# Question on proof in Evans PDE

This is on page 542 of Evans PDE book. The last inequality states that

$$\int_{U}{C(|Du|+1)|u|dx} \leq \frac{1}{2}\int_{U}|Du|^2dx + C\int_{U}{|u|^2+1 \ dx}$$

Where is this coming from? I think this is just young's inequality and then holder applied to $|Du|$ (since $u$ is assumed to be in $H_0^1[U]$) but why write it in such a weird way?

-

So we have the inequality $$\int_U |Du |^2 + \mu |u|^2 dx \leqslant \frac{1}{2} \int_U |Du|^2 dx + C \int_U |u|^2 + 1 dx .$$ Now assume that $\mu$ is sufficiently large, for example let $\mu > C + \frac{1}{2}$ . Then this inequality becomes \begin{eqnarray*} \frac{1}{2}\int_U |Du|^2 dx &\leqslant& (C - \mu ) \int_U |u|^2 dx + C \int_U dx \\ &\leqslant & - \frac{1}{2} \int_U |u|^2 dx + C \int_U dx \end{eqnarray*} or $$\frac{1}{2} \int_U |Du|^2 + |u|^2 dx \leqslant C \int_U dx .$$ Thus we can see that $$\| u \|_{H_0^1 (U)} = \int_U |Du|^2 + |u|^2 dx \leqslant 2C \int_U dx \leqslant C' < \infty$$ holds.
What?? $u$ is assumed to be in $H_0^1$ from the start. – Euler....IS_ALIVE Dec 12 '12 at 20:17
@Euler....IS_ALIVE Yes, we assumed $u \in H_0^1$, but we could not know the dependence of $\lambda$. What I wrote means that C' does not depend on $\lambda$ so that the set K:= $\{ u \in H_0^1 | u = \lambda A[u] \; \text{for some} \; 0 \leqslant \lambda \leqslant 1 \}$ is bounded in $H_0^1$. – Leun Kim Dec 13 '12 at 10:26
@Euler....IS_ALIVE If that constant $C'$ depends on $\lambda$, we cannot say the boundedness of the set. Because if $C'$ depends on $\lambda$ and if there are infinitely many $\lambda's$(denote$\lambda_i, i=1,2,\cdots$), then of course for every $\lambda_i$'s, the corresponding $u_i$ satisfies $\| u_i \|_{H_0^1} < \infty$. But $\lim_{i \to \infty} \| u_i \|_{H_0^1}$may diverge. – Leun Kim Dec 13 '12 at 10:29
@Euler....IS_ALIVE So if $C'$ does not depend on $\lambda_i$, then we can say that for all $i=1,2,\cdots$, $\| u_i \|_{H_0^1} \leqslant C'$ uniformly which means that the set $K$ is bounded by $C'$ in $H_0^1$ with sufficiently large $\mu$. – Leun Kim Dec 13 '12 at 10:38