Show that the function 1/t is not in L2 (0,1]

Question

Need some help getting started with this problem: $$f(t) = \frac{1}{t}$$ Show that $f(t)$ is not in $L_2(0,1]$, but that it is in the Hilbert space $L_{2}w(0,1)$ where the inner product is given by $$\langle x,y\rangle = \int(x(t)\overline{y(t)}w(t)dt$$ where $w(t)=t^2$.

I am thinking that I have to show that the function is continuous over the interval.

Answer 1

First prove that $f \not\in L_2(0,1]$. Let's try to calculate $\int_{0}^{1}f^2(t)dt$:

$$\int_{0}^{1}\frac{1}{t^{2}}dt=\left.-\frac{1}{t}\right|_{0}^{1}=-\infty$$

But $f \in L_2(0,1]$ iff $|\int_{0}^{1}|f(t)|^2dt|<\infty$

Next prove that $f \in L_2w(0,1]$, so calculate integral $\int_{0}^{1}|f(t)|^2w(t)dt$:

$$\int_{0}^{1}\frac{1}{t^{2}}\cdot t^2 dt=1<\infty$$

So $f \in L_2w(0,1]$.

Answer 2

SOME HELP: $\int 1/t^2\,dt$ is divergent on $(0,1]$.