I want to prove that two norms are equivalent but I am struggling with an upper bound

Question

I want to compare the usual norm on $L^2(-1,1)$ with the following:

$$ \Vert f \Vert_H^{2} = \int_{-1}^1 \vert f(x) \vert^2 \frac{1}{1+x^2}dx $$

Now, for sure I have this:$$\Vert f \Vert_H \leq \Vert f \Vert_{L^2} $$ Because $\frac{1}{1+x^2} \leq 1$

I want to now find some M such that $$\Vert f \Vert_{L^2} \leq M \Vert f \Vert_{H}$$

But I am not sure how to go from here.

EDIT: Forgot to say that we can assume f is measurable.

Answer 1

We simply have $$\|f\|_{L^2}^2 =\int_{-1}^1 |f|^2 \le 2\int_{-1}^1 |f(x)|^2 \frac1{1+x^2}=2\|f\|_H^2 $$ because $\frac1{1+x^2}\ge\frac12$ for all $x\in[-1,1]$.

Answer 2

Here is an abstract approach:

Let us define a new measure $d \lambda = \frac{1}{1 + x^2} d\mu$ meaning: $ \forall E\, measurable\; \lambda(E)=\int_{E}\frac{1}{1+x^2}d \mu$ where $\mu$ is the usual Lebesgue measure. Note that since the integral $\int_{-1}^{1}\frac{1}{1+x^2}d \mu$ is finite ,$\lambda$ is a finite measure, and thus a function is in $L^2(\lambda)$ iff it is in $L^2(\mu)$ and $$\Vert f \Vert_{L^2(\lambda)}=\int \vert f \vert ^2d\lambda = \int_{-1}^{1}\vert f(x) \vert^2 \frac{1}{1+x^2}d \mu = \Vert f \Vert_H^2$$ Now by your first inequality you proved the identity function $id: L^2(\mu) \rightarrow L^2(\lambda)$ is bounded (can you see why?). By the open mapping theorem ($L^2$ are Banach spaces) one necessarily has $id:L^2(\lambda) \rightarrow L^2(\mu)$ is bounded, which gives the converse bound.