A C₀-Semigroup of Ulam Unstable Operators

Abstract

The Ulam stability of the composition of two Ulam stable operators has been investigated by several authors. Composition of operators is a key concept when speaking about $C_0$-semigroups. Examples of $C_0$-semigroups formed with Ulam stable operators are known. In this paper, we construct a $C_0$-semigroup ${\left(R_t\right)}_{t\ge 0}$ on $C[0,1]$ such that for each $t>0$, $R_t$ is Ulam unstable. Moreover, we compute the central moments of $R_t$ and establish a Voronovskaja-type formula. This enables to prove that $C^2[0,1]$ is contained in the domain $D\left(A\right)$ of the infinitesimal generator of the semigroup. We raise the problem to fully characterize the domain $D\left(A\right)$.

1. Introduction

Let X be a Banach space and $L:X\to X$ a bounded linear operator. We say that L is Ulam stable if there exists a constant $K>0$ such that for each $\epsilon >0$ and $x\in X$ with $\parallel Lx\parallel \le \epsilon$ there is a $u\in X$ with $Lu=0$ and $\parallel u-x\parallel \le K\epsilon$. If L is injective, then it is Ulam stable if and only if $L^{-1}$ is bounded.

More general definitions and more general results can be found, for example, in Reference [1] [Section 2.3] and the references therein. The Ulam stability of the composition of two Ulam stable operators is investigated in Reference [2,3,4,5,6]. In Reference [2] [Example 2], a $C_0$-semigroup of operators ${\left(T\left(t\right)\right)}_{t\ge 0}$ is presented such that for each $t>0$, $T\left(t\right)$ is Ulam stable. The problem of finding a $C_0$-semigroup with Ulam unstable operators was raised in Reference [2] [Section 5]. In this paper, we construct such a $C_0$-semigroup ${\left(R_t\right)}_{t\ge 0}$ on the Banach space $C[0,1]$ of all real-valued continuous functions defined on $[0,1]$, endowed with the sup-norm. It is related to some operators introduced and investigated in Reference [7]. We prove that for each $t>0$, $R_t$ is injective and $R_t^{-1}$ is unbounded, and thus $R_t$ is Ulam unstable. This is done in Section 2, Section 3 and Section 4. The moments and the central moments of $R_t$ are calculated in Section 5; this enables to prove a Voronovskaja-type formula and consequently to describe the infinitesimal generator of ${\left(R_t\right)}_{t\ge 0}$ on $C^2[0,1]$. In Section 6, using some expansions in terms of Chebyshev polynomials of first kind, we find the images of some particular functions under $R_t$. Section 7 is devoted to other properties of the semigroup. We find ${\displaystyle \underset{t\to \infty }{lim}{R}_{t}f}$, $f\in C[0,1]$. Using this result, we prove that for each $t>0$, $R_t$ has the same unique invariant probability measure. Section 8 contains some conclusions and projects for further work. For the general theory of $C_0$-semigroup see, for example, Reference [8]. We use the notation $e_k\left(t\right)=t^k$, $k\ge 0$, $t\in [0,1]$, and consider an empty product to be equal to 1.

2. The Operators

For $n\ge 1$, $f\in C[0,1]$, $x\in [0,1]$, let us denote

$$M_{n}f\left(x\right):=\sqrt{{\displaystyle \frac{n}{\pi }}}{\int }_{\mathbb{R}}f\left(si{n}^2(s+arcsin\sqrt{x})\right)e^{-n{s}^2}ds.$$

(1)

The operators $M_n:C[0,1]\to C[0,1],n\ge 1$, were introduced and investigated in Reference [7]. Generalizing (1), let us introduce the operators $R_t:C[0,1]\to C[0,1],t>0$,

$$R_{t}f\left(x\right):={\displaystyle \frac{1}{\sqrt{\pi t}}}{\int }_{\mathbb{R}}f\left(si{n}^2(s+arcsin\sqrt{x})\right)e^{-s^2/t}ds.$$

(2)

It is easy to see that $R_t$ is a positive linear operator and $R_t{e}_0=e_0,t>0$.

