How to reduce the calculation of $\Vert f \Vert$ to $\Vert u \Vert$

Question

I know that for every $f \in H^{-1}(a,b)$ there exists a function $u_f \in L^2(a,b)$ such that $$\langle f,v\rangle = -∫_a^bu_f(x)v'(x) \ dx, \ \ \ v\in H_0^1(a,b).$$

How can I reduce the calculation of $\|f\|_{-1,2}$ to the calculation of $\|u_f\|_{H_0^2(a,b)}$ where $\|f \|_{-1,2}:= \sup_{v \in H^1_0(a,b)\backslash\{0\}}\frac{|\langle f,v\rangle|}{\vert v\vert_{H_0^1(a,b)}}$?

Answer 1

To ensure the uniqueness of $u_f$ you need some additional condition (as $u_f$ is defined up to a constant).

Let us take $\int_a^b u_f(x) dx=0$.

• Then using C-S, for all $v \in H^1_0$: $$\left|\langle f,v \rangle \right| =\left|\int_a^b u_f(x) v'(x) dx \right| \leq \|u_f\|_{L^2} \|v'\|_{L^2} =\|u_f\|_{L^2} \|v\|_{H^1_0} $$ so taking the $\sup$: $$\|f\|_{-1,2} \leq \|u_f\|_{L^2}$$

• To show the equality let $v \in H^1_0$ such that: $$v'=u_f$$ (here the condition $\int_a^b u_f(x) dx=0$ is important to have the right boundaries conditions, and if your functions are complex consider $v'=\overline{u_f}$). Then: $$\left|\langle f,v \rangle \right|=\int_a^b |u_f|^2(x) dx =\|u_f\|^2_{L^2}$$ and: $$\|v\|_{H^1_0}=\|v'\|_{L^2}=\|u_f\|_{L^2}$$ so: $$\frac{\left|\langle f,v \rangle \right|}{\|v\|_{H^1_0}}=\|u_f\|_{L^2}$$

To conclude you have: $$\|f\|_{-1,2}=\|u_f\|_{L^2}$$ where $u_f$ is chosen such that $\int_a^b u_f(x) dx=0$ or for any $u_f$: $$\|f\|_{-1,2}=\left\|u_f-\int_a^b u_f(x) dx\right\|_{L^2}$$

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Edit: why we can take $\int_a^b u_f(x) dx=0$.

Let $u_f \in L^2$ such that $$\langle f,v\rangle = -\int_a^b u_f(x)v'(x) \ dx, \ \ \forall v\in H_0^1(a,b) \tag{1}$$ and let $C$ be any constant.

Then for all $v \in H_0^1(a,b)$ you have: $$\int_a^b Cv'(x) dx=v(b)-v(a)=0$$ so: $$-\int_a^b (u_f(x)+C) v'(x) dx=-\int_a^b u_f(x) v'(x) dx-\int_a^b C v'(x) dx=\langle f ,v \rangle +0$$ i.e $u_f +C$ also satisfies $(1)$.

With the particular choice of $C=-\frac{1}{b-a}\int_a^b u_f(x) dx$: $$\tilde{u_f}=u_f-\frac{1}{b-a}\int_a^b u_f(x) dx$$ you obtain:

• $\tilde{u_f} \in L^2$ and $$\langle f,v\rangle = -\int_a^b \tilde{u_f}(x)v'(x) \ dx, \ \ \forall v\in H_0^1(a,b) $$

• and: \begin{align}\int_a^b \tilde{u_f}(x) dx&=\int_a^b \left( u_f(x) -\frac{1}{b-a}\int_a^b u_f(s) ds\right) dx \\ &=\int_a^b u_f(x) dx-(b-a) \frac{1}{b-a}\int_a^b u_f(s) ds =0\end{align}

so $\tilde{u_f}$ is the same as $u_f$ and has a zero mean.