Lecture 6. Gagliardo-Nirenberg inequalities in Dirichlet spaces

Let (X,μ,ℰ,ℱ) be a Dirichlet space and {P_t}_t∈[0,∞) denote the associated Markovian semigroup. Throughout the lecture, we shall assume that P_t admits a measurable heat kernel p_t(x,y) satisfying, for some C > 0 and β > 0,

$p_{t}(x,y)\leq C t^{-\beta}$

for μ × μ-a.e. (x,y) ∈ X × X, and for each t ∈ (0,+∞). We also assume stochastic completeness, which means that P_t1 = 1 for every t ≥ 0.

The goal of the lecture is to prove the Gagliardo-Nirenberg inequalities in that setting. The techniques we use come from a paper by Bakry-Coulhon-Ledoux-Saloff-Coste. For another approach to the Sobolev inequality, we refer to the post.

Preliminary lemmas
Sobolev inequality
Gagliardo-Nirenberg inequalities

Preliminary lemmas

Lemma. For every f ∈ ℱ, t ≥ 0,

$|| P_t f -f ||_{L^2(X,\mu)} \le C \sqrt{t} \mathcal{E}(f,f)^{1/2}.$

Proof: Let Δ be the generator of the semigroup (P_t)_t≥0. We use spectral theorem to represent Δ as a multiplier in some L²(Ω,ν) space:

$U^{-1} A U g (x)=-\lambda(x) g(x),$

so that

$|| P_t f -f ||_{L^2(X,\mu)}^2=\int_{\Omega} \left( 1-e^{-t\lambda(x)}\right)^2(U^{-1} f)^2(x) d\nu(x)$

$\le C t \int_{\Omega} \lambda(x) (U^{-1} f)^2(x) d\nu(x)$

$= Ct \mathcal{E}(f,f).$

Lemma. For every f ∈ ℱ,

$\lim_{t \to 0} t^{-1}\int_X \int_X |f (x)-f(y)|^2 p_t(x,y) d\mu(x)d\mu(y) =2 \mathcal{E}(f,f).$

Proof: Using P_t1 = 1 we have

$\int_X \int_X |f (x)-f(y)|^2 p_t(x,y) d\mu(x)d\mu(y)$

$\int_X \int_X( f (x)^2-2f(x)f(y)+f(y)^2) p_t(x,y) d\mu(x)d\mu(y)$

$=2\int_X f(x)^2 d\mu(x)-2\int_X f(x)(P_t f) (x)d\mu(x)$

$=2 \int_X (f(x)-P_tf(x)) f(x) d\mu(x)$

and we conclude thanks to a previous result.

Sobolev inequality

Lemma. Let 1 ≤ q < +∞. There exists a constant C > 0 such that for every f ∈ ℱ ∩ L^q(X,μ) and s ≥ 0,

$\sup_{s \ge 0} s^{1 +\frac{q}{2\beta}} \mu \left( \{ x \in X\, :\, | f(x) | > s \} \right)^{\frac{1}{2}}\le C \mathcal{E}(f,f)^{1/2} || f ||_{L^q(X,\mu)}^{\frac{q}{2\beta}}.$

Proof: Let f ∈ ℱ and denote

$F(s)=\mu \left( \{ x \in X\, :\, | f(x) | > s \} \right).$

We have then

$F(s) \le \mu \left( \{ x \in X\, :\, | f(x) -P_t f (x) | > s/2 \} \right)+\mu \left( \{ x \in X\, :\, | P_t f (x) | > s/2 \} \right).$

Now, from the heat kernel upper bound p_t(x,y) ≤ C t^-β, t > 0, one deduces, for g ∈ L¹(X,μ), that

$|P_t g (x) | \le Ct^{-\beta} \| g \|_{L^1(X,\mu)}.$

Since P_t is a contraction in L^∞(X,μ), by the Riesz-Thorin interpolation one obtains

$| P_t f (x) | \le \frac{C^{1/q}}{t^{\beta /q}} \| f \|_{L^q(X,\mu)}.$

Therefore, for

$s= 2 \frac{C^{1/q}}{t^{\frac{\beta}{ q}}} \| f \|_{L^q(X,\mu)},$

one has

$\mu \left( \{ x \in X\, :\, | P_t f (x) | > s/2 \} \right)=0.$

On the other hand, from previous lemma,

$\mu \left( \{ x \in X\, :\, | f(x) -P_t f (x) | > s/2 \} \right) \le C s^{-2} t \mathcal{E}(f,f).$

We conclude that

$F(s)^{1/2} \le C s^{-1 -\frac{q}{2\beta}} \mathcal{E}(f,f)^{1/2} \| f \|_{L^q(X,\mu)}^{\frac{q}{2\beta}}.$

Lemma. Assume β > 1. There exists a constant C > 0 such that for every f ∈ ℱ,

$\sup_{s \ge 0}\, s\, \mu \left( \{ x \in X\, :\, | f(x) | \ge s \} \right)^{\frac{1}{q}} \le C \mathcal{E}(f,f)^{1/2},$

where q = 2β/(β – 1).