Considering the Fourier coefficients of the function ${sin}^{2m}x,m\in \mathbb{N}$, we obtain the formula

$${\displaystyle {sin}^{2m}x={\displaystyle \frac{1}{4^m}}\left(\genfrac{}{}{0pt}{}{2m}{m}\right)+{\displaystyle \frac{2}{4^m}}\sum _{j=1}^m{(-1)}^j\left(\genfrac{}{}{0pt}{}{2m}{m-j}\right)cos2jx.}$$

(3)

Set $\theta =arcsin\sqrt{x}$. Using (3), we get

$$\begin{array}{c}\hfill R_t{e}_m\left(x\right)={\displaystyle \frac{1}{4^m}}\left(\genfrac{}{}{0pt}{}{2m}{m}\right)+{\displaystyle \frac{2}{4^m\sqrt{\pi t}}}\sum _{j=1}^m{(-1)}^j\left(\genfrac{}{}{0pt}{}{2m}{m-j}\right)(cos2j\theta ){\int }_{\mathbb{R}}{e}^{-\frac{s^2}{t}}cos2jsds.\end{array}$$

(4)

Combined with

$${\displaystyle {\int }_{\mathbb{R}}{e}^{-\frac{s^2}{t}}cos2jsds=\sqrt{\pi t}{e}^{-t{j}^2},}$$

(5)

the relation (4) leads to

$$R_t{e}_m\left(x\right)={\displaystyle \frac{1}{4^m}}\left(\genfrac{}{}{0pt}{}{2m}{m}\right)+{\displaystyle \frac{2}{4^m}}\sum _{j=1}^m{(-1)}^j\left(\genfrac{}{}{0pt}{}{2m}{m-j}\right)e^{-t{j}^2}cos2j\theta .$$

(6)

From $cos2j\theta +isin2j\theta ={(cos\theta +isin\theta )}^{2j}$ we obtain

$${\displaystyle cos2j\theta =\sum _{r=0}^j\left(\genfrac{}{}{0pt}{}{2j}{2r}\right){(1-{sin}^2\theta )}^{j-r}{\left({sin}^2\theta \right)}^r{(-1)}^r,}$$

which implies

$${\displaystyle cos(2jarcsin\sqrt{x})=\sum _{r=0}^j\left(\genfrac{}{}{0pt}{}{2j}{2r}\right){(-1)}^r{x}^r{(1-x)}^{j-r}.}$$

(7)

From (6) and (7), we infer

$$R_t{e}_m\left(x\right)={\displaystyle \frac{1}{4^m}}\left(\genfrac{}{}{0pt}{}{2m}{m}\right)+{\displaystyle \frac{2}{4^m}}\sum _{j=1}^m{(-1)}^j\left(\genfrac{}{}{0pt}{}{2m}{m-j}\right)e^{-t{j}^2}\sum _{r=0}^j\left(\genfrac{}{}{0pt}{}{2j}{2r}\right){(-1)}^r{x}^r{(1-x)}^{j-r}.$$

(8)

In particular,

$$\begin{array}{cc}\hfill R_t{e}_1\left(x\right)& ={\displaystyle \frac{1}{2}}+\left(x-{\displaystyle \frac{1}{2}}\right)e^{-t},\hfill \end{array}$$

(9)

$$\begin{array}{cc}\hfill R_t{e}_2\left(x\right)& ={\displaystyle \frac{1}{8}}\left(3+4(2x-1)e^{-t}+(8x^2-8x+1)e^{-4t}\right),\hfill \\ \hfill R_t{e}_3\left(x\right)& ={\displaystyle \frac{1}{32}}\left(10+15e^{-t}(2x-1)+6e^{-4t}(8x^2-8x+1)+e^{-9t}(32x^3-48x^2+18x-1)\right),\hfill \\ \hfill R_t{e}_4\left(t\right)& ={\displaystyle \frac{1}{128}}\left(35+56e^{-t}(2x-1)+28e^{-4t}(8x^2-8x+1)+8e^{-9t}\left(32x^3-48x^2+18x-1\right)\right.\hfill \\ \hfill & +e^{-16t}\left(128x^4-256x^3+160x^2-32x+1\right).\hfill \end{array}$$

(10)

Now (9), (10) and $R_t{e}_0=e_0$, together with Korovkin’s theorem, yield

Theorem 1.

