Let $Tf(t)= \int_0^tf(s)ds$ find $\alpha$ such that $||T|| \leq \alpha$

Question

On the space of $L^2([0,1])$ consider the following operator; $$Tf(t)= \int_0^tf(s)ds$$ Find $\alpha < 1$ such that $||T|| \leq \alpha$

I try to use Fubini's theorem to compute $||Tf(t)||$ but I not sure how work with the integration limits or how is the right way to mayorate these. Any hint or help I will very grateful

$\alpha =1$ works! But I made a mistake and write $\alpha \leq 1$ but the problem require that $\alpha <1$ — Nick, Jul 26 '22 at 09:56

score 1 · Accepted Answer · answered Jul 26 '22 at 09:57

1

$\int_0^{1}|\int_0^{t} f(s)ds|^{2}dt\leq \int_0^{1} \|f\|^{2}t dt=\frac 1 2 \|f\|^{2}$. So we can take $\alpha=\frac 1 {\sqrt 2}$.

answered Jul 26 '22 at 09:57

Kavi Rama Murthy

359,332

@geetha290km really thanks!! I only have one question why $\left|\int_0^t f(s)ds\right|^2\leq ||f||^2 t$ ? – Nick Jul 26 '22 at 09:59
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By Holder's ineqiuliaty the left side does not exceed $(\int_0^{t} |f|^{2}))(\int_0^{t} 1)dt$ – Kavi Rama Murthy Jul 26 '22 at 10:07

Let $Tf(t)= \int_0^tf(s)ds$ find $\alpha$ such that $||T|| \leq \alpha$

1 Answers1