I'm doing an exercise where I have to prove the derivative operator on the space of differentiable functions in $L^2(-∞,∞)$ where their derivative is in $L^2(-∞,∞)$ is unbounded using the standard $L^2$ norm $$\|f\|^2 = \int_{-∞}^{∞}|f(t)|^{2}dt$$ (which I did by means of counterexample). I then had to prove that introducing a new norm $\|f\|_D^2 = \|f\|^2 + \|f'\|^2$ makes this linear operator bounded.
My initial approach was to use integration by parts combined with the fact that the limit of a function in $L^2(-∞,∞)$ as $t$ tends to $\pm \infty$ has to be $0$ but I can't get this approach to work. How would I show that this norm makes the differentiation operator bounded?
Suppose one has a linear operator $T$ between normed spaces $(X,\|\ \|_1)$ and $(Y,\|\ \|_2)$. Then $T$ may not be a bounded operator, but one can make it so by changing the norm on $X$. Define $\|x\|_1'=\|x\|_1+\|Tx\|_2$. This is a norm on $X$ and now $T$ is bounded as a map from $(X,\|\ \|_1')$ to $(Y,\|\ \|_2)$. Indeed $\|Tx\|_2\le\|x\|_1$.
In this case $X$ is the subset of functions in $L^2$ with differentiable derivative, $Y$ is $L^2$ and $\|\ \|_1$ and $\|\ \|_2$ are the $L^2$ norms. Also $T$ is differentiation, and there is a slightly different recipe to create a new norm on $X$, but the principle still applies.