Finiteness of the functional $\int_0^1 \frac{\phi(x)}{\sqrt{x}}dx$ in $L^2$

Question

Consider the following functional for functions in $L^2([0,1])$ $$\phi\to\int_0^1 \frac{\phi(x)}{\sqrt{x}}dx$$ Is this always finite? My guess is no, but the obvious check of $x^a$ for $a>-\frac12$ is indeed finite.

Can anyone prove this or, more likely, provide a counter example?

Answer 1

As suggested in the comment, from general theory one knows that the integral $\int_0^1 g(x) \phi(x) \, dx$ will be finite for all $\phi \in L^2[0,1]$ only when $g$ itself is in $L^2[0,1]$. And of course your function $g(x) = \frac{1}{\sqrt{x}}$ is not in $L^2[0,1]$, so just as you suspect, your functional cannot be finite for all $\phi$ in $L^2$.

The general theory does not, of course, immediately provide a specific example of such a $\phi$. In particular, because $\frac{1}{\sqrt{x}}$ is not itself in $L^2[0,1]$, the function $\frac{1}{\sqrt{x}}$ is not itself an example of such a $\phi$. And as you point out in your attempt, your functional does happen to be finite on those functions $x^a$ that do happen to lie in $L^2[0,1]$. So one has to look elsewhere in $L^2[0,1]$ for a specific example that is easy to compute.

But consider e.g. $\phi: [0,1] \to \mathbb{R}$ given by $\phi(x) = \frac{1}{\sqrt{x} \ln(x)}$ if $0 < x < 1/2$ and $\phi(x) = 0$ otherwise.

Using the "substitution" "$u = \ln(x)$" it is possible to see that $-\frac{1}{\ln(x)}$ is an antiderivative of $\frac{1}{x (\ln(x))^2}$ on $(0,1)$, and hence that $\int_0^1 \phi^2(x) \, dx = 1/\ln(2)$ is finite, so that $\phi$ is in $L^2[0,1]$.

The same substitution helps to see that $\ln|\ln(x)|$ is an antiderivative of $\frac{1}{x \ln(x)}$ on $(0,1)$, which can be used to show that $$ \int_0^1 \frac{1}{\sqrt{x}} \phi(x) \, dx = \int_0^{1/2} \frac{1}{x \ln(x)} \, dx = +\infty. $$

Answer 2

Your guess is right. Since you are in the critical exponent case, you have to use logarithms. Take $\phi(x) = 1/(\sqrt x\ln x)$ on $(0,1/2)$ and $0$ outside. Then $$ \int_0^{1/2} \frac{\mathrm d x}{x\,\ln x} = -\infty $$ since this is the integral of the derivative of $\ln \ln x$, but $\phi\in L^2$ since $$ \int_0^{1/2} \frac{\mathrm d x}{x\, \left|\ln x\right|^2} = \left[\frac{-1}{\ln x}\right]_0^{1/2} = \frac{1}{\ln 2}. $$