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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.

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Original MSE question: math.stackexchange.com/questions/412134. –  user17240 Jun 11 '13 at 15:56
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2 Answers

The answer to your modified question is still no. Rather than just give an example, let me explain in a conceptual way why it cannot be right. Your conditions on $f$ imply by interpolation that $\|f\|_q \le 1 $, where $1/p +1/q =1$. So you need $\|f\|_q$ close to one and (by uniform convexity) $g$ to be an approximately norming functional for $f$, which is to say (assuming WLOG $f \ge 0$) that $g$ is close to $f^{q-1}$ in the $L_p$ norm. So if $f$ is close to $g$ in $L_p$, it is also close to $f^{q-1}$, which forces $f$ to be close to a characteristic function of a set, and of course the set should then have measure one.

So you only need to build $f\ge 0$ and $g$ to satisfy the hypotheses with $f$ not close to a characteristic function. This you obviously cannot do if the measure space has measure at most one, but it is pretty clear that you can if the measure of the whole space is larger than one; in particular, if it is $[-1,1]$. The simplest thing is to have $f = a 1_A + b 1_B$ with $A$ and $B$ disjoint and $a,b$ distinct. You need $0 < a,b \le 1$ and $a^p m(A) + b^p m(B)= 1 = a^q m(A) + b^q m(B)$. Fixing $a$ and $b$ you have unique positive solutions for $m(A)$ and $m(B)$, and it is evident that both of these will be less than one if $a$ and $b$ are chosen sufficiently large.

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$f_n= 1_{(0,2/n)} (n/2)^{1/p}$, $g_n= 1_{(0,1/n)} n^{1/p}$.

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@Bill: I forgot to add one crucial assumption (which is present in the L^1 counterexample): one of the functions (f, say) is bounded by 1. Corrected, thanks. –  user17240 Jun 11 '13 at 17:26
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