Let $T_f:(C([a,b],\mathbb C), \lVert \cdot \rVert_1) \to \mathbb C$ with $g\mapsto \int_a^b f(x)g(x) dx$ for any given $f\in C([a,b],\mathbb C)$. I have to find the norm of $T_f$. I started with: $|T_fg| = \left|\int_a^b f(x)g(x)dx\right| \leqslant\int_a^b |f(x)g(x)|dx \leqslant\lVert g \rVert _1 \lVert f \rVert _\infty.$ For the last step I used Hölder. Now all I need is an example $g\in C([a,b],\mathbb C)$ with $\lVert g \rVert_1 = \int_a^b |g(x)|dx = 1$ and $|T_fg| = \lVert f \rVert _\infty$ (I suppose $\lVert f \rVert _\infty$ is the operator norm here). I have tried several functions, but none worked, maybe $\lVert T_f \rVert < \lVert f \rVert _\infty$?
Norm of operator $g\mapsto \int fg$
1 Answers
Assume that $\max_{0\leqslant t\leqslant 1}|f(t)|=f(x_0)$ for some $x_0\in [0,1]$. I assume that $t\in (0,1)$, otherwise the argument can be adapted. Let $g_n$ the function such that $g_n=1$ on $(x_0-n^{-1},x_0+n^{-1})$, $0$ on $[0,x_0-2n^{-1})$ and $[x_0+2n^{-1},1)$, and piecewise linear. Then $T_f(g_n)=2n^{-1}f(x_0)+\int_{x_0-2n^{-1}}^{x_0-n^{—1}}f(t)g_n(t)dt+\int_{x_0+n^{-1}}^{x_0+2n^{—1}}f(t)g_n(t)dt.$ We have $\lVert g_n\rVert=3n^{-1}$, and $T_f(g_n)=3n^{-1}f(x_0)-\int_{x_0-2n^{-1}}^{x_0-n^{—1}}(f(t)-f(x_0)g_n(t)dt+\int_{x_0+n^{-1}}^{x_0+2n^{—1}}(f(t)-f(x_0))g_n(t)dt\\ .$ By continuity of $f$, $\lim_{n\to +\infty}\frac{\int_{x_0+n^{-1}}^{x_0+2n^{—1}}(f(t)-f(x_0))g_n(t)dt-\int_{x_0-2n^{-1}}^{x_0-n^{—1}}(f(t)-f(x_0)g_n(t)dt}{\lVert g_n\rVert}=0$ so we get the result.
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0Oh I think I see it, thank you! – 2012-11-06