If $f:C[-1,1]\rightarrow \mathbb{R} $ is given by $$f(u)=\int_{-1}^{1} (1-t^2) u(t) \ dt .$$ I managed to bound $||f||$ above by $2$ and so want to find a function such that $|f(u)|=2 $ or some sequence such that $|f(u_n)| \rightarrow 2 $. Any ideas? Is the norm of $f$ even 2 because if not then obviously this won't work?
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2Should that be $u(t)$ in the integral? – G. Chiusole Apr 28 '21 at 20:28
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1It seems $|f|\leq\frac{4}{3}$ – Nick Castillo Apr 28 '21 at 21:05
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1assuming you are using sup-norm – Nick Castillo Apr 28 '21 at 21:05
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Which is attained for the constant function $u=1$. – Nicolas Apr 28 '21 at 21:09
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On one hand, $$ \begin{align} |f(u)|&=\biggl|\int_{-1}^{1}(1-t^2)u(t)\,dt\biggr|\\ &\leq \int_{-1}^{1}|1-t^2||u(t)|\,dt\\ &\leq ||u||_{\infty}\int_{-1}^{1}(1-t^2)\,dt=\frac{4}{3}||u||_{\infty}. \end{align} $$ And therefore, $||f||_{\infty}\leq 4/3$. And on the other hand, for $u(t)=1$ for $t\in [-1,1]$ we obtain $$f(u)=\int_{-1}^{1}(1-t^2)\,dt=\frac{4}{3},$$ thereby, $\frac{4}{3}=|f(u)|\leq ||f||_{\infty}\cdot ||u||_{\infty}=||f||_{\infty}$, since $||u||_{\infty}=1$ when $u(t)=1$ for all $t$. So, at the end we end up with $||f||_{\infty}=4/3$.
ChrisNick92
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