For each $f\in C[0,1]$, we have

$${\displaystyle \underset{t\to 0}{lim}{R}_{t}f\left(x\right)=f\left(x\right),}$$

uniformly on $x\in [0,1]$.

3. The Semigroup

Consider the Chebyshev polynomials of first kind on $[0,1]$, namely

$$T_m\left(x\right):=cos(marccos(2x-1)),m\ge 0,x\in [0,1].$$

(11)

Then

$$T_m\left({sin}^{2}y\right)={(-1)}^{m}cos2my,$$

(12)

which implies

$$T_m\left(x\right)={(-1)}^{m}cos(2marcsin\sqrt{x}),$$

(13)

and moreover,

$$R_t{T}_m\left(x\right)={\displaystyle \frac{{(-1)}^m}{\sqrt{\pi t}}}cos(2marcsin\sqrt{x}){\int }_{\mathbb{R}}{e}^{-\frac{s^2}{t}}cos2msds.$$

Using (13) and (5) we get

$$R_t{T}_m\left(x\right)=e^{-t{m}^2}{T}_m\left(x\right),m\ge 0.$$

(14)

Let $t>0,s>0$. Then (14) shows that

$$R_t{R}_s{T}_m=R_t\left(e^{-s{m}^2}{T}_m\right)=e^{-(s+t)m^2}{T}_m=R_{t+s}{T}_m,$$

for all $m\ge 0$.

This entails $R_{t+s}{T}_m=R_t{R}_s{T}_m$, $m\ge 0$, and moreover $R_{t+s}p=R_t{R}_{s}p$, for each polynomial p. Since the space of polynomial functions is dense in $C[0,1]$ and $\parallel R_t\parallel =1,t>0$, we conclude that $R_{t+s}=R_t{R}_s$ for all $t>0$ and $s>0$.

Define $R_0$ to be the identity operator on $C[0,1]$. Thus ${\left(R_t\right)}_{t\ge 0}$ is a semigroup of operators on $C[0,1]$. Using Theorem 1, we get:

Theorem 2.

${\left(R_t\right)}_{t\ge 0}$ is a $C_0$-semigroup of operators on $C[0,1]$.

4. ${\mathbf{R}}_{\mathbf{t}}$ Is Ulam-Unstable

Lemma 1.

For each $t>0$, $R_t$ is injective.

Proof. 

Let $t>0$ and $f\in C[0,1]$ such that $R_{t}f=0$, that is,

$${\displaystyle {\int }_{-\infty }^{+\infty }f\left({sin}^2(s+arcsin\sqrt{x})\right)e^{-s^2/t}ds=0,x\in [0,1].}$$

(15)

Set $g:\mathbb{R}\to \mathbb{R}$, $g=f\circ {sin}^2$. Then g is even and $\pi$-periodic. Define

$${\displaystyle K\left(a\right):={\int }_{-\infty }^{+\infty }g(a+s)e^{-s^2/t}ds,a\in \mathbb{R}.}$$

(16)

Replacing s by $-s$, we get

$${\displaystyle K\left(a\right)={\int }_{-\infty }^{+\infty }g(a-s)e^{-s^2/t}ds,a\in \mathbb{R}.}$$

(17)

As a consequence of (15), we have

$$K\left(a\right)=0,a\in \left[0,\frac{\pi }{2}\right].$$

(18)

From (16), it follows that

$$\begin{array}{cc}\hfill K(\pi -a)& {\displaystyle ={\int }_{-\infty }^{+\infty }g(\pi -a+s)e^{-s^2/t}ds={\int }_{-\infty }^{+\infty }g(s-a)e^{-s^2/t}ds}\hfill \\ \hfill & {\displaystyle ={\int }_{-\infty }^{+\infty }g(a-s)e^{-s^2/t}ds.}\hfill \end{array}$$

