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## Return to Answer

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The existence of such functions being established by now, there remains a curiosity for the concrete situations that may generate them. Here is one, which is also interesting as an example of a $C^\infty$, non-analytic solution, of a simple first order linear DDE.

Let's consider the problem

$$f'(x)=\lambda f(3x),\qquad x < \frac{1}{2} \$$ where we seek for a function $f\in C^1(\mathbb{R})$ with support in $I:=[0,1]$, and with the condition $f(x)=f(1-x)$.

I will show that it has a simple eigenvalue $\lambda= \frac{9}{2}$, corresponding to an eigenfunction with $f(\frac{1}{2})=1$ (see the details below). Then, it is very easy to check that $f$ is locally polynomial on the complement of the Cantor set; and it is, of course, $C^\infty$.

Indeed, since $f (3x)=0$ for $x > \frac{1}{3}$, from the equation we have that $f'(x)$ vanishes identically on the interval $(\frac{1}{3},\frac{2}{3})$, so that $f$ is the constant $1$ therein. By induction, it is easy to see that $f$ is polynomial of degree $k-1$ on each component of the complement of the closed set $F_k$ in $I$, inductively defined by $F_0:=I$ and $F_{k+1}:= \frac{1}{3}F_k \cup\left(\frac{1}{3}F_k + \frac{2}{3}\right)$. Note that $\{F_k\} _ {k\in\mathbb{N}}$ is the well-known nested sequence of closed sets whose intersection is the Cantor set.

$$*$$ Construction. Consider the linear operator $T$ on $L^\infty(0,1)$ (thought as subspace of $L^1(\mathbb{R})$ via extension by zero), defined by

$$Tu(x)= \cases{ \int_0^{3x}u(t)\ dt\qquad \qquad 0 < x < \frac{1}{3}\\ 0 \qquad \qquad \qquad \qquad \frac{1}{3} < x < \frac{2}{3}\\ \int_0^{3(1-x)}u(t)\ dt \qquad\frac{2}{3}< x < 1.}$$ Since $T$ and $T ^ 2$ are integral operators with non-negative kernels, their norms are attained on the constant function $1$ on $[0,1]$. We have, by simple computations $T1(x)=3\min(x,1-x)\chi _ { F _ 1 }(x)$, $\|T\| = \|T1\| _ \infty=1$ and $\|T^2\|=\|T(T1)\|_\infty=\frac{1}{3}$. Therefore $\rho(T)\le\frac{1}{\sqrt 3} < \frac{2}{3}$ and $I-\frac{3}{2}T$ is invertible. The function

$$f:= \left(I-\frac{3}{2}T\right) ^ {-1}\chi _ {[\frac{1}{3},\frac{2}{3}]} = \sum _ {k=0} ^ \infty \left( \frac{3}{2}\right) ^ k T ^ k \chi _ { [\frac{1}{3}, \frac{2}{3}] }$$

verifies

$$f(x)=\frac{3}{2} \int_0^{3x}f(t)\ dt\qquad \mathrm{for\ all}\quad 0 < x < \frac{1}{3}\ ,$$
$$f(x)=1\qquad \mathrm{for\ all}\quad \frac{1}{3} < x < \frac{2}{3}\ ,$$

and $$f(x)=f(1-x)\qquad \mathrm{for\ all}\quad x\in\mathbb{R} \ .$$ The only points in $\mathbb{R}$ where the continuity of $f$ is not immediate and has to be checked, are $\frac{1}{3}$ and $\frac{2}{3}$. We have

$$\int_ 0 ^ 1 f(x)dx=\int _ 0 ^ \frac{1}{3} f(x)dx+\int _ \frac{1}{3} ^ \frac{2}{3} f(x)dx+\int _ \frac{2}{3} ^ 1 f(x)dx= 2\int _ 0 ^ \frac{1}{3} f(x)dx+ \frac{1}{3}\ ,$$

and

$$2 \int _ 0 ^ \frac{1}{3} f(x)dx = 3\int _ 0 ^ \frac{1}{3} \int _ 0 ^ {3x} f(t) dt dx= \int _ 0 ^ 1 \int _ 0 ^ x f(t) dt dx=$$

$$= \int _ 0 ^ 1 (1-t) f(t) dt = \int _ 0 ^ 1 tf(1-t)dt= \int _ 0 ^ 1 tf(t)dt =$$

$$=\frac{1}{2}\left(\int _ 0 ^ 1 (1-t)f(t)+ tf(t)dt\right) =\frac{1}{2} \int _ 0 ^ 1 f(t)dt\ .$$

