2
$\begingroup$

Let $a,b,g,h$ be real numbers. How to prove that the functional $F\colon L^2 [a,b]\to \mathbb{R}$, given by $F(u)=\int_a^b (u^2(x)-gu(x)-h)\,dx$ is continuous?

Thank you

  • 1
    @MaxTilt: you should probably post your request for clarification and follow-ups as a comment _to the answer itself_, rather than as a comment to the question. By commenting _on the answer_ the user (in this case Jose27) will get notified of your request/comment. By leaving your comment on the main question the user will not know you have a follow-up question.2011-09-14

1 Answers 1

10

You can decompose your functional $F=F_1-F_2-F_3$ where $F_1(u)=\| u\|_{L^2}^2$, $F_2(u)=(g,u)_{L^2}$ and $F_3(u)=h(b-a)$. Continuity of the first follows from that of the norm, the second is a bounded linear functional by Cauchy-Schwarz inequality and the third is constant.

  • 0
    @MaxTilt Yes: Using the triangle inequality and Cauchy-Schwarz we get $F_1(u)-F_2(u)\geq \| u\|_{L^2}^2 -|(g,u)_{L^2}| \geq \| u\|_{L^2}(\| u\|_{L^2} - |g|)$ then clearly $(F_1-F_2)(u)\to \infty$ if $\| u\| \to \infty$, since $F_3$ is constant then $F(u)\to\infty$ as well.2011-09-14