Show that given two norms are equivalent 
I have proved that $\sigma$ is a scaler product on $H$. So question (a) is done.
We have $N_\sigma^2(f)=\lVert f\rVert^2=\sigma(f,f)=f(1)^2+\int_0^1 \frac{e^x(f'(x))^2}{1+x^2}\,dx$.
I am unable to estimate the inequalities. Please help.
 A: Well, you have
$$\begin{aligned}N_\sigma^2 (f) &= f^2(1)+\int_0^1 \frac{e^x \left(f^\prime(x)\right)^2}{1+x^2} \ dx\\ &\le f^2(1) + e \int_0^1 \left(f^\prime(x)\right)^2 \ dx\\
&\le e\left(\Vert f \Vert_\infty^2+ \int_0^1 \left(f^\prime(x)\right)^2 \ dx\right)
\end{aligned}$$
and therefore $N_\sigma(f) \le \sqrt e N^\prime(f)$.
Now using the inequality
$$\vert f(x) \vert \le \vert f(1) \vert + \left(\int_0^1 \left(f^\prime(x)\right)^2 \ dx\right)^{1/2}$$ provided in the hint of the question you get
$$\Vert f \Vert_\infty \le \vert f(1) \vert + \left(\int_0^1 \left(f^\prime(x)\right)^2 \ dx\right)^{1/2} $$ and
$$\Vert f \Vert_\infty^2 \le 2 \left(\vert f(1) \vert^2 + \int_0^1 \left(f^\prime(x)\right)^2 \ dx\right) $$ as $2 ab \le a^2+b^2$ for any $a,b \in \mathbb R$.
By adding $\int_0^1 \left(f^\prime(x)\right)^2 \ dx$ on both sides of this last inequality, you obtain
$$\begin{aligned}\Vert f \Vert_\infty^2 +\int_0^1 \left(f^\prime(x)\right)^2 \ dx &\le 2 \vert f(1) \vert^2 + 3\int_0^1 \left(f^\prime(x)\right)^2 \ dx\\
&\le 6 \left( \vert f(1) \vert^2 + \int_0^1 \frac{e^x \left(f^\prime(x)\right)^2}{1+x^2} \ dx\right)
\end{aligned}$$ as $\frac{1}{2} \le \frac{e^x}{1+x^2}$ for $x \in [0,1]$.
Finally $$(1/ \sqrt 6) N^\prime(f) \le N_\sigma(f).$$
