Strict version of Hölder's inequality when $p=1$ and $q=\infty$ How do I prove that there is no function $h : \mathbb{R} \to \mathbb{R}$ of unit norm on $L^1$ such that
$\int_{\mathbb{R}} f h = \lVert f \rVert_{\infty} $
Whenever $f$ is a function (like $\tan^{-1}(x)$)  which is everywhere strictly less that its $L^{\infty}$-norm?
At first I thought that this would be a straightforward application of Hölder's inequality with $p=1$ and $q=\infty$, but that would only get me so far as
$\int_{\mathbb{R}} f h \leq \lVert h \rVert_1 \lVert f \rVert_{\infty} \\
\qquad \, = \lVert f \rVert_{\infty}$
which unfortunately doesn't rule out equality, which is what I'm trying to achieve.
 A: Assume that  $\|f\|_\infty -|f(x)|>0$ a.e.
Let $$A_n=\left \{x\in \mathbb{R}\,:\, \|f\|_\infty -|f(x)|\ge {1\over n}\right\}$$
By assumption the Lebesgue measure of
$$\mathbb{R}\setminus \bigcup_{n=1}^\infty A_n$$
is equal $0.$
For $h\in L^1$ and $\|h\|_1=1$ we thus have
$$\lim_{n\to \infty}\int\limits_{A_n}|h(x)|\,dx =\int\limits_{\mathbb{R}}|h(x)|\,dx=1$$
Hence there exists $n$ such that
$$\int\limits_{A_n}|h(x)|\,dx\ge {1\over 2}$$
We have
$$\left |\int\limits_{\mathbb{R}}f(x)h(x)\,dx\right |\le 
\int\limits_{\mathbb{R}}|f(x)|\,|h(x)|\,dx\\
=
\int\limits_{\mathbb{R}\setminus A_n}|f(x)|\,|h(x)|\,dx+
 \int\limits_{A_n}|f(x)|\,|h(x)|\,dx\\
\le \int\limits_{\mathbb{R}\setminus A_n}|f(x)|\,|h(x)|\,dx+
\int\limits_{ A_n}\left (\|f\|_\infty-{1\over n}\right )\,|h(x)|\,dx\\
\le \|f\|_\infty \int\limits_{\mathbb{R}\setminus A_n}|h(x)|\,dx+ \left (\|f\|_\infty-{1\over n}\right )\int\limits_{ A_n}|h(x)|\,dx\\
= \|f\|_\infty \|h\|_1-{1\over n}\int\limits_{ A_n}|h(x)|\,dx\le \|f\|_\infty -{1\over 2n}$$
Thus $$\left |\int\limits_{\mathbb{R}}f(x)h(x)\,dx\right |<\|f\|_\infty $$
