# The functional $I[w] = \int_U \frac{1}{2} |Dw|^2 - fw \, dx$ is weakly lower semicontinuous

I am studying calculus of variation, and I need to prove that

$I[w] = \int_U \frac{1}{2} |Dw|^2 - fw \, dx$ with $f \in L^2(U)$

is weakly lower semicontinuous on $H_0^1(U)$.

In classes, I only learned that

Assume L is bounded below, and in addition $L(p,z,x)$ is convex in $p$, for each $z \in \mathbb{R}$, $x \in U$. Then $I[.]$ is weakly lower semicontinuous on $W^{1,q}(U)$, where $I[w] = \int_U L(Dw, w, x)$.

And I think my $L$ isn't bounded below.

I don't know how I can solve the problem.

• Actually, it is bounded below. Use the Poincare inequality and Holder's inequality. And I'm pretty sure this was asked before. – user147263 Jun 5 '15 at 21:08
• Thank you! And I'm sorry, I searched and I didn't find. But this argument is to prove that $L$ is bounded below or $I$? – Daniela Jun 5 '15 at 21:45
• I see. But for convex $I$, a simpler argument exists. I'll post as an answer. – user147263 Jun 5 '15 at 22:11

It can be proved from basic principles that $$\text{(norm lower semicontinuity)} + \text{(convexity)} \implies \text{(weak lower semicontinuity)}$$ Indeed,

1. For any topological space $X$ the lower semicontinuity of a function $f:X\to\mathbb{R}$ is equivalent to the epigraph $\{(x,z):z\ge f(x)\}\subset X\times \mathbb{R}$ being closed in the product topology.

2. In a normed space, a convex closed subset is weakly closed.

The above applies to $I$, which is convex and continuous with respect to the norm on $H_0^1$. Indeed, $$|I[v]-I[w]|\le \frac12\int_U (|Dv+Dw|\,|Dv-Dw| + |f||v-w|) \\ \le \frac12(\|v\|_{H^1}+\|w\|_{H^1}) \|v-w\|_{H^1} + \frac12 \|f\|_{L^2}\|v-w\|_{L^2}$$

For completeness, I also quote the comment by Chee Han which shows the boundedness of $I$ from below (this does not help with lower semicontinuity, but is needed for the existence of minimizer).

Poincaré's inequality tells you that the $L^2$ norm of the gradient is equivalent to the full $H^1$-norm, so you can bound the first term below by the $H^1$ norm. For the second term, apply Cauchy-Schwarz, then use Young's inequality (some called this inequality by other name but my lecturer always calls it this way) $2ab\le a^2/2\varepsilon + \varepsilon b^2/2$. Then choose $\varepsilon$ so that the constant associated to the $H^1$ norm of $w$ is positive.

• I still don't understand why $I$ is convex :\ – Daniela Jun 6 '15 at 7:42
• It's the sum of two things: one is a linear functional of $w$, the other is the integral of a convex function of $w$. – user147263 Jun 8 '15 at 3:01