Proof: Let f ∈ ℱ be a non-negative function. For k ∈ ℤ, we denote

$f_k=(f-2^k)_+ \wedge 2^k.$

Observe that f_k ∈ L²(X,μ) and ||f_k||_L²(X,μ) ≤ ||f||_L²(X,μ). Moreover, for every x, y ∈ X, |f_k(x) – f_k(y)| ≤ |f(x) – f(y)| and so ℰ(f_k,f_k) ≤ ℰ(f,f). We also note that f_k ∈ L¹(X,μ), with

$||f_k||_{L^1(X,\mu)} =\int_X |f_k| d\mu \le 2^k \mu (\{ x \in X\, :\, f(x) \ge 2^k \}).$

We now use the previous lemma to deduce:

$\sup_{s \ge 0} s^{1 +\frac{1}{2\beta}} \mu \left( \{ x \in X\, :\, f_k(x) > s \} \right)^{\frac{1}{2}} \le C \mathcal{E}(f_k,f_k)^{1/2} || f_k \|_{L^1(X,\mu)}^{\frac{1}{2\beta}}$

$\le C \mathcal{E}(f_k,f_k)^{1/2} \left( 2^k \mu (\{ x \in X\, :\, f(x) \ge 2^k \})\right)^{\frac{1}{2\beta}}$

In particular, by choosing s = 2^k we obtain

$2^{k \left(1 +\frac{1}{2\beta} \right)} \mu \left( \{ x \in X\, :\, f(x) \ge 2^{k+1} \} \right)^{\frac{1}{2}} \le C \mathcal{E}(f_k,f_k)^{1/2} \left( 2^k \mu (\{ x \in X\, :\, f(x) \ge 2^k \})\right)^{\frac{1}{2\beta}}$

Let

$M(f)=\sup_{k \in \mathbb{Z}} 2^k \mu (\{ x \in X\, :\, f(x) \ge 2^k \})^{1/q}$

where q = 2β/(β – 1). Using the fact that 1/q = 1/2 – 1/(2β) and the previous inequality we obtain:

$2^{k} \mu \left( \{ x \in X\, :\, f(x) \ge 2^{k+1} \} \right)^{\frac{1}{2}} \le C 2^{ -\frac{kq}{2\beta}}\mathcal{E}(f,f)^{1/2} M(f)^{\frac{q}{2\beta}}$

and

$2^k \mu \left( \{ x \in X\, :\, f(x) \ge 2^{k+1} \} \right)^{\frac{1}{q}} \le C^{\frac{2}{q} } \mathcal{E}(f,f)^{1/q}M(f)^{\frac{1}{\beta}}.$

Therefore

$M(f)^{1-\frac{1}{\beta}} \le 2 C^{\frac{2}{q} } \mathcal{E}(f,f)^{1/q}$

and one concludes

$M(f) \le 2^{q/2} C \mathcal{E}(f,f)^{1/2}.$

This easily yields:

$\sup_{s \ge 0} s \mu \left( \{ x \in X\, :\, f(x) \ge s \} \right)^{\frac{1}{q}} \le 2^{1+q/2} C \mathcal{E}(f,f)^{1/2}$

Let now f ∈ ℱ, which is not necessarily non-negative. From the previous inequality applied to |f| we deduce

$\sup_{s \ge 0}\, s\, \mu \left( \{ x \in X\, :\, |f(x) | \ge s \} \right)^{\frac{1}{q}}$

$\le 2^{1+q/2} C \mathcal{E}(|f|,|f|)^{1/2}$

$\le 2^{1+q/2} C \mathcal{E}(f,f)^{1/2}.$

Theorem. (Sobolev inequality) Assume β > 1. There exists a constant C > 0 such that for every f ∈ ℱ,

$\| f \|_{L^q(X,\mu)} \le C \mathcal{E}(f,f)^{1/2}$

where q = 2β/(β – 1).

To show that the weak type inequality implies the desired Sobolev inequality, we will need another slicing argument and the following lemma is needed.

