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Let $u_k , u \in L^2 (\Bbb R^n)$ for $k \in \Bbb N$. Assume that $f : \Bbb R^n \to \Bbb R$ is continuous and $|f (u_k) | \leqslant M$ , $|f(u) | \leqslant M$ for some $M >0$. If $u_k$ converges to $u$ in $L^2$ ( $\| u_k \|_{L^2{\Bbb R^n}} \to \| u \|_{L^2{(\Bbb R^n)}} $) then can I conclude that $$ \| f(u_k) u_k \|_{L^2 (\Bbb R^n)} \to \| f(u)u \|_{L^2{\Bbb (\Bbb R^n)}}\;\;(k \to \infty)\;\;?$$

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up vote 0 down vote accepted

By the triangle inequality, \begin{gather} \| f(u_k) u_k - f(u)u \|=\| f(u_k) u_k - f(u_k)u +f(u_k)u-f(u)u \| \leqslant \\ \leqslant \| f(u_k) u_k - f(u)u \|=\| f(u_k) u_k - f (u_k)u \|+\|f(u_k)u-f(u)u \| = \\ =\|f(u_k)(u_k-u) \| +\|u(f(u_k)-f(u)) \| \leqslant \\ \leqslant M \|u_k -u \|+ \|u(f(u_k)-f(u)) \| \, . \end{gather} Let $\varepsilon >0$ be an arbitrary. Because $f$ is continuous, for this $\varepsilon$ there exists $\delta>0$ such that $\|u_k-u\|<\delta$ implies $|f(u_k)-f(u)|<\varepsilon $. Because $\|u_k-u\|\underset{k\to \infty}{\to {0}},$ there exists $k_\delta: \quad \|u_k-u\|<\delta \quad \forall k>k_\delta.$ Therefore, $$\|u(f(u_k)-f(u)) \| \leqslant {\varepsilon}\|u\|$$ If we take $k^*$ such that $\|u_k-u\|<\varepsilon_1=\min(\varepsilon, \,\delta) \quad(\forall k>k^*)$ then $$ M \|u_k -u \| < \varepsilon_1 M\leqslant\varepsilon M$$ and $$ \|u(f(u_k)-f(u)) \| \leqslant {\varepsilon}\|u\|.$$ Finally, $$\| f(u_k) u_k - f(u)u \| \leqslant {\varepsilon}M + {\varepsilon}\|u\| ={\varepsilon}(M+\|u\|).$$

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Thank you very much. Then we need more assumption that "u is bounded" in the third line -> fourth line. – Ann Nov 7 '12 at 12:37
@Ann Thanks for your remark. I was mistaken in the fourth line because boundedness of $u$ is not given and corrected my answer. – M. Strochyk Nov 7 '12 at 13:12

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