Prove or find a counterexample to each statement:
\(L^2(\mathbb R)\subset L^1(\mathbb R)\);
\(L^1(\mathbb R)\subset L^2(\mathbb R)\);
\(L^2([0,1])\subset L^1([0,1])\);
\(L^1([0,1])\subset L^2([0,1])\).
Solution
The answers are: (a) false, (b) false, (c) true, and (d) false.
1. Statement (a) is false.
Let \[ f(x)=\frac{\mathbf 1_{[1,\infty)}(x)}{x}. \] Then \[ \int_1^\infty |f(x)|^2\,dx=\int_1^\infty x^{-2}\,dx<\infty, \] so \(f\in L^2(\mathbb R)\), whereas \[ \int_1^\infty |f(x)|\,dx=\int_1^\infty x^{-1}\,dx=\infty. \] Thus \(f\notin L^1(\mathbb R)\).
2. Statement (b) is false.
Let \[ g(x)=x^{-2/3}\mathbf 1_{(0,1)}(x). \] Then \[ \int_0^1 |g(x)|\,dx=\int_0^1x^{-2/3}\,dx<\infty, \] so \(g\in L^1(\mathbb R)\), but \[ \int_0^1 |g(x)|^2\,dx=\int_0^1x^{-4/3}\,dx=\infty. \] Hence \(g\notin L^2(\mathbb R)\).
3. Statement (c) is true.
By Cauchy–Schwarz, \[ \|f\|_{L^1([0,1])} =\int_0^1|f(x)|\cdot1\,dx \le \|f\|_{L^2([0,1])}\|1\|_{L^2([0,1])} =\|f\|_2. \] Thus every \(L^2([0,1])\) function belongs to \(L^1([0,1])\).
4. Statement (d) is false.
The same function \[ g(x)=x^{-2/3},\qquad 0<x<1, \] belongs to \(L^1([0,1])\) but not to \(L^2([0,1])\), by the calculation in 2.