# Non-linear functional on $L^2$

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

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Thank you Jose27. Is it clear that if $\|u_n\|\to \infty$, then $F(u_n)\to\infty$ as well? –  MaxTilt Sep 14 '11 at 14:34
@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. –  Willie Wong Sep 14 '11 at 14:49

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.
Thank you. Btw. is it clear that if $\|u_n\|\to \infty$, then $F(u)\to \infty$ as well? –  MaxTilt Sep 14 '11 at 14:58
@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. –  Jose27 Sep 14 '11 at 16:21