Let $f, g :(0,1) \to \mathbb{R}\,\,$ s.t. $\lVert g \rVert_2 =3$ and $\lVert f \rVert_1 = e\,\,$, show that $|\int_{0}^{1} g \sqrt{\log(f)}dx| \le 3$

Question

Let $g\in L^2((0,1))$ s.t. $\lVert g \rVert_2 =3\,\,\,$ and $f \ge1, f \in L^1((0,1))$ s.t. $\lVert f \rVert_1 = e\,\,$, prove that:

$$ \Biggl|\int_{0}^{1} g \sqrt{\log(f)}\,dx \Biggr|\,\,\le\,\,3$$

My solution was:

$$ \Biggl|\int_{0}^{1} g \sqrt{\log(f)}\,dx \Biggr|\le \int_{0}^{1} \biggl|g \sqrt{\log(f)}\biggr|\,dx \le \biggl(\int_{0}^{1} |g|^2 \,dx\biggr)^{\frac{1}{2}}\biggl(\int_{0}^{1} \log(f)\,dx\biggr)^{\frac{1}{2}} =3\biggl(\int_{0}^{1} \log(f)\,dx\biggr)^{\frac{1}{2}}$$

the temptation is to say that $(\sup_{x\in(0,1)}|f|)\le\lVert f \rVert_1=e\,\,$ to conclude, but I don't know if it is true in this case. Any suggestion?

Answer 1

It is not true that $|f(x)| \leq \|f\|_1$ for all $x$ so your argument does not hold. Natural logarithm is concave so Jensen's inequaity gives $\int_0^{1} \log f \leq log \int_0^{1} f=1$.