Now, (17) and (18) show that $K(\pi -a)=K\left(a\right)=0,a\in \left[0,\frac{\pi }{2}\right]$. It follows immediately that $K\left(a\right)=0,a\in [0,\pi ]$. Since g is $\pi$-periodic, K is also $\pi$-periodic, and therefore $K\left(a\right)=0,a\in \mathbb{R}$. On the other hand, K is the convolution of g and the function $s\to e^{-s^2/t},s\in \mathbb{R}$. Titchmarsch Theorem implies that $g=0$ on $\mathbb{R}$, and so $f=0$ on $\mathbb{R}$. □

Theorem 3.

For each $t>0$, $R_t$ is Ulam unstable.

Proof. 

From Lemma 1 and (14), we deduce that $R_t^{-1}{T}_m=e^{t{m}^2}{T}_m$, $m=0,1,\cdots$. It follows that $R_t^{-1}$ is unbounded, and so $R_t$ is Ulam unstable. □

5. The Moments of the Operators $R_t$

Remember that $R_t:C[0,1]\to C[0,1]$, $t>0$,

$$R_{t}f\left(x\right)={\displaystyle \frac{1}{\sqrt{\pi t}}}{\int }_{\mathbb{R}}f\left(si{n}^2(s+arcsin\sqrt{x})\right)e^{-s^2/t}ds.$$

Let $M_{t,m}\left(x\right)$ be the central moment of order $m\ge 0$ of the operator $R_t$, that is,

$$\begin{array}{cc}\hfill M_{t,m}\left(x\right)& :=R_t{(e_1-x{e}_0)}^m\left(x\right)={\displaystyle \frac{1}{\sqrt{\pi t}}}{\int }_{\mathbb{R}}{\left(si{n}^2(s+arcsin\sqrt{x})-x\right)}^m{e}^{\frac{-s^2}{t}}ds\hfill \\ \hfill & {\displaystyle ={\displaystyle \frac{1}{\sqrt{\pi t}}}{\int }_{\mathbb{R}}{\left((1-2x)sins+2\sqrt{x(1-x)}coss\right)}^m\left(si{n}^{m}s\right)e^{-\frac{s^2}{t}}ds}\hfill \\ \hfill & {\displaystyle ={\displaystyle \frac{1}{\sqrt{\pi t}}}\sum _{k=0}^m\left(\genfrac{}{}{0pt}{}{m}{k}\right){(1-2x)}^k{\left(2\sqrt{x(1-x)}\right)}^{m-k}{\int }_{\mathbb{R}}{sin}^{m+k}s{cos}^{m-k}s{e}^{-\frac{s^2}{t}}ds.}\hfill \end{array}$$

We need also the formula

$${\displaystyle {\displaystyle \frac{1}{\sqrt{\pi t}}}{\int }_{\mathbb{R}}{s}^{2p}{e}^{-\frac{s^2}{t}}ds=(2p-1)!!{\left({\displaystyle \frac{t}{2}}\right)}^p,t>0,p\ge 0.}$$

(19)

For $m=2j$, we get

$${\displaystyle M_{t,2j}\left(x\right)=\sum _{k=0}^{2j}\left(\genfrac{}{}{0pt}{}{2j}{k}\right){(1-2x)}^k{\left(2\sqrt{x(1-x)}\right)}^{2j-k}{\displaystyle \frac{1}{\sqrt{\pi t}}}{\int }_{\mathbb{R}}{sin}^{2j+k}s{cos}^{2j-k}s{e}^{-\frac{s^2}{t}}ds.}$$

(20)

