# Estimation of the $L^2$ norm of a function

given the function $$g(x,t)=\frac{e^{\frac{Y}{2}}}{2 \epsilon [1+e^{\frac{Y}{2}}]^3}$$ with $$Y=\frac{x-\frac{t}{2}}{\epsilon}$$ where $$x\in \mathbb{R}$$ and $$t>0$$.

I'm looking for $$C_{\epsilon}>0$$ where $$||g(.,t)||_2\leq C_{\epsilon}$$ where $$C_{\epsilon} \to 0$$ as $$\epsilon \to 0^+.$$

My idea, I used the following variable change $$u=x/\epsilon$$ and I ve got:

$$||g(.,t)||_2^2=\frac{e^{\frac{-t}{2 \epsilon}}}{4 \epsilon} \int_{\mathbb{R}}\frac{ e^u}{[1+e^{\frac{-t}{4 \epsilon}} e^{\frac{u}{2}}]^6}du$$

Now I only have to prove that the integral in bounded by a constant that doesn't depend on $$\epsilon$$

So I used the following variable change: $$z=e^{-\frac{t}{4 \epsilon}}e^{\frac{u}{2}}$$ but I only got:

$$||g(.,t)||_2^2=\frac{1}{2 \epsilon}\int_0^{+\infty}\frac{z}{(1+z)^6}dz$$

Integration-by-parts results in

$$\int\frac{z}{(1+z)^6}\text{d}z = - \frac{z}{5(1+z)^5} + \frac{1}{5}\int\frac{1}{(1+z)^5}\text{d}z,$$

where the second integral is straightforward,

$$\int\frac{1}{(1+z)^5}\text{d}z = -\frac{1}{4(1+z)^4}.$$

Taking limits, the improper integral is

$$\int_{0}^{+\infty}\frac{z}{(1+z)^6}\text{d}z = \frac{1}{20}.$$

• That s not what I’m asking , I’m asking to prove that the $L^2$ norm of $g$ converges to $0$ as $\epsilon$ goes to $0$ Jan 2, 2020 at 14:34
• @AnasBOUALII One of the main issues is the change of variables. If you want to prove that $||g||_2 \rightarrow 0$ as $\epsilon \rightarrow 0$, the variable of transformation, named either z or u, shouldn't contain the variable $\epsilon$. I'll think about the problem and I'll edit my comment if I solve it.
– V.J.
Jan 2, 2020 at 22:15