Lemma. For f ∈ ℱ, f ≥ 0, denote f_k = (f – 2^k)₊ ∧ 2^k, k ∈ ℤ. There exists a constant C > 0 such that for every f ∈ ℱ,

$\sum_{k \in \mathbb{Z}} \mathcal{E} (f_k,f_k) \le C \mathcal{E} (f,f).$

Proof: Let p_t(x,y) denote the heat kernel of the semigroup P_t. We first observe that, once we prove

$\sum_{k \in \mathbb{Z}} \int_X \int_X |f_{k} (x)-f_{k}(y)|^2 p_t(x,y) d\mu(x)d\mu(y) \le C\int_X \int_X |f (x)-f(y)|^2 p_t(x,y) d\mu(x)d\mu(y)$

where C > 0 is independent from t, then

$\liminf_{t \to 0^+} \sum_{k \in \mathbb{Z}} t^{-1 } \!\!\!\int_X \int_X |f_{\rho} (x)-f_{\rho}(y)|^2 p_t(x,y)d\mu(x)d\mu(y)$

$\le C\liminf_{t \to 0^+} t^{-1 } \!\!\!\int_X \int_X |f (x)-f(y)|^p p_t(x,y) d\mu(x)d\mu(y),$

and, using the superadditivity of the liminf, one concludes

$\sum_{k \in \mathbb{Z}} \liminf_{t \to 0^+} t^{-1 } \int_X \int_X |f_{k} (x)-f_{k}(y)|^2 p_t(x,y) d\mu(x)d\mu(y)$

$\le C \liminf_{t \to 0^+} t^{-1 } \int_X \int_X |f (x)-f(y)|^2 p_t(x,y) d\mu(x)d\mu(y)$

which yields

$\sum_{k \in \mathbb{Z}} \mathcal{E} (f_k,f_k) \le C \mathcal{E} (f,f).$

We therefore aim to prove the inequality above. For each k ∈ ℤ, set B_k = {x ∈ X : 2^k < f ≤ 2^k+1}. In this way, the external integral on the left-hand side is decomposed it into an integral over B_k and B_k^c. For the integrals over B_k, since the mapping f → f_k is a contraction, it follows that

$\sum_{k\in\mathbb{Z}}\int_{B_k} \int_X |f_{k} (x)-f_{k}(y)|^2 p_t(x,y)d\mu(x)d\mu(y)\leq \int_X \int_X |f (x)-f(y)|^2 p_t(x,y)d\mu(x)d\mu(y).$

To perform the integrals over B_k^c, we decompose them as

$\sum_{k\in\mathbb{Z}}\int_{B_k^c} \int_{B_k} |f_{k} (x)-f_{k}(y)|^2 p_t(x,y) d\mu(x)d\mu(y) +\sum_{k\in\mathbb{Z}}\int_{B^c_k} \int_{B_k^c}|f_{k} (x)-f_{k}(y)|^2 p_t(x,y)d\mu(x)d\mu(y)$

$=:\sum_{k\in\mathbb{Z}}J_1(k)+\sum_{k\in\mathbb{Z}}J_2(k).$

Again, the contraction property of f → f_k yields

$\sum_{k\in\mathbb{Z}}J_1(k)\leq \sum_{k\in\mathbb{Z}} \int_X \int_{B_k}|f_{k} (x)-f_{k}(y)|^2 p_t(x,y)d\mu(x)d\mu(y)$

$\leq \int_X \sum_{k\in\mathbb{Z}} \int_{B_k}|f_{k} (x)-f_{k}(y)|^2 p_t(x,y)d\mu(x)d\mu(y)\leq \int_X \int_X |f (x)-f(y)|^2 p_t(x,y) d\mu(x)d\mu(y).$

On the other hand, notice that for any (x,y) ∈ B_k^c × B_k^c we have |f_k(x) – f_k(y)| ≠ 0 only if

$(x,y)\in \{f(x)\leq 2^k<f(y)/2 \}\cup\{f(y)\leq 2^k<f(x)/2\}=:Z_k\cup Z_k^*.$

Also, |f_k(x) – f_k(y)| = 2^k for (x,y) ∈ Z_k ∪ Z_k^*. Thus,

$\sum_{k\in\mathbb{Z}}J_2(k)\leq \sum_{k\in \mathbb{Z}}\int_X\int_X\big(\mathbf{1}_{Z_k}(x,y)+\mathbf{1}_{Z^*_k}(x,y)\big)|f_k(x)-f_k(y)|^2 p_t(x,y)d\mu(x)d\mu(y)$

$=\int_X\int_X\sum_{k\in\mathbb{Z}}\big(\mathbf{1}_{Z_k}(x,y)+\mathbf{1}_{Z^*_k}(x,y)\big)2^{2k} p_t(x,y) d\mu(x)d\mu(y).$

One can see that

$\sum_{k\in\mathbb{Z}}\mathbf{1}_{Z_k}(x,y) 2^{2k}\leq 2 |f(x)-f(y)|^2$

and the same holds for Z_k^*, hence

$\sum\limits_{k\in\mathbb{Z}}J_1(k)+\sum\limits_{k\in\mathbb{Z}}J_2(k)\leq 5 \int_X\int_X|f(x)-f(y)|^2 p_t(x,y) d\mu(x)d\mu(y).$

Adding to these the term above finally yields the result

We can now conclude the proof of the Sobolev inequality.