Using the McLaurin series for ${sin}^{2j+k}s{cos}^{2j-k}s$, we have

$${\displaystyle {\int }_{\mathbb{R}}{sin}^{2j+k}s{cos}^{2j-k}s{e}^{-\frac{s^2}{t}}ds={\int }_{\mathbb{R}}\left(s^{2j+k}+\mathrm{terms}\mathrm{with}\mathrm{higher}\mathrm{powers}\mathrm{of}s\right)e^{-\frac{s^2}{t}}ds.}$$

(21)

From (19)–(21), it follows that

$$M_{t,2j}\left(x\right)={\left(2\sqrt{x(1-x)}\right)}^{2j}(2j-1)!!{\left({\displaystyle \frac{t}{2}}\right)}^j+\mathrm{terms}\mathrm{with}\mathrm{higher}\mathrm{powers}\mathrm{of}t.$$

This yields

$${\displaystyle \underset{t\to 0}{lim}{t}^{-j}{M}_{t,2j}\left(x\right)=2^j(2j-1)!!{\left(x(1-x)\right)}^j.}$$

(22)

Similarly, for $m=2j-1$ we have

$$\begin{array}{cc}\hfill M_{t,2j-1}\left(x\right)& {\displaystyle =\sum _{k=1}^{2j-1}\left(\genfrac{}{}{0pt}{}{2j-1}{k}\right){(1-2x)}^k{\left(2\sqrt{x(1-x)}\right)}^{2j-1-k}}\hfill \\ \hfill & {\displaystyle \times {\displaystyle \frac{1}{\sqrt{\pi t}}}{\int }_{\mathbb{R}}{sin}^{2j-1+k}s{cos}^{2j-1-k}s{e}^{-\frac{s^2}{t}}ds}\hfill \\ \hfill & =(2j-1)(1-2x){\left(2\sqrt{x(1-x)}\right)}^{2j-2}{\left({\displaystyle \frac{t}{2}}\right)}^j(2j-1)!!+\mathrm{terms}\mathrm{with}\mathrm{higher}\mathrm{powers}\mathrm{of}t.\hfill \end{array}$$

Consequently,

$${\displaystyle \underset{t\to 0}{lim}{t}^{-j}{M}_{t,2j-1}\left(x\right)=(2j-1)(2j-1)!!2^{j-2}(1-2x){\left(x(1-x)\right)}^{j-1}.}$$

(23)

By a direct calculation,

$$M_{t,0}\left(x\right)=1,$$

(24)

$$M_{t,1}\left(x\right)={\displaystyle \frac{1}{2}}(1-e^{-t})(1-2x),$$

(25)

$$M_{t,2}\left(x\right)=(2e^{-t}-e^{-4t}-1)x(1-x)+{\displaystyle \frac{1}{8}}(3-4e^{-t}+e^{-4t}).$$

(26)

From (25) and (26), or from (22) and (23) with $j=1$, we derive

$${\displaystyle \underset{t\to 0}{lim}{\displaystyle \frac{M_{t,1}\left(x\right)}{t}}={\displaystyle \frac{1}{2}}(1-2x),}$$

(27)

$${\displaystyle \underset{t\to 0}{lim}{\displaystyle \frac{M_{t,2}\left(x\right)}{t}}=2x(1-x).}$$

(28)

Using Sikkema’s classical result [9], combined with (22)–(24), (27), (28), we get:

Theorem 4.

For each $f\in C^2[0,1]$, the following Voronovskaja type formula is valid

$${\displaystyle \underset{t\to 0}{lim}{\displaystyle \frac{R_{t}f\left(x\right)-f\left(x\right)}{t}}=x(1-x)f^{″}\left(x\right)+\left({\displaystyle \frac{1}{2}}-x\right)f^{\prime }\left(x\right),}$$

(29)

uniformly on $x\in [0,1]$.

Let $(A,D(A\left)\right)$ be the infinitesimal generator of the semigroup ${\left(R_t\right)}_{t\ge 0}$. From Theorem 4, we infer:

Corollary 1.

