Given are $ g_n ~and~ k_n$ two step functions, such that for $f\geq0$, which is Riemann-integrable the following holds: $0\leq{g_n}\leq{f}\leq{k_n}$ We diefined Riemann's integrability of $f$ in our lecture as a function, having such a property: for every $n\in\mathbb{N}$ there are 2 step functions $ g_n ~and~ k_n$ (as given above) such that:
$\int_a^b \! (k_n(x)-g_n(x)) \, \mathrm{d}x<1/n$
Now we defined $h(x):=\sqrt{g_n(x)+1/n}-1/n$ and $s(x):=\sqrt{k_n(x)+1/n}$
I was able to prove, that $h(x)\leq{\sqrt{f}}\leq{s(x)}$
AND HERE IS MY QUESTION:
I want to prove, that: $\int_a^b \! (s(x)-h(x)) \, \mathrm{d}x<(b-a+1)\sqrt{n}$
I have tried to work with the integrand by using one of the bonominal formulas, but then I just came to a conclusion, that:
$\int_a^b \! (s(x)-h(x)) \, \mathrm{d}x\leq\int_a^b \! ((\sqrt{s(x)}-\sqrt{h(x)}+\sqrt{1/n}) \, \mathrm{d}x$ Now here I do not know, how to move further. Thank you
EDIT:
I made a bit more and figured out, that I have to prove this inequality:
$\int_a^b \! \frac{k_n(x)-g_n(x)}{\sqrt{k_n(x)+\frac{1}{n}}+\sqrt{g_n(x)+\frac{1}{n}}} \, \mathrm{d}x<\frac{1}{\sqrt{n}}$
But I see no way at the moment how to move on. Thank you
It seems the following.
Since the functions $g_n$ and $k_n$ are non-negative, we have $$\sqrt{k_n(x)+\frac{1}{n}}+\sqrt{g_n(x)+\frac{1}{n}}\ge 2\frac{1}{\sqrt{n}}.$$ Hence $$\int_a^b \! \frac{k_n(x)-g_n(x)}{\sqrt{k_n(x)+\frac{1}{n}}+\sqrt{g_n(x)+\frac{1}{n}}} \, dx\le$$ $$\int_a^b \! \frac{k_n(x)-g_n(x)}{2\frac{1}{\sqrt{n}}}dx=$$ $$\frac{\sqrt{n}}{2}\int_a^b \! (k_n(x)-g_n(x)) \,dx< \frac{\sqrt{n}}{2}\frac 1n=\frac{1}{2\sqrt{n}}.$$