Proving the uniform convergence of a function

Question

Let $u\in C^2(\Omega)\cap C^0(\overline{\Omega})$, $\Omega\subset\mathbb{R}^n$ open and bounded and $\lambda>0$ sufficiently small so that $2\lambda u<1$. Define $w$ by $w\leq\frac{1}{\lambda}$ and $$ u(x)=w(x)-\frac{\lambda}{2}w(x)^2. $$ I have to prove that $w\rightarrow u$ uniformly as $\lambda\rightarrow0$.

My first attempt. $$ |w(x)-u(x)|=\left|u(x)-\frac{\lambda}{2}w(x)^2-u(x)\right|=\left|-\frac{\lambda}{2}w(x)^2\right|\leq\frac{\lambda}{2}\frac{1}{\lambda^2}=\frac{1}{2\lambda}\rightarrow\infty $$ as $\lambda\rightarrow0$. What is the mistake? Some help to prove this uniform convergence?

My second attempt. Note that $w$ can be written in terms of $u$ as $$ w(x)=\frac{1}{\lambda}(1-\sqrt{1-2\lambda u}). $$ In this way, by taking the limit of $w$ for $\lambda\rightarrow 0$, it follows that $w\rightarrow u$. Indeed $$ \lim_{\lambda\rightarrow0}\frac{1}{\lambda}(1-\sqrt{1-2\lambda u})=\lim_{\lambda\rightarrow0}\frac{1}{\lambda}\frac{(1-\sqrt{1-2\lambda u})(1+\sqrt{1-2\lambda u})}{(1+\sqrt{1-2\lambda u})}=\ldots=u. $$

Is it now correct? Is it sufficient to prove the uniform convergence?

Thank you

Answer 1

I don't like using the explicit solution given by the quadratic formula; what if $u$ and $w$ are related in a more complicated way? So I found a different proof that does not rely on the solution formula: it fixes your first attempt.

The reason your first attempt fails is because we only use $w<\frac1\lambda$ there(nevermind if we know that $|w|<\frac 1\lambda$...), which is a bad bound for $\lambda\ll 1$. But we are trying to prove that $\|w-u\|_{C^0} \to 0$, which would imply $\|w\|_{C^0} ≤ \|u\|_{C^0}(1+o(1))$ as $\lambda \to 0$, which is much better.

In fact, a bound like $\|w\|_{C^0} < C$ for a constant $C$ independent of $\lambda$ is enough to fix the first attempt: $$ \|u-w\|_{C^0} = \frac{\lambda}{2} \|w\|_{C^0}^2 ≤ C^2 \frac{\lambda}2 \to 0 $$ We now prove that such a bound exists: $|u| = |w-\frac\lambda 2 w^2|$ implies $$ |w| = \frac{|u|}{|1-\lambda w/2|}$$ we can also compute $$\lambda w < 1 ⟹ \frac{\lambda w}{2}< \frac 12 ⟹ \frac{\lambda w}{2} - 1< \frac {-1}2 ⟹ \left|1-\frac{\lambda w}{2}\right| > \frac12 $$ Hence, $$\|w\|_{C^0} ≤ 2\|u\|_{C^0}$$ as claimed, and the result follows.

Answer 2

Quadratic formula gives $w = \frac{1 ±\sqrt{1-2\lambda u} }{\lambda}$. Imposing $w≤1/\lambda$ forces the choice of sign, $$w = \frac{1 -\sqrt{1-2\lambda u} }{\lambda} = \frac{2\lambda u}{\lambda(1+\sqrt{1-2\lambda u})} = \frac{2}{1+\sqrt{1-2\lambda u}}u$$ So we are done if we can show $$ \frac{2}{1+\sqrt{1-2\lambda u}} \to 1 \quad \text{ uniformly}$$

this is true as $\|u\|_{C^0} < \infty$, so $2\lambda u \to 0$ uniformly as $\lambda \to 0$, implying the result: For $\lambda \ll 1$, $\frac{2}{1+\sqrt{1-2\lambda u}} \in (1-\epsilon,1+\epsilon)$, so that $$ |w(x) - u(x)| ≤ \epsilon |u(x)| ≤ \epsilon \|u\|_{C^0} $$ which is a bound independent of $x$. Hence $$\|w-u\|_{C^0} ≤ \epsilon \|u\|_{C^0} $$ as needed.