$C^2[0,1]\subset D\left(A\right)$, and

$$Af\left(x\right)=x(1-x)f^{″}\left(x\right)+\left({\displaystyle \frac{1}{2}}-x\right)f^{\prime }\left(x\right),f\in C^2[0,1],x\in [0,1].$$

6. Applications

According to (11),

$$cos(2karcsin\sqrt{x})={(-1)}^k{T}_k\left(x\right).$$

(30)

Let $f\in C[0,1]$. Then $f\circ {sin}^2$ is even and $\pi$-periodic on $\mathbb{R}$.

Theorem 5.

Suppose that $f\circ {sin}^2$ has the Fourier expansion

$${\displaystyle f\left({sin}^2\alpha \right)=\sum _{k=0}^{\infty }{a}_{2k}cos2k\alpha ,\alpha \in \mathbb{R}.}$$

Then ${\displaystyle R_{t}f\left(x\right)=\sum _{k=0}^{\infty }{(-1)}^k{a}_{2k}{e}^{-t{k}^2}{T}_k\left(x\right),x\in [0,1].}$

Proof. 

Setting $x={sin}^2\alpha$, we get

$${\displaystyle f\left(x\right)=\sum _{k=0}^{\infty }{a}_{2k}cos(2karcsin\sqrt{x}),x\in [0,1].}$$

Now, (30) implies

$${\displaystyle f\left(x\right)=\sum _{k=0}^{\infty }{(-1)}^k{a}_{2k}{T}_k\left(x\right),x\in [0,1].}$$

(31)

Remember also that (see (14)),

$$R_t{T}_k=e^{-t{k}^2}{T}_k,t\ge 0,k\ge 0.$$

(32)

From (31) and (32), we infer

$${\displaystyle R_{t}f\left(x\right)=\sum _{k=0}^{\infty }{(-1)}^k{a}_{2k}{e}^{-t{k}^2}{T}_k\left(x\right),x\in [0,1],}$$

(33)

and this concludes the proof. □

Example 1.

Let $f\left(x\right)=e_{1/2}\left(x\right)=\sqrt{x}$. Then we can use the Fourier expansion

$${\displaystyle f\left({sin}^{2}x\right)=|sinx|={\displaystyle \frac{2}{\pi }}-{\displaystyle \frac{4}{\pi }}\sum _{k=1}^{\infty }{\displaystyle \frac{1}{4k^2-1}}cos2kx.}$$

We get from (33)

$${\displaystyle R_t{e}_{1/2}\left(x\right)={\displaystyle \frac{2}{\pi }}-{\displaystyle \frac{4}{\pi }}\sum _{k=1}^{\infty }{\displaystyle \frac{{(-1)}^k}{4k^2-1}}{e}^{-t{k}^2}{T}_k\left(x\right).}$$

(34)

For $t=0$, (34) reduces to

$${\displaystyle \sqrt{x}={\displaystyle \frac{2}{\pi }}-{\displaystyle \frac{4}{\pi }}\sum _{k=1}^{\infty }{\displaystyle \frac{{(-1)}^k}{4k^2-1}}{T}_k\left(x\right),x\in [0,1].}$$

Example 2.

Let $f\left(x\right)=arcsin{e}_{1/2}\left(x\right)$. Then, for $x\in [0,1]$,

$$f\left({sin}^{2}x\right)=arcsin|sinx|={\displaystyle \frac{\pi }{4}}-{\displaystyle \frac{2}{\pi }}\sum _{k=1}^{\infty }{\displaystyle \frac{1}{{(2k-1)}^2}}cos2(2k-1)x.$$

Now, (33) yields

$${\displaystyle R_{t}arcsin{e}_{1/2}\left(x\right)={\displaystyle \frac{\pi }{4}}+{\displaystyle \frac{2}{\pi }}\sum _{k=1}^{\infty }{\displaystyle \frac{1}{{(2k-1)}^2}}{e}^{-t{(2k-1)}^2}{T}_{2k-1}\left(x\right),}$$

(35)

and for $t=0$, (35) reduces to

$$arcsin\sqrt{x}={\displaystyle \frac{\pi }{4}}+{\displaystyle \frac{2}{\pi }}\sum _{k=1}^{\infty }{\displaystyle \frac{1}{{(2k-1)}^2}}{T}_{2k-1}\left(x\right),x\in [0,1].$$

Example 3.

