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Let the operator $T:L^2(0,+\infty )\to L^2(1,+\infty )$ be defined by $$Tf(s)= \int_{s^2}^{\infty }\frac{f(t)}{t}dt$$ To check continuity I decided to find the norm $\| T\| =\text{sup}_{\| f\| =1}\| Tf\| $.

$$\begin{split} \left\| \int_{s^2}^{\infty }\frac{f(t)}{t}dt \right\| & =\left( \int_{1}^{\infty }\left( \left| \int_{s^2}^{\infty }\frac{f(t)}{t}dt) \right| \right)^2 \right)^{1/2}\\ &\overset{\text{Hölder}}{\le} \left( \int_{1}^{\infty }\left(\int_{s^2}^{\infty }\left|f(t)\right|^2dt)\right)^{1/2}\left(\int_{s^2}^{\infty }\left|\frac{1}{t}\right|^2dt) \right)^{1/2}d\mu \right)^{1/2} \\ &\le \left[ \int_{1}^{\infty }\left( \frac{1}{s^2} \right)^{1/2}ds \right]^{1/2}=\infty \end{split}$$

But I can't find a lower bound, can anyone suggest how to find it?

Bruno B
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stboy
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2 Answers2

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The operator $T$ can be considered as $T:L^2(1,\infty)\to L^2(1,\infty)$ as it vanishes on functions $f$ whose support is contained in $(0,1).$ The operator is of the form $$(Tf)(s)=\int\limits_1^\infty K(s,t)f(t)\,dt$$ where $K(s,t)={1\over t}{\bf 1}_{[s^2,\infty)}(t).$ It is well known that $T$ is bounded if $K(s,t)$ is square integrable on $[1,\infty)\times [1,\infty)$ and $$\|T\|\le \|K\|_2:=\left (\int\limits_1^\infty \int\limits_1^\infty|K(s,t)|^2\,dt\,ds\right )^{1/2}$$ The proof can be made by applying the Cauchy-Schwarz inequality two times. In our case we have $$\|K\|_2^2=\int\limits_1^\infty\int\limits_{s^2}^\infty {1\over t^2}\,dt\,ds= \int\limits_1^\infty{1\over s^2}=1$$ Thus $\|T\|\le 1.$

Remark The operator $T$ belongs to the Hilbert-Schmidt class of compact operators. Its norm is actually strictly less then $1.$ To this end we can show that the adjoint operator $T^*$ corresponds to $K^*(s,t)=K(t,s).$ Furthermore, the operator $T^*T$ corresponds to $$\widetilde{K}(s,t)=\int\limits_1^\infty K^*(s,x)K(x,t)\,dx=\int\limits_1^\infty K(x,s)K(x,t)\,dx\\ ={1\over st}[\min(t^{1/2},s^{1/2})-1] $$ We have $$\int\limits_1^\infty\int\limits_1^\infty {1\over s^2t^2}[\min(t^{1/2},s^{1,2})-1]^2\,ds\,dt \\ =2\int\limits_1^\infty\int\limits_t^\infty {1\over s^2t^2}[t^{1,2}-1]^2\,ds\,dt =2\int\limits_1^\infty{1\over t^3}[t^{1/2}-1]^2\,dt={1\over 3}$$ Thus $$\|T\|=\|T^*T\|^{1/2}\le {1\over \sqrt{3}}$$

Ryszard Szwarc
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The operator $T$ in the OP can be viewed as a composition of operators:

  1. the adjoint operator of the Hardy operator $H:L_p(0,\infty)\rightarrow L_p(0,\infty)$, $1<p<\infty$, defined as $(Hf)(x)=\frac1x\int^x_0 f(s)\,ds$, namely $$(H^*f)(t)=\int^\infty_t \frac{f(y)}{y}\,dy$$ It is known that $$\|H\|_2=\|H^*\|_2=2$$

  2. $A:L_p(0,\infty)\rightarrow L_p(1,\infty)$ define as $(Af)(s)=\mathbb{1}_{(1,\infty)}(s)f(s^2)$. Notice that $$\|Af\|^2_{L_2(1,\infty)}=\int^\infty_1|f(s^2)|^2\,ds=\frac12\int^\infty_1|f(u)|^2\frac{1}{\sqrt{u}}\,du\leq\frac12\|f\|^2_{L_2(0,\infty)} $$

Observe that $T=A\circ H^*$. Hence $\|T\|\leq\|A\|\,\|H^*\|$.

Mittens
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