# Norm of operator $g\mapsto \int fg$

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 $\lVert T_f \rVert < \lVert f \rVert _\infty$?

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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.
Would it be easier to show that the integrals turn to 0 because $f\cdot g$ is bounded by $\lVert f \rVert _\infty$ and thus for $\lim\limits_{n\to\infty}$ the integrals go to 0? –  Stefan Nov 6 '12 at 21:53
Yes, it converges to $0$, but we have to divide by the norm of $g_n$. –  Davide Giraudo Nov 6 '12 at 22:00
Oh, of course. And how do we get this limit exactly? If I divide by $\frac 3 n$ then it isn't obvious (to me) that it still converges to 0. –  Stefan Nov 6 '12 at 22:10
We use continuity of $f$ at $x_0$ and $\varepsilon$'s (I didn't give the details as it's a homework question). –  Davide Giraudo Nov 6 '12 at 22:11