Return to Answer

It is probably best to do this by hand, by looking at the domains described in the usual way as those $f$ with $Lf\in L^2$ that satisfy certain boundary conditions at $x=0,1$ (in fact, there will be no boundary condition at $x=0$ when $\nu^2\ge 1$).

But we can also avoid dealing with boundary conditions explicitly, as follows: if we had $D(L)=D(L')$, then $D(L)\subseteq D(M)$, with $M$ denoting the (self-adjoint) operator of multiplication by $1/x^2$ on its natural domain. This implies, by the closed graph theorem, that $M$ is $L$-bounded, that is, $$ \|Mf\|\lesssim \|f\| +\|Lf\| \quad\quad\quad\quad (1) $$ for all $f\in D(L)$.

To refute this, fix an eigenfunction $g$ of $L$ and let $f(x)=\varphi(x-\delta) g(x)$, with $\varphi\in C^{\infty}$, $\varphi=1$ on $x<0$, $\varphi=0$ for $x>1/2$. From the behavior of the Bessel functions near $x=0$, we obtain $g(x)\simeq x^{\alpha}$, $g'(x)\simeq x^{\alpha-1}$ with $\alpha=\nu+1/2$ (and I exclude the trivial case $\nu=-1/2$ from the discussion here).

Then $Lf=-\varphi''g -2\varphi'g'$, so $\|Lf\|^2\lesssim \max\{ 1, \delta^{2\alpha-2}\}$ as $\delta\to 0+$. On the other hand, $|Mf|\gtrsim\delta^{\alpha-2}$ on $(0,\delta)$, so $\|Mf\|^2\gtrsim \delta^{2\alpha-3}$.

This shows that (1) cannot hold when $\alpha<3/2$ or $\nu<1$.

For $\nu\ge 1$ this argument no longer works, and we also transition to limit point case at $x=0$ at exactly this value of $\nu$. This does not feel like a coincidence, though I'd have to think about this more carefully to take a meaningful guesses.

It is probably best to do this by hand, by looking at the domains described in the usual way as those $f$ with $Lf\in L^2$ that satisfy certain boundary conditions at $x=0,1$ (in fact, there will be no boundary condition at $x=0$ when $\nu^2\ge 1$).

But we can also avoid dealing with boundary conditions explicitly, as follows: if we had $D(L)=D(L')$, then $D(L)\subseteq D(M)$, with $M$ denoting the (self-adjoint) operator of multiplication by $1/x^2$ on its natural domain.

But from the behavior of the Bessel functions near $x=0$, we also know that the eigenfunctions of $L$ satisfy $\psi(x)\simeq x^{\nu+1/2}$, so $\psi\in D(M)$ only if $\nu> 1$.

If indeed $\nu\ge 1$, then we transition to limit point case at $x=0$, and no boundary condition is imposed here. It seems possible or even plausible that $D(L)\subseteq D(M)$ now. If this is correct, then $M$ is $L$-bounded, and the bound would drop to values $<1$ if we multiply $M$ by a small constant. Moreover, it cannot get larger if we make $\nu$ larger. That would mean that $D(L)$ is actually constant in the range $\nu>1$.

It is probably best to do this by hand, by looking at the domains described in the usual way as those $f$ with $Lf\in L^2$ that satisfy certain boundary conditions at $x=0,1$ (in fact, there will be no boundary condition at $x=0$ when $\nu^2\ge 1$).

But we can also avoid dealing with boundary conditions explicitly, as follows: if we had $D(L)=D(L')$, then $D(L)\subseteq D(M)$, with $M$ denoting the (self-adjoint) operator of multiplication by $1/x^2$ on its natural domain. This implies, by the closed graph theorem, that $M$ is $L$-bounded, that is, $$ \|Mf\|\lesssim \|f\| +\|Lf\| \quad\quad\quad\quad (1) $$ for all $f\in D(L)$.

To refute this, consider the solutions of the ODE $Lg=0$, $L=-d^2/dx^2+c/x^2$. This an Euler equation that we can solve explicitly by $x^{1/2\pm s}$, $s=\sqrt{1/4+c}$ (as an aside, note that the large near $x=0$ solution stops being square integrable when $c\ge 3/4$; this explains the transition alluded to above).

Now let's take $f=\varphi g$, with $$ g(x)=x^{1/2}\left( \left( \frac{x}{2\delta}\right)^{-s} - \left( \frac{x}{2\delta}\right)^s \right) $$ and a $\varphi\in C_0^{\infty}(0,1)$ with $\varphi=1$ on $[2\delta,4\delta]$ and $\varphi$ supported by $[\delta,5\delta]$. Notice that $g(2\delta)=0$, $|g'(2\delta)|\lesssim \delta^{-1/2}$, so we can keep $|\varphi'|\lesssim \delta^{-1/2}$, $|\varphi''|\lesssim \delta^{-3/2}$.

We then have $Lf=\varphi Lg-\varphi'' g -2\varphi' g'=-\varphi'' g-2\varphi'g'$. So $|(Lf)(x)|\lesssim \delta^{-1}$ on two intervals of length $\delta$ each and $Lf=0$ otherwise. On the other hand, $|(Mf(x)|\gtrsim\delta^{-3/2}$ on $[3\delta,4\delta]$, so (1) isn't working.

It is probably best to do this by hand, by looking at the domains described in the usual way as those $f$ with $Lf\in L^2$ that satisfy certain boundary conditions at $x=0,1$ (in fact, there will be no boundary condition at $x=0$ when $\nu^2\ge 1$).