So

$$\int _ 0 ^ 1f(t)dt = \frac{2}{3}\ ,$$ which implies the continuity of $f$ at $\frac{1}{3}$ (and by symmetry at $\frac{2}{3}$ ), since
$$f(\frac{1}{3}-)=\frac{3}{2}\int_ 0 ^ 1 f(x)dx =1= f(\frac{1}{3}+) \ .$$ As a consequence, $f$ actually satisfies $$f(x)=\frac{3}{2} \int_0^{3x}f(t)\ dt$$ for all $x < \frac{2}{3}\ ,$ and we conclude that $f$ is in $C ^ 1(\mathbb{R})$, and has the stated properties.

Here is a Pudding Function $f$, and its integral function, $F(x):=\int_0^x f(t)dt$. Note how the latter coincides with the first third of the former, up to rescaling. Simple considerations based on odd and even symmetry show that $F(1)=\frac{2}{3}$.

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The existence of such functions being established by now, there remains a curiosity for the concrete situations that may generate them. Here is one, which is also interesting as an example of a $C^\infty$, non-analytic solution, of a simple first order linear DDE.

Let's consider the problem

$$f'(x)=\lambda f(3x),\qquad x < \frac{1}{2} \$$ where we seek for a function $f\in C^1(\mathbb{R})$ with support in $I:=[0,1]$, and with the condition $f(x)=f(1-x)$.

I will show that it has a simple eigenvalue $\lambda= \frac{9}{2}$, corresponding to an eigenfunction with $f(\frac{1}{2})=1$ (see the details below). Then, it is very easy to check that $f$ is locally polynomial on the complement of the Cantor set; and it is, of course, $C^\infty$.

Indeed, since $f (3x)=0$ for $x > \frac{1}{3}$, from the equation we have that $f'(x)$ vanishes identically on the interval $(\frac{1}{3},\frac{2}{3})$, so that $f$ is the constant $1$ therein. By induction, it is easy to see that $f$ is polynomial of degree $k-1$ on each component of the complement of the closed set $F_k$ in $I$, inductively defined by $F_0:=I$ and $F_{k+1}:= \frac{1}{3}F_k \cup\left(\frac{1}{3}F_k + \frac{2}{3}\right)$. Note that $\{F_k\} _ {k\in\mathbb{N}}$ is the well-known nested sequence of closed sets whose intersection is the Cantor set.

$$*$$ Construction. Consider the linear operator $T$ on $L^\infty(0,1)$ (thought as subspace of $L^1(\mathbb{R})$ via extension by zero), defined by

$$Tu(x)= \cases{ \int_0^{3x}u(t)\ dt\qquad \qquad 0 < x < \frac{1}{3}\\ 0 \qquad \qquad \qquad \qquad \frac{1}{3} < x < \frac{2}{3}\\ \int_0^{3(1-x)}u(t)\ dt \qquad\frac{2}{3}< x < 1.}$$ Since $T$ and $T ^ 2$ are integral operators with non-negative kernels, their norms are attained on the constant function $1$ on $[0,1]$. We have, by simple computations $T1(x)=3\min(x,1-x)\chi _ { F _ 1 }(x)$, $\|T\| = \|T1\| _ \infty=1$ and $\|T^2\|=\|T(T1)\|_\infty=\frac{1}{3}$. Therefore $\rho(T)\le\frac{1}{\sqrt 3} < \frac{2}{3}$ and $I-\frac{3}{2}T$ is invertible. The function

