), t\geq 0$ is strongly continuous? Is $\{T(t):t\geq 0\}$ a contraction semi-group?

I have got $T(0)=I, T(t+s)=T(s)T(t)$, but how to prove:

(1)$\Vert T(t)f-f\Vert \rightarrow 0,$ which implies $T(t)$ strongly continuous.

(2)$\vert T(t) \Vert \leq 1, $which implies $T(t)$ contraction.

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I think these properties of $G(\theta, t)$ are useful:

• $G(\theta, t)$ is continuous
• $G(\theta, t)\geq 0, \forall t\geq 0$

In fact, $T(t)f$ gives the solution $u(\cdot, t)$ to the heat equation on a circle.

Answer 1

I misread your question earlier. Everything appears to correct. $$ G(\theta-\xi,t)=1+2\sum_{n=1}^{\infty}e^{-n^2t}(\cos(\theta)\cos(\xi)+\sin(\theta)\sin(\xi)) $$ From this, $$ (T(t)f)(\theta)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(\xi)d\xi+\sum_{n=1}^{\infty}\left[\frac{1}{\pi}\int_{-\pi}^{\pi}f(\xi)\cos\xi d\xi\cos\theta+\frac{1}{\pi}\int_{-\pi}^{\pi}f(\xi)\sin\xi d\xi\sin\theta\right]e^{-n^2 t} $$ As an identity in $L^2$, $T(0)f=f$ because $T(0)f$ is the Fourier series for $f$. If you let $c_0=\frac{1}{\sqrt{2\pi}}$, $c_n=\frac{1}{\sqrt{\pi}}\cos(n\theta)$ and $s_n=\frac{1}{\sqrt{\pi}}\sin(n\theta)$, then $\{c_0,c_1,s_1,c_2,s_2,\cdots\}$ is the standard Fourier orthonormal basis for $L^2[-\pi,\pi]$, and $$ T(t)f =\langle f,c_0\rangle c_0 + \sum_{n=1}^{\infty}e^{-n^2t}(\langle f,c_n\rangle c_n + \langle f,s_n\rangle s_n). $$ By Parseval's equality, \begin{align} \|T(t)f\|^2&=|\langle f,c_0\rangle|^2+\sum_{n=1}^{\infty}e^{-2n^2t}(|\langle f,c_n\rangle|^2+|\langle f,s_n\rangle|^2) \\ &\le |\langle f,c_0\rangle|^2+\sum_{n=1}^{\infty}(|\langle f,c_n\rangle|^2+|\langle f,s_n\rangle|^2) \\ &= \|f\|^2. \end{align} Strong continuity of $f$ follows from the crude estimate: $$ \|T(t)f-T(s)f\|^2=\sum_{n=1}^{\infty}(e^{-n^2t}-e^{-n^2s})^2(|\langle f,c_n\rangle|^2+|\langle f,s_n\rangle|^2) \\ \le (e^{-t}-e^{-s})^2\|f\|^2. $$