But we can also avoid dealing with boundary conditions explicitly, as follows: if we had $D(L)=D(L')$, then $D(L)\subseteq D(M)$, with $M$ denoting the (self-adjoint) operator of multiplication by $1/x^2$ on its natural domain. This implies, by the closed graph theorem, that $M$ is $L$-bounded, that is, $$ \|Mf\|\lesssim \|f\| +\|Lf\| \quad\quad\quad\quad (1) $$ for all $f\in D(L)$.

To refute this, fix an eigenfunction $g$ of $L$ and let $f(x)=\varphi(x-\delta) g(x)$, with $\varphi\in C^{\infty}$, $\varphi=1$ on $x<0$, $\varphi=0$ for $x>1/2$. From the behavior of the Bessel functions near $x=0$, we obtain $g(x)\simeq x^{\alpha}$, $g'(x)\simeq x^{\alpha-1}$ with $\alpha=\nu+1/2$ (and I exclude the trivial case $\nu=-1/2$ from the discussion here).

Then $Lf=-\varphi''g -2\varphi'g'$, so $\|Lf\|^2\lesssim \max\{ 1, \delta^{2\alpha-2}\}$ as $\delta\to 0+$. On the other hand, $|Mf|\gtrsim\delta^{\alpha-2}$ on $(0,\delta)$, so $\|Mf\|^2\gtrsim \delta^{2\alpha-3}$.

This shows that (1) cannot hold when $\alpha<3/2$ or $\nu<1$.

For $\nu\ge 1$ this argument no longer works, and we also transition to limit point case at $x=0$ at exactly this value of $\nu$. This does not feel like a coincidence, though I'd have to think about this more carefully to take a meaningful guesses.

It is probably best to do this by hand, by looking at the domains described in the usual way as those $f$ with $Lf\in L^2$ that satisfy certain boundary conditions at $x=0,1$ (in fact, there will be no boundary condition at $x=0$ when $\nu^2\ge 1$).

But we can also avoid dealing with boundary conditions explicitly, as follows: if we had $D(L)=D(L')$, then $D(L)\subseteq D(M)$, with $M$ denoting the (self-adjoint) operator of multiplication by $1/x^2$ on its natural domain. This implies, by the closed graph theorem, that $M$ is $L$-bounded, that is, $$ \|Mf\|\lesssim \|f\| +\|Lf\| \quad\quad\quad\quad (1) $$ for all $f\in D(L)$.

To refute this, consider the solutions of the ODE $Lg=0$, $L=-d^2/dx^2+c/x^2$. This an Euler equation that we can solve explicitly by $g=x^{\lambda}$, $\lambda=1/2\pm \sqrt{1/4+c}$ (as an aside, note that the large near $x=0$ solution stops being square integrable when $c\ge 3/4$; this explains the transition alluded to above).

Now let's take $f=\varphi g$, with $\lambda<0$ and a $\varphi\in C_0^{\infty}(0,1)$ with $\varphi=1$ on $(2\delta,(N+2)\delta)$ and $\varphi$ supported by $[\delta,(N+3)\delta]$. We then have $Lf=\varphi Lg-\varphi'' g -2\varphi' g'=-\varphi'' g-2\varphi'g'$. We can keep $\varphi'\lesssim 1/\delta$, $\varphi''\lesssim 1/\delta^2$. Then $|(Lf)(x)|\lesssim \delta^{\lambda-2}$ on two intervals of length $\delta$ each and $Lf=0$ otherwise. Since $|(Mf(x)|\gtrsim\delta^{\lambda-2}$ also, but on an interval of length $N\delta$, we see that (1) isn't working.

It is probably best to do this by hand, by looking at the domains described in the usual way as those $f$ with $Lf\in L^2$ that satisfy certain boundary conditions at $x=0,1$ (in fact, there will be no boundary condition at $x=0$ when $\nu^2\ge 1$).

But we can also avoid dealing with boundary conditions explicitly, as follows: if we had $D(L)=D(L')$, then $D(L)\subseteq D(M)$, with $M$ denoting the (self-adjoint) operator of multiplication by $1/x^2$ on its natural domain. This implies, by the closed graph theorem, that $M$ is $L$-bounded, that is, $$ \|Mf\|\lesssim \|f\| +\|Lf\| \quad\quad\quad\quad (1) $$ for all $f\in D(L)$.

To refute this, consider the solutions of the ODE $Lg=0$, $L=-d^2/dx^2+c/x^2$. This an Euler equation that we can solve explicitly by $x^{1/2\pm s}$, $s=\sqrt{1/4+c}$ (as an aside, note that the large near $x=0$ solution stops being square integrable when $c\ge 3/4$; this explains the transition alluded to above).

Now let's take $f=\varphi g$, with $$ g(x)=x^{1/2}\left( \left( \frac{x}{2\delta}\right)^{-s} - \left( \frac{x}{2\delta}\right)^s \right) $$ and a $\varphi\in C_0^{\infty}(0,1)$ with $\varphi=1$ on $[2\delta,4\delta]$ and $\varphi$ supported by $[\delta,5\delta]$. Notice that $g(2\delta)=0$, $|g'(2\delta)|\lesssim \delta^{-1/2}$, so we can keep $|\varphi'|\lesssim \delta^{-1/2}$, $|\varphi''|\lesssim \delta^{-3/2}$.

We then have $Lf=\varphi Lg-\varphi'' g -2\varphi' g'=-\varphi'' g-2\varphi'g'$. So $|(Lf)(x)|\lesssim \delta^{-1}$ on two intervals of length $\delta$ each and $Lf=0$ otherwise. On the other hand, $|(Mf(x)|\gtrsim\delta^{-3/2}$ on $[3\delta,4\delta]$, so (1) isn't working.