Proof of the Sobolev inequality: Let f ∈ ℱ. We can assume f ≥ 0. As before, denote f_k = (f – 2^k)₊ ∧ 2^k, k ∈ ℤ. From Lemma 13.2 applied to f_k, we see that

$\sup_{s \ge 0} s \mu \left( \{ x \in X\, :\, | f_k(x) | \ge s \} \right)^{\frac{1}{q}} \le C \mathcal{E}(f_k,f_k)^{1/2}.$

In particular for s = 2^k, we get

$2^k \mu \left( \{ x \in X\, :\, f (x) \ge 2^{k+1} \} \right)^{\frac{1}{q}} \le C \mathcal{E}(f_k,f_k)^{1/2}$

Therefore,

$\sum_{k \in \mathbb{Z}} 2^{k q} \mu \left( \{ x \in X\, :\, f (x) \ge 2^{k+1} \} \right) \le C^q \sum_{k \in \mathbb{Z}} \mathcal{E}(f_k,f_k)^{q/2}.$

Since q ≥ 2, one has

$\sum_{k \in \mathbb{Z}} \mathcal{E}(f_k,f_k)^{q/2} \le \left( \sum_{k \in \mathbb{Z}} \mathcal{E}(f_k,f_k) \right)^{q/2}$

Thus, from the previous lemma

$\sum_{k \in \mathbb{Z}} 2^{k q} \mu \left( \{ x \in X\, :\, f (x) \ge 2^{k+1} \} \right)\le C \mathcal{E}(f,f)^{q/2}.$

Finally, we observe that

$\sum_{k \in \mathbb{Z}} 2^{k q} \mu \left( \{ x \in X\, :\, f (x) \ge 2^{k+1} \} \right) \ge \frac{q}{2^{q+1}-2^q} \sum_{k \in \mathbb{Z}}\int_{2^{k+1}}^{2^{k+2}} s^{q-1} \mu \left( \{ x \in X\, :\, f (x) \ge s \} \right)ds$

$\ge \frac{1}{2^{q+1}-2^q} \| f \|_{L^q(X,\mu)}^q.$

The proof is thus complete.

Gagliardo-Nirenberg inequalities

Using similar methods (see the paper) in the general case β > 0 one can get the family of Gagliardo-Nirenberg inequalities.

Theorem. Let q = 2β/(β – 1) with the convention that q = ∞ if β = 1. Let r,s ∈ (0,+∞] and θ ∈ (0,1] satisfying

$\frac{1}{r}=\frac{\theta}{q}+\frac{1-\theta}{s}.$

If β = 1, we assume r < +∞. Then, there exists a constant C > 0 such that for every f ∈ ℱ,

$\| f \|_{L^r(X,\mu)} \le C \mathcal{E}(f,f)^{\theta/2}\|f\|_{L^s(X,\mu)}^{1-\theta}.$

We explicitly point out some particular cases of interest.

Assume that β > 1. If r = s, then r = 2β/(β – 1) and above recovers the Sobolev inequality
$\| f \|_{L^r(X,\mu)} \le C \mathcal{E}(f,f)^{1/2}.$
Assume that β > 1. If s = +∞ and r ≥ 2β/(β – 1), then above yields
$\| f \|_{L^r(X,\mu)} \le C \mathcal{E}(f,f)^{\theta/2} \| f \|^{1-\theta}_{L^\infty(X,\mu)}$

with θ = 2β/(r(β – 1)).
If r = 2 and s = 1, then above yields the Nash inequality
$\| f \|_{L^2(X,\mu)} \le C \mathcal{E}(f,f)^{\theta/2} \| f \|^{1-\theta}_{L^1(X,\mu)}$

with θ = β/(1 + β).

In the case β = 1 one obtains the Trudinger-Moser inequalities.

Corollary . Assume that β = 1. Then, there exist constants c,C > 0 such that for every f ∈ ℱ with ℰ(f,f) = 1,

$\int_X \left( \exp \left( c |f|^2 \right)-1 \right) d\mu \le C \| f \|^2_{L^2(X,\mu)}.$