$$f:= \left(I-\frac{3}{2}T\right) ^ {-1}\chi _ {[\frac{1}{3},\frac{2}{3}]} = \sum _ {k=0} ^ \infty \left( \frac{3}{2}\right) ^ k T ^ k \chi _ { [\frac{1}{3}, \frac{2}{3}] }$$

verifies

$$f(x)=\frac{3}{2} \int_0^{3x}f(t)\ dt\qquad \mathrm{for\ all}\quad 0 < x < \frac{1}{3}\ ,$$
$$f(x)=1\qquad \mathrm{for\ all}\quad \frac{1}{3} < x < \frac{2}{3}\ ,$$

and $$f(x)=f(1-x)\qquad \mathrm{for\ all}\quad x\in\mathbb{R} \ .$$ The only points in $\mathbb{R}$ where the continuity of $f$ is not immediate and has to be checked, are $\frac{1}{3}$ and $\frac{2}{3}$. We have

$$\int_ 0 ^ 1 f(x)dx=\int _ 0 ^ \frac{1}{3} f(x)dx+\int _ \frac{1}{3} ^ \frac{2}{3} f(x)dx+\int _ \frac{2}{3} ^ 1 f(x)dx= 2\int _ 0 ^ \frac{1}{3} f(x)dx+ \frac{1}{3}\ ,$$

and

$$2 \int _ 0 ^ \frac{1}{3} f(x)dx = 3\int _ 0 ^ \frac{1}{3} \int _ 0 ^ {3x} f(t) dt dx= \int _ 0 ^ 1 \int _ 0 ^ x f(t) dt dx=$$

$$= \int _ 0 ^ 1 (1-t) f(t) dt = \int _ 0 ^ 1 tf(1-t)dt= \int _ 0 ^ 1 tf(t)dt =$$

$$=\frac{1}{2}\left(\int _ 0 ^ 1 (1-t)f(t)+ tf(t)dt\right) =\frac{1}{2} \int _ 0 ^ 1 f(t)dt\ .$$

So

$$\int _ 0 ^ 1f(t)dt = \frac{2}{3}\ ,$$ which implies the continuity of $f$ at $\frac{1}{3}$ (and by symmetry at $\frac{2}{3}$ ), since
$$f(\frac{1}{3}-)=\frac{3}{2}\int_ 0 ^ 1 f(x)dx =1= f(\frac{1}{3}+) \ .$$ As a consequence, $f$ actually satisfies $$f(x)=\frac{3}{2} \int_0^{3x}f(t)\ dt$$ for all $x < \frac{2}{3}\ ,$ and we conclude that $f$ is in $C ^ 1(\mathbb{R})$, and has the stated properties.

Here is a Pudding Function $f$, and its integral function $F(x):=\int_0^x f(t)dt$. Note how the latter coincides with the first third of the former, up to rescaling. Simple considerations based on odd and even symmetry show that $F(1)=\frac{2}{3}$.

5 m

The existence of such functions being established by now, there remains a curiosity for the concrete situations that may generate them. Here is one, which is also interesting as an example of a $C^\infty$, non-analytic solution, of a simple first order linear DDE.

Let's consider the problem

$$f'(x)=\lambda f(3x),\qquad x < \frac{1}{2} \$$ where we seek for a function $f\in C^1(\mathbb{R})$ with support in $I:=[0,1]$, and with the condition $f(x)=f(1-x)$.

I will show that it has a simple eigenvalue $\lambda= \frac{9}{2}$, corresponding to an eigenfunction with $f(\frac{1}{2})=1$ (see the details below). Then, it is very easy to check that $f$ is locally polynomial on the complement of the Cantor set; and it is, of course, $C^\infty$.

