# Clarification on the proof of Poincaré's inequality

I am following the proof of the following Poincaré's inequality from the Lieb and Loss textbook

8.12 Let $$Ω∈\mathbb{R}^n$$ be a bounded, connected open set that has the cone property for some $$θ$$ and $$r.$$ Let $$1≤p≤∞$$, and let $$g$$ be a function in $$L^{p′}(Ω)$$, s.t. $$\int_{Ω}g=1.$$ Let $$1 < q < np/(n - p)$$ when $$p < n,$$ $$q <\infty$$, when $$p = n,$$ and $$1 < q <\infty$$ when $$p > n.$$ Then there is a finite number $$S>0,$$ which depends on $$Ω,g,p,q$$ such that $$$$\lvert|f−\int_Ωfg\rvert|_{L^q(\Omega)}≤S||∇f||_{L^p(\Omega)},\quad \quad\quad \quad (1)$$$$ $$∀f∈W^{1,p}(Ω).$$

They begin the proof by assuming $$q\geq p$$. Then they assume, by contradiction, that the Poincaré inequality $$(1)$$ is false for every $$S>0$$.

I don't understand the next part: Then there is a sequence of functions $$f_j$$, such that the left side of the inequality $$(1)$$ equals to $$1$$, for all $$j$$, while the right side tends to zero as $$j\rightarrow \infty$$.

Next they let $$h_j=f_j+\int_{\Omega}gf_j$$, such that the gradient of $$f_j$$ equals the gradient of $$h_j$$. (Why?) Then the sequence $$h_j$$ is bounded in $$W^{1,p}(\Omega).$$

The rest of the proof I think makes sense and follows from Rellich-Kondrachov theorem, but I was wondering if someone can please ellaborate on the bolded steps?

If the Poincaré inequality is false for every $$S$$, then for each $$j$$, we may take $$S=j$$ to see that there exists a function $$f_j$$ for which $$$$\lvert|f_j−\int_Ωf_j g\rvert|_{L^p(\Omega)}> j ||∇f_j||_{L^p(\Omega)},\quad \quad\quad \quad (1)$$$$ Both sides are homogeneous in $$f$$, so by replacing $$f_j$$ with $$c_j f_j$$ for some constant $$c_j$$, we can assume the left side equals 1. Therefore we have $$||∇f_j||_{L^p(\Omega)} < 1/j$$ which tends to 0 as $$j \to \infty$$.
Replacing $$f_j$$ with $$h_j$$ by adding the constant $$\int g f_j$$ is just a convenience. The left side now equals $$1 = \|h_j\|_{L^p}$$ and the right side still equals $$\|\nabla f_j\|_{L^p} = \|\nabla h_j\|_{L^p} < 1/j$$. Then the $$W^{1,p}$$ norm of $$h_j$$ is comparable to $$\|h_j\|_{L^p} + \|\nabla h_j\|_{L^p} = 1 + 1/j \le 2$$.
Some authors would summarize this step just by saying: suppose without loss of generality that $$\int g f_j = 0$$.
• Thank you that is much clearer now! I have one more question; they state next that $\int_{\Omega} hg=0$ which seem to follow from the fact that $\nabla h=0$, and therefore $h$ is a constant function. Are you able to clarify? Commented Dec 12, 2022 at 17:19
• No, the equality $\int hg = 0$ is nothing to do with $\nabla f$, and at this point there is no reason for $h$ to be constant. It's just simple algebra. Plug in $h(x) = f(x) - \int f(y)g(y)\,dy$, and distribute over the multiplication by $g(x)$. The first term is $\int f(x) g(x)\,dx$. The second term is now $\int \int f(y) g(y) dy \, g(x) dx$. The inner integral does not depend on $x$, so this equals $\left(\int f(y)g(y)\,dy\right) \left(\int g(x)\,dx\right)$. But $\int g = 1$ so this just equals $\int f(y)g(y)\,dy$, which exactly cancels the first term. Commented Dec 15, 2022 at 2:11