Define $$ I : L^2(\mathbb R) \to L^2(\mathbb R), (If)(y) = \int_0^\infty \frac{f(x+y)}{\sqrt{x}} dx. $$

Why is this well-defined and a bounded operator?

Since $\int_1^\infty \frac 1 {\sqrt{x}} dx$ diverges and the function obviously isn't $L^2$ either, most of the usual methods (say via Cauchy-Schwartz) one could use here don't seem to apply. Even Minkowski inequalities or convolution/Fourier transform tricks don't really seem to work out. Since this is a minor exercise in a relatively early part of a functional analysis course, it should be doable with elementary methods but I just ran out of ideas.


1 Answer 1


Let $f_n(x)={\bf 1}_{[0,n]}(x).$ Then $\|f\|_2^2={n}.$ On the other hand for $0\le y\le n$ we get $$(If_n)(y)=\int\limits_0^\infty {f_n(x+y)\over \sqrt{x}}\,dx =\int\limits_y^\infty{f_n(x)\over \sqrt{x-y}}\,dx\\ =\int\limits_y^n{1\over \sqrt{x-y}}\,dx=2\sqrt{n-y}$$ Thus $$\|If_n\|_2^2 \ge 4\int\limits_0^n(n-y)\,dy=2n^2$$ Therefore the operator is unbounded.

The operator is not well defined, i.e. the integral defining $(If)(y)$ may be divergent. Indeed let $f(x)=x^{-1/2}\log^{-1}(x),$ for $x\ge 2$ and $f(x)=0$ otherwise. Then $f\in L^2. $ However for $y\ge 2$ we have
$$\int\limits_0^\infty {f(x+y)\over \sqrt{x}}\,dx= \int\limits_y^\infty {f(x)\over \sqrt{x-y}}\,dx\\ =\int\limits_y^\infty {1\over \sqrt{x}\sqrt{x-y}\log x}\,dx\ge \int\limits_y^\infty {1\over x\log x}\,dx=+\infty$$


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