Indeed, since $f (3x)=0$ for $x > \frac{1}{3}$, from the equation we have that $f'(x)$ vanishes identically on the interval $(\frac{1}{3},\frac{2}{3})$, so that $f$ is the constant $1$ therein. By induction, it is easy to see that $f$ is polynomial of degree $k-1$ on each component of the complement of the closed set $F_k$ in $I$, inductively defined by $F_0:=I$ and $F_{k+1}:= \frac{1}{3}F_k \cup\left(\frac{1}{3}F_k + \frac{2}{3}\right)$. Note that $\{F_k\} _ {k\in\mathbb{N}}$ is the well-known nested sequence of closed sets whose intersection is the Cantor set.

$$*$$ Construction. Consider the linear operator $T$ on $L^\infty(0,1)$ (thought as subspace of $L^1(\mathbb{R})$ via extension by zero), defined by

$$Tu(x)= \cases{ \int_0^{3x}u(t)\ dt\qquad \qquad 0 < x < \frac{1}{3}\\ 0 \qquad \qquad \qquad \qquad \frac{1}{3} < x < \frac{2}{3}\\ \int_0^{3(1-x)}u(t)\ dt \qquad\frac{2}{3}< x < 1.}$$ Since $T$ and $T ^ 2$ are integral operators with non-negative kernels, their norms are attained on the constant function $1$ on $[0,1]$. We have, by simple computations $T1(x)=3\min(x,1-x)\chi _ { F _ 1 }(x)$, $\|T\| = \|T1\| _ \infty=1$ and $\|T^2\|=\|T(T1)\|_\infty=\frac{1}{3}$. Therefore $\rho(T)\le\frac{1}{\sqrt 3} < \frac{2}{3}$ and $I-\frac{3}{2}T$ is invertible. The function

$$f:= \left(I-\frac{3}{2}T\right) ^ {-1}\chi _ {[\frac{1}{3},\frac{2}{3}]} = \sum _ {k=0} ^ \infty \left( \frac{3}{2}\right) ^ k T ^ k \chi _ { [\frac{1}{3}, \frac{2}{3}] }$$

verifies

$$f(x)=\frac{3}{2} \int_0^{3x}f(t)\ dt\qquad \mathrm{for\ all}\quad 0 < x < \frac{1}{3}\ ,$$
$$f(x)=1\qquad \mathrm{for\ all}\quad \frac{1}{3} < x < \frac{2}{3}\ ,$$

and $$f(x)=f(1-x)\qquad \mathrm{for\ all}\quad x\in\mathbb{R} \ .$$ The only points in $\mathbb{R}$ where the continuity of $f$ is not immediate and has to be checked, are $\frac{1}{3}$ and $\frac{2}{3}$. We have

$$\int_ 0 ^ 1 f(x)dx=\int _ 0 ^ \frac{1}{3} f(x)dx+\int _ \frac{1}{3} ^ \frac{2}{3} f(x)dx+\int _ \frac{2}{3} ^ 1 f(x)dx= 2\int _ 0 ^ \frac{1}{3} f(x)dx+ \frac{1}{3}\ ,$$

and

$$2 \int _ 0 ^ \frac{1}{3} f(x)dx = 3\int _ 0 ^ \frac{1}{3} \int _ 0 ^ {3x} f(t) dt dx= \int _ 0 ^ 1 \int _ 0 ^ x f(t) dt dx=$$

$$= \int _ 0 ^ 1 (1-t) f(t) dt = \int _ 0 ^ 1 tf(1-t)dt= \int _ 0 ^ 1 tf(t)dt =$$

$$=\frac{1}{2}\left(\int _ 0 ^ 1 (1-t)f(t)+ tf(t)dt\right) =\frac{1}{2} \int _ 0 ^ 1 f(t)dt\ .$$

So

$$\int _ 0 ^ 1f(t)dt = \frac{2}{3}\ ,$$ which implies the continuity of $f$ at $\frac{1}{3}$ (and by symmetry at $\frac{2}{3}$ ), since
$$f(\frac{1}{3}-)=\frac{3}{2}\int_ 0 ^ 1 f(x)dx =1= f(\frac{1}{3}+) \ .$$ As a consequence, $f$ actually satisfies $$f(x)=\frac{3}{2} \int_0^{3x}f(t)\ dt$$ for all $0 < x < \frac{2}{3}\ ,$ and we conclude that $f$ is in $C ^ 1(\mathbb{R})$, and has the stated properties.

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