If $f \in H^{1}(I)$, then $\|f\|_{H^{-1}} = \|f\|_{H^{1}}$ with $|I| < \infty$.

Question

Let $f \in H^{1} = H^{1}(I)$ with $|I| < \infty$. Then, $$ \|f\|_{H^{-1}} = \|f\|_{H^{1}} $$ where $H^{-1}(I) = (H_{0}^{1})^{\prime}$ and $\|f\|_{H^{1}} = \|f\|_{L^{2}(I)} + \|f^{\prime}\|_{L^{2}(I)}$.

I know that $L^{2}(I) \subset H^{-1}(I)$ with continuous injection, then there is $C_{1}>0$ such that $\|f\|_{H^{-1}} \leq C \|f\|_{L^{2}(I)} \leq C_{2}\|f\|_{H^{1}}$ for $C_{2} > 0$.

On the other hand, since $f \in H^{-1}$ then there is $f_{0} \in L^{2}(I)$ such $$ \|f\|_{H^{-1}} = \|f_{0}\|_{L^{2}(I)}. $$ But, how to show that there is $C_{3} > 0$ such that $\|f\|_{H^{-1}} \geq C_{3}\|f\|_{H^{1}}$?

Answer 1

Let $f\in H^1$, and note that $f\in H^{-1}$ can be represented by an operator $T_f:H^{1}_0\rightarrow \mathbb{R}$ such that $T_f(g)=\left<f,g \right>_{H^{-1}\times H_0^1}=\left< f,g \right>_{H_0^1}$.(For example, see Riesz representation theorem.) Then observe that \begin{align*} \left| \left< f,g \right>_{H^{-1}\times H^{-1}} \right|\leq \|f\|_{H_0^1}\|g\|_{H_0^1}. \end{align*}

It implies that $\|f\|_{H^{-1}}\leq \|f\|_{H_0^1}$.

Also, in the case when $g=f$, we have \begin{align*} |T_f(f)|=|\left<f,f \right>_{H^{-1}\times H_0^1}| = \|f\|_{H_0^1}^2 \end{align*} which implies that $\|f\|_{H_0^1}=\|f\|_{H^{-1}}$.

Here, I used $$\|f\|_{H_0^1}^2 = \int |\nabla f|^2$$ which is equivalent to with your norm by Poincare inequality.

Thus, in your representation, it would be $\|f\|_{H^{-1}}< \|f\|_{H^1}$ unless $f=0$ almost everywhere.

Answer 2

Before I give my answer, the short answer is: a dual space is not always canonically identified with a function space, so you have to choose how you make the identification. You should be aware of this.

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I think you are having some confusion on the space $H^{-1}$. If you define it as the dual of $H^1_0(I)$, that is just a vector space of linear functionals on $H^1_0$, it is not a space of functions, or of distributions. Then, it is then not immediate how you identify vectors in $H^{-1}$ with functions/distributions on $I$. There is more than one way to make this identification, and the answer to your question depends on the identification.

1. You use Riesz representation theorem (morally, $H^1_0$ is a Hilbert space, so it coincides with its dual), in that case $H^{-1}$ is isomorphic to $H^1_0$ as a Hilbert space, and the statement you want to prove is obvious. The identification is given by $$ g\in H^1_0(I)\mapsto \phi\in H^{-1}(I) $$ such that $$ \phi(f)=\int (fg+f’g’)dx=\langle f,g\rangle_{H^1_0\times H^1_0}. $$ Note that in this case, it makes no sense to write that $\|f\|_{H^{-1}}\leq C\|f\|_{L^2}$, this is simply not true with the above identification, because the norm of $H^{-1}$ essentially coincides with the $H^1$ norm, so the inequality should be reversed.

2. The second way is as follows. You identify vectors in $H^{-1}(I)$ as distributions by using the scalar product $$ \int fg\,\,dx=\langle g,f\rangle_{H^{-1}\times H^1_0}. $$ The identification is then given by $$ g\in H^1_0(I)\mapsto \phi\in H^{-1}(I) $$ such that $$ \phi(f)=\int fg\,\,dx=\langle f,g\rangle_{H^{-1}\times H^1_0}. $$ If you make this identification, you clearly have $\|f\|_{H^{-1}}\leq C\|f\|_{L^2}$. But at this point, $H^{-1}$ is muuuuuuch larger than $L^2$, which in turn is much larger than $H^1_0$. In particular, it does not hold $$ \|f\|_{H^1_0}\leq C\|f\|_{H^{-1}}.$$ It is simply false.

So, to summarize, what you wrote in your question is contradictory and the question itself is not clear. In particular, it is not clear how you identify vectors of $(H^1_0)’$ with functions, or distributions (in different parts of the question, you are using different identifications). If you make the identification (1), what you want to prove follows by Riesz representation theorem and you don’t need to do anything else: it is just a property of Hilbert spaces that can be identified with their dual spaces, with equality of norms. If you make identification (2), then the statement is simply not true, and it is not even true that there exists $C$ such that $$ \|f\|_{H^1}\leq C\|f\|_{H^{-1}}. $$

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So, every time you deal with a dual space, you need to really think about what you are doing, otherwise you end up having an appearant paradox.