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

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 $x\in[-1,1]$.

Tags:

Inequality

Functional Analysis

Lp Spaces

Upper Lower Bounds

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