I recently asked this question on Math StackExchange and someone suggested that it would probably be more suited for Math Overflow. Since it still has not been answered, here it goes:

If two functions are far from being orthogonal, their difference cannot be too large in $L^2$. A precise statement - easily verified with the Pythagorean theorem - is as follows: let $f,g:[-1,1]\rightarrow [0,\infty)$ be (Lebesgue) measurable functions such that $\|f\|_2=1=\|g\|_2$. If

$$\int_{-1}^1 f(t)g(t)dt\geq \theta^2$$ for some $0<\theta<1$, then

$$\|f-g\|_2\leq\sqrt{2}(1-\theta^2)^{1/2}.$$

A similar statement cannot hold in $L^1$: take, for instance, $f=\chi_{[-\frac{1}{2},\frac{1}{2}]}$ and $0\leq g\in C_0^\infty(\mathbb{R})$ supported on a small neighborhood of the origin (say, on $[-{\delta}/{2},{\delta}/{2}]$ for some $\delta \ll1$) and such that $\|g\|_1=1$. Then $\int fg=\int g=1$ but $\|f-g\|_1\geq1-\delta$.

My question is: **What happens in $L^p$ for $1< p<2$?**

More precisely: let $d\in\mathbb{N}$, let $p\in(1,2)$, and let $f,g$ be measurable functions defined on a compact subset $K\subset\mathbb{R}^d$. Assume $\|f\|_p=1=\|g\|_p$ and suppose additionally that $|f|\leq 1$ on $K$ (in particular, the integral $\int fg$ converges absolutely). If $$\int_K fg\geq\theta^p$$ for some $0<\theta<1$, does there exist a constant $C_p<\infty$ such that $$\|f-g\|_p\leq C_p (1-\theta^p)^{1/p}?$$

Thank you.