For $f\in C^{1}(\mathbb{R}^{n})$, Gross's logarithmic Sobolev inequality says that

$$\int f^{2} \log f^{2}\,d\mu -\int f^{2}\,d\mu \log\left(\int f^{2}\,d\mu\right)\leq \frac{2}{c}\int |\nabla f|^{2}d\mu,$$

where $d\mu=\frac{1}{\pi^{n/2}}e^{-|x|^{2}}dx$ for some $c>0$.

Any suggestions on a variational type proof? Can someone explain the variational argument proposed in the linked paper at pg 1268?

Attempt

Proof

We first prove it for $n=1$ and then finish by induction. We may assume $\int u^{2}d\mu=1$ and let $f(t)=v e^{t^{2}/2}$ then it suffices to prove

$$J(v):=\int_{0}^{\infty} \left(\frac{|\nabla v|^{2}}{2}-v^{2}\ln(|v|)\right)\,dt\geq \frac{\sqrt{\pi}}{4}$$

constrained to $\int_{0}^{\infty} v^{2}\,dt=\frac{\sqrt{\pi}}{2}$. But $\Delta v+2v\ln(v)+(\lambda+1)v=0$, where $\lambda$ is the Lagrange multiplier, doesn't look easy to solve.

In the paper below, a different variational calculus argument is proposed, which I still don't understand.

Adams, R. A.; Clarke, Frank H. Gross's logarithmic Sobolev inequality: a simple proof. Amer. J. Math. 101 (1979), no. 6, 1265–1269. MR 548880 DOI 10.2307/2374139

Thank you

For $f\in C^{1}(\mathbb{R}^{n})$, Gross's logarithmic Sobolev inequality says that

$$\int f^{2} \log f^{2}\,d\mu -\int f^{2}\,d\mu \log\left(\int f^{2}\,d\mu\right)\leq \frac{2}{c}\int |\nabla f|^{2}d\mu,$$

where $d\mu=\frac{1}{\pi^{n/2}}e^{-|x|^{2}}dx$ for some $c>0$.

Any suggestions on a variational type proof? Can someone explain the variational argument proposed in the linked paper at pg 1268?

Attempt

Proof

We first prove it for $n=1$ and then finish by induction. We may assume $\int u^{2}d\mu=1$ and let $f(t)=v e^{t^{2}/2}$ then it suffices to prove

$$J(v):=\int_{0}^{\infty} \left(\frac{|\nabla v|^{2}}{2}-v^{2}\ln(|v|)\right)\,dt\geq \frac{\sqrt{\pi}}{4}$$

constrained to $\int_{0}^{\infty} v^{2}\,dt=\frac{\sqrt{\pi}}{2}$. But $\Delta v+2v\ln(v)+(\lambda+1)v=0$, where $\lambda$ is the Lagrange multiplier, doesn't look easy to solve.

In the paper below, a different variational calculus argument is proposed, which I still don't understand.

Adams, R. A.; Clarke, Frank H. "Gross's logarithmic Sobolev inequality: a simple proof." Amer. J. Math. 101 (1979), no. 6, 1265–1269. MR 548880 DOI 10.2307/2374139

Thank you

For $f\in C^{1}(\mathbb{R}^{n})$ Gross's sobolev inequality says that

$$\int f^{2} log f^{2}d\mu -\int f^{2}d\mu log(\int f^{2}d\mu)\leq \frac{2}{c}\int |\triangledown f|^{2}d\mu,$$

where $d\mu=\frac{1}{\pi^{n/2}}e^{-|x|^{2}}dx$ for some $c>0$.

Any suggestions on a variational type proof? Can someone explain the variational argument proposed in the linked paper at pg 1268?

Attempt

Proof

We first prove it for n=1 and then finish by induction. We may assume $\int u^{2}d\mu=1$ and let $f(t)=v e^{t^{2}/2}$ then it suffices to prove

$J(v):=\int_{0}^{\infty} |\triangledown v|^{2}/2-v^{2}ln(|v|)dt\geq \frac{\sqrt{\pi}}{4}$

constrained to $\int_{0}^{\infty} v^{2}dt=\frac{\sqrt{\pi}}{2}$. But $\triangle v+2vln(v)+(\lambda+1)v=0$ ,where $\lambda$ is the lagrange multiplier, doesn't look easy to solve.

In the paper below, a different variational calculus argument is proposed, which I still don't understand.

Adams, R. A.; Clarke, Frank H. Gross's logarithmic Sobolev inequality: a simple proof. Amer. J. Math. 101 (1979), no. 6, 1265–1269. MR 548880

Thank you

For $f\in C^{1}(\mathbb{R}^{n})$, Gross's logarithmic Sobolev inequality says that

$$\int f^{2} \log f^{2}\,d\mu -\int f^{2}\,d\mu \log\left(\int f^{2}\,d\mu\right)\leq \frac{2}{c}\int |\nabla f|^{2}d\mu,$$

where $d\mu=\frac{1}{\pi^{n/2}}e^{-|x|^{2}}dx$ for some $c>0$.

Any suggestions on a variational type proof? Can someone explain the variational argument proposed in the linked paper at pg 1268?

Attempt

Proof

We first prove it for $n=1$ and then finish by induction. We may assume $\int u^{2}d\mu=1$ and let $f(t)=v e^{t^{2}/2}$ then it suffices to prove

$$J(v):=\int_{0}^{\infty} \left(\frac{|\nabla v|^{2}}{2}-v^{2}\ln(|v|)\right)\,dt\geq \frac{\sqrt{\pi}}{4}$$

constrained to $\int_{0}^{\infty} v^{2}\,dt=\frac{\sqrt{\pi}}{2}$. But $\Delta v+2v\ln(v)+(\lambda+1)v=0$, where $\lambda$ is the Lagrange multiplier, doesn't look easy to solve.

In the paper below, a different variational calculus argument is proposed, which I still don't understand.

Adams, R. A.; Clarke, Frank H. Gross's logarithmic Sobolev inequality: a simple proof. Amer. J. Math. 101 (1979), no. 6, 1265–1269. MR 548880 DOI 10.2307/2374139

Thank you

For $f\in C^{1}(\mathbb{R}^{n})$ Gross's sobolev inequality says that

$$\int f^{2} log f^{2}d\mu -\int f^{2}d\mu log(\int f^{2}d\mu)\leq \frac{2}{c}\int |\triangledown f|^{2}d\mu,$$

where $d\mu=\frac{1}{\pi^{n/2}}e^{-|x|^{2}}dx$ for some $c>0$.

Any suggestions on a variational type proof? Can someone explain the variational argument proposed in the linked paper at pg 1268?

Attempt

Proof

We first prove it for n=1 and then finish by induction. We may assume $\int u^{2}d\mu=1$ and let $f(t)=v e^{t^{2}/2}$ then it suffices to prove

$J(v):=\int_{0}^{\infty} |\triangledown v|^{2}/2-v^{2}ln(|v|)dt\geq \frac{\sqrt{\pi}}{4}$

constrained to $\int_{0}^{\infty} v^{2}dt=\frac{\sqrt{\pi}}{2}$. But $\triangle v+2vln(v)+(\lambda+1)v=0$ ,where $\lambda$ is the lagrange multiplier, doesn't look easy to solve.

In the paper below, a different variational calculus argument is proposed, which I still don't understand.

Adams "Gross's Logarithmic Sobolev Inequality: A Simple Proof" 1979

Thank you

For $f\in C^{1}(\mathbb{R}^{n})$ Gross's sobolev inequality says that

$$\int f^{2} log f^{2}d\mu -\int f^{2}d\mu log(\int f^{2}d\mu)\leq \frac{2}{c}\int |\triangledown f|^{2}d\mu,$$

where $d\mu=\frac{1}{\pi^{n/2}}e^{-|x|^{2}}dx$ for some $c>0$.

Any suggestions on a variational type proof? Can someone explain the variational argument proposed in the linked paper at pg 1268?

Attempt

Proof

We first prove it for n=1 and then finish by induction. We may assume $\int u^{2}d\mu=1$ and let $f(t)=v e^{t^{2}/2}$ then it suffices to prove

$J(v):=\int_{0}^{\infty} |\triangledown v|^{2}/2-v^{2}ln(|v|)dt\geq \frac{\sqrt{\pi}}{4}$

constrained to $\int_{0}^{\infty} v^{2}dt=\frac{\sqrt{\pi}}{2}$. But $\triangle v+2vln(v)+(\lambda+1)v=0$ ,where $\lambda$ is the lagrange multiplier, doesn't look easy to solve.

In the paper below, a different variational calculus argument is proposed, which I still don't understand.

Adams, R. A.; Clarke, Frank H. Gross's logarithmic Sobolev inequality: a simple proof. Amer. J. Math. 101 (1979), no. 6, 1265–1269. MR 548880

Thank you