I have been trying to check whether the topologies induced by two norms on Hilbert space are different. So I thought by checking (topological) equivalence of the two norms, I could solve the problem: if the norms aren't equivalent, then the topologies would be different.
Let there be given two norms $\left\|\cdot\right\|_2$, $\left\|\cdot\right\|^*_2$ defined on the Hilbert product space $H=H^2_0(0,1)\times L^2(0,1)$ by
$$
\left\|\left(\begin{matrix}
f\\
g
\end{matrix}\right)\right\|_2:=\int_0^1(\left|f''\right|^2+\left|g\right|^2)\,dx,\quad \left\|\left(\begin{matrix}
f\\
g
\end{matrix}\right)\right\|^*_2:=\int_0^1(\left|f''\right|^2+a\left|f'\right|^2+\left|g\right|^2)\,dx,\quad \left(\begin{matrix}
f\\
g
\end{matrix}\right)\in H,
$$
respectively ($a>0$ some constant). Here $H^2_0(0.1):=\left\{f\in H^2(0,1):f(0)=f'(0)=0\right\}$ and $H^2(0,1)$ denotes the Sobolev space of order 2 associated with $L^2(0,1)$.
I suspect the topologies induced by these two norms are different, which I would like to prove (or disprove).
Any suggestions how to proceed with this?
Best Answer
The two norms are equivalent. First remember that $\|\cdot\|_2$ and $\|\cdot\|_2^* $ are equivalent if and only if there exist some positive constants $c_1$ and $c_2$ such that for all $h\in H$, $c_1\|h\|_2^*\le \|h\|_2\le c_2\| h\|_2^* $.
Now, if we denote by $\|f\|^2 := \int_0^1 |f(x)|^2dx $ the $L^2(0,1)$ norm, we have that for all $(f,g)\in H$ that
$$\|(f,g)\|_2^2 = \|f''\|^2 + \|g\|^2,\quad \big[\|(f,g)\|_2^*\big]^2 = \|f''\|^2 + a\|f'\|^2 + \|g\|^2 $$
It is pretty clear that $ \|(f,g)\|_2\le \|(f,g)\|_2^*$, so now we only need to find a positive constant $C$ such that $\|(f,g)\|_2^*\le C\|(f,g)\|_2$ for all $(f,g)\in H$.
If $(f,g)\in H$ then by definition $f\in H_0^2(0,1)$ and in particular $f'\in H_0^1(0,1)$, so $f'$ has a continuous representative and we can apply the fundamental theorem of calculus : for all $x\in[0,1], f'(x) = f'(0) + \int_{0}^x f''(t)\ dt = \int_{0}^x f''(t)\ dt$.
Hence by Cauchy-Schwarz we have for all $x\in[0,1]$ $$ |f'(x)|^2 =\left| \int_{0}^x 1\cdot f''(t) \ dt \right|^2\le x\cdot \int_{0}^x |f''(t)|^2 \ dt\le \|f''\|^2 $$
And so by integrating we find that $$\|f'\|^2 := \int_0^1 |f'(x)|^2\ dx \le\int_0^1 \|f''\|^2\ dx = \|f''\|^2 $$
And finally $$\|f''\|^2 + a\|f'\|^2 + \|g\|^2\le (1+a)\|f''\|^2 +\|g\|^2\le (1+a)\left(\|f''\|^2 +\|g\|^2\right) $$
So the desired inequality holds with $C:=\sqrt{1+a}$, and we can conclude that the two norms are indeed equivalent.