# Let $X$ be dense in $Y$. Looking for an example such that a bounded bilinear form fulfills an inf-sup condition on $X\times Y$ but not on $X\times X$

Let $X$ and $Y$ be two separable Hilbert spaces such that $X \subsetneq Y$ densely, continuously and the norms are related through $\|x\|_X^2=|x|_X^2+\|x\|_Y^2$ for all $x \in X$; e.g. take $X=H^1$, $Y=L^2$. Here, $|\cdot|_X$ denotes a semi norm on $X$. Further, let $$a: X \times Y \to \mathbb{R}$$ be a bounded bilinear form, i.e. there is a $c$ such that $|a(x,y)|\leq c \|x\|_X \|y\|_Y$ for all $x \in X$, $y \in Y$. Additionally $a(\cdