I'm a little bit stuck with the following little exercise:
If I know $\mathcal{F}$ is sequentially weakly lower semi-continous in $\{u \in W^{1,1}, u(0) = a, u(1) = b\}$, then I know that $F(tp_1 + (1-t)p_2) \leq tF(p_1) + (1-t)F(p_2)$ is fine if $tp_1 + (1-t)p_2 = b-a$. But is $F$ convex in $p$?
Thanks in advance for your help!
Edit: I'm talking about $\mathcal{F} = \int F(x,z,p) {\rm d}x$ with $z = u(x)$ and $p = \nabla u(x)$.
Yes. For simplicity I consider a functional of the form $\int F(\nabla u)$ below; the dependence on $u(x)$ and $x$ does not change things much, but muddles the water.
Let $u:[0,1]\to\mathbb R$ be such that $$u(0)=a, \quad u(1/2)=a+\frac12 (tp_1+(1-t)p_2),\quad u(1)=b \tag1$$ and make $u$ affine in between these points. Observe that $u'=tp_1+(1-t)p_2$ on $(0,1/2)$. Hence, $$\int_0^1 F(u) = \frac{1}2 F(tp_1+(1-t)p_2)+ \int_{1/2}^1 F(u') \tag2$$
Next, divide $[0,1/2]$ into $n$ equal subintervals $[\alpha_k,\beta_k]$. Define $u_n$ so that for each $k$ $$u_n(\alpha_k)=u(\alpha_k),\quad u_n(\alpha_k+t/(2n)) = u(\alpha_k)+tp_1/(2n), \quad u_n(\beta_k)=u(\beta_k) \tag3$$ and $u_n=u$ on $[1/2,1]$. Observe that $u_n'=p_1$ on $(\alpha_k,\alpha_k+t/(2n))$. Also, the average of $u_n'$ over $[\alpha_k,\beta_k]$ is the same as for $u$, namely, $tp_1+(1-t)p_2$. Hence $u_n'=p_2$ on the interval $(\alpha_k+t/(2n),\beta_k)$.
Thus, $$\int_0^1 F(u_n) = \frac{t}2 F(p_1)+\frac{1-t}2 F(p_2) + \int_{1/2}^1 F(u') \tag4$$
Letting $n\to \infty$, we infer from weak lsc property and the comparison of (2) and (4) that $$\frac{1}2 F(tp_1+(1-t)p_2)\le \frac{t}2 F(p_1)+\frac{1-t}2 F(p_2) \tag5$$ that is, $F$ is convex.
The above essentially uses the fact that the domain is one-dimensional. In higher dimensions weak lsc property does not imply convexity; it still implies rank-one convexity though.