Let $f\left(x\right)=arccos(2x-1),x\in [0,1]$. Using the Fourier expansion

$${\displaystyle \theta ={\displaystyle \frac{\pi }{2}}-{\displaystyle \frac{4}{\pi }}\sum _{n=0}^{\infty }{\displaystyle \frac{cos(2n+1)\theta }{{(2n+1)}^2}},\theta \in [0,\pi ],}$$

with $\theta =arccos(2x-1)$, we get

$${\displaystyle f\left(x\right)={\displaystyle \frac{\pi }{2}}-{\displaystyle \frac{4}{\pi }}\sum _{n=0}^{\infty }{\displaystyle \frac{T_{2n+1}\left(x\right)}{{(2n+1)}^2}}.}$$

Together with (32), this shows that

$$R_{t}f\left(x\right)={\displaystyle \frac{\pi }{2}}-{\displaystyle \frac{4}{\pi }}\sum _{n=0}^{\infty }{e}^{-t{(2n+1)}^2}{\displaystyle \frac{T_{2n+1}\left(x\right)}{{(2n+1)}^2}}.$$

Example 4.

Let $f\left(x\right)=arccos(2x-1)$, $g\left(x\right)=arcsin(2x-1),x\in [0,1]$. Then $f\left(x\right)+g\left(x\right)={\displaystyle \frac{\pi }{2}}$, and using the result from Example 3, we obtain

$$R_{t}g\left(x\right)={\displaystyle \frac{4}{\pi }}\sum _{n=0}^{\infty }{e}^{-t{(2n+1)}^2}{\displaystyle \frac{T_{2n+1}\left(x\right)}{{(2n+1)}^2}}.$$

Example 5.

Let $f\left(x\right)=\sqrt{x(1-x)},x\in [0,1]$. Then

$$f\left(x\right)={\displaystyle \frac{1}{\pi }}-{\displaystyle \frac{2}{\pi }}\sum _{n=1}^{\infty }{\displaystyle \frac{T_{2n}\left(x\right)}{4n^2-1}}.$$

Therefore,

$$R_{t}f\left(x\right)={\displaystyle \frac{1}{\pi }}-{\displaystyle \frac{2}{\pi }}\sum _{n=1}^{\infty }{e}^{-4t{n}^2}{\displaystyle \frac{T_{2n}\left(x\right)}{4n^2-1}}.$$

Example 6.

For $f\left(x\right)=arctan(2x-1),x\in [0,1]$, we get

$$f\left(x\right)=2\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n{(\sqrt{2}-1)}^{2n+1}}{2n+1}}{T}_{2n+1},x\in [0,1].$$

Then

$$R_{t}f\left(x\right)=2\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^n{(\sqrt{2}-1)}^{2n+1}}{2n+1}}{e}^{-t{(2n+1)}^2}{T}_{2n+1}\left(x\right).$$

7. Properties of the Semigroup ${\left(R_t\right)}_{t\ge 0}$

Theorem 6.

For each $f\in C[0,1]$,

$${\displaystyle \underset{t\to \infty }{lim}{R}_{t}f={\displaystyle \frac{2}{\pi }}\left({\int }_0^{1}f\left(x\right)darcsin\sqrt{x}\right)e_0.}$$

(36)

Proof. 

It suffices to prove (36) for $f=T_m,m\ge 0$. For $m=0$, it reduces to $e_0=e_0$. For $m\ge 1$,

$${\displaystyle \underset{t\to \infty }{lim}{R}_t{T}_m=\underset{t\to \infty }{lim}{e}^{-t{m}^2}{T}_m=0,}$$

and

$${\displaystyle {\displaystyle \frac{2}{\pi }}{\int }_0^1{T}_m\left(x\right)darcsin\sqrt{x}={\displaystyle \frac{1}{\pi }}{\int }_0^1{\displaystyle \frac{T_m\left(x\right)T_0\left(x\right)}{\sqrt{x(1-x)}}}dx=0.}$$

□

Theorem 7.

The probability measure μ on $[0,1]$ given by

$$d\mu \left(x\right)={\displaystyle \frac{1}{\pi }}{\displaystyle \frac{1}{\sqrt{x(1-x)}}}dx$$

(37)

is invariant with respect to all $R_t$, $t>0$. It is the unique probability measure with this property.

Remark 1.

Theorem 7 asserts that for each $t>0$ the dynamical system $(C[0,1],R_t)$ is uniquely ergodic, that is, it has a unique invariant probability measure μ, described by (37).

Proof. 

We have to prove that

$${\displaystyle {\int }_0^1{R}_{t}f\left(x\right)d\mu \left(x\right)={\int }_0^{1}f\left(x\right)d\mu \left(x\right),f\in C[0,1].}$$

(38)

It suffices to prove (38) for $f=T_m,m\ge 0$. For $m=0$, (38) follows from $R_t{T}_0=T_0$. For $m\ge 1$, we have

$${\displaystyle {\int }_0^1{R}_t{T}_m\left(x\right)d\mu \left(x\right)={\displaystyle \frac{e^{-m^{2}t}}{\pi }}{\int }_0^1{\displaystyle \frac{T_m\left(x\right)T_0\left(x\right)}{\sqrt{x(1-x)}}}dx=0={\int }_0^1{T}_m\left(x\right)d\mu \left(x\right).}$$

Thus (38) is proved. If $\nu$ is also a probability measure invariant with respect to all $R_t,t>0$, then using Theorem 6 we can write

$$\begin{array}{cc}\hfill {\displaystyle {\int }_0^{1}f\left(x\right)d\nu \left(x\right)}& {\displaystyle =\underset{t\to \infty }{lim}{\int }_0^1{R}_{t}f\left(x\right)d\nu \left(x\right)={\int }_0^1\left(\underset{t\to \infty }{lim}{R}_{t}f\left(x\right)\right)d\nu \left(x\right)}\hfill \\ \hfill & {\displaystyle ={\int }_0^1{\displaystyle \frac{2}{\pi }}\left({\int }_0^{1}f\left(s\right)darcsin\sqrt{s}\right)d\nu \left(x\right)={\displaystyle \frac{2}{\pi }}{\int }_0^{1}f\left(s\right)darcsin\sqrt{s}}\hfill \\ \hfill & {\displaystyle ={\int }_0^{1}f\left(x\right)d\mu \left(x\right).}\hfill \end{array}$$

Thus $\nu =\mu$, and the proof is complete. □

8. Conclusions and Further Work

Composition of operators is a core concept when speaking about $C_0$-semigroups. The Ulam stability of the composition of two Ulam stable operators was studied in several papers. Some problems were raised in Reference [2]. In this paper, we solve one of these problems. More precisely, we construct a $C_0$-semigroup ${\left(R_t\right)}_{t\ge 0}$ on $C[0,1]$ such that for each $t>0$, $R_t$ is Ulam unstable. Moreover, we investigate the properties of this $C_0$-semigroup. In particular, we consider its infinitesimal generator $(A,D(A\left)\right)$ and prove that $C^2[0,1]$ is included in the domain $D\left(A\right)$. This means that if $f\in C^2[0,1]$, then the Cauchy problem

$$\left\{\begin{array}{c}{\displaystyle \frac{\partial u(t,x)}{\partial t}}=Au(t,x),t\ge 0,\hfill \\ u(0,x)=f\left(x\right),x\in [0,1],\hfill \end{array}\right.$$

has a unique solution, namely $u(t,x)=R_{t}f\left(x\right)$. Of course, this is true for an arbitrary $f\in D\left(A\right)$ (see, e.g., Reference [8]) and from this point of view, it would be useful to have an explicit description of the domain $D\left(A\right)$. To this end we intend to consider the $C_0$-semigroup ${\left(R_t\right)}_{t\ge 0}$ in the more general framework presented in Reference [7].