1. ## heat equation problem

Hi!

I've got the following problem:

$\displaystyle f:\mathbb{R}\rightarrow\mathbb{R}$ a $\displaystyle C^2$-function such that $\displaystyle f$ is convex and $\displaystyle f(0)=f'(0)=0$. $\displaystyle g\in C^{\infty}([0,\infty),\mathscr{S}(\mathbb{R}^d))$ a real-valued solution to the heat equation $\displaystyle g_t=\Delta g$. Using this show that $\displaystyle F\in C^1$ and $\displaystyle F$ decrasing, where:

$\displaystyle F(t)=\int_{\mathbb{R}^d}f(g(x,t))dx$.

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'I started by just diff. $\displaystyle F$ and I got

$\displaystyle F'(t) = \frac{d}{dt}\int_{\mathbb{R}^d}f(g(x,t))dx = \int_{\mathbb{R}^d} \frac{d}{dt}f(g(x,t))dx = \int_{\mathbb{R}^d}\frac{df(g)}{dg}g_tdx = \int_{\mathbb{R}^d} \frac{df}{dg}\Delta gdx$.

What next?

2. ## Re: heat equation problem

Notice

$\displaystyle F'(t)= \int_{ \mathbb{R}^d } f'(g)\Delta g dx = - \int_{ \mathbb{R}^d } \nabla \left( f'(g) \right) \cdot \nabla (g)dx = -\int_{ \mathbb{R}^d } f''(g)|\nabla(g)|^2 dx\leq0$

where we first integrated by parts and then used that $\displaystyle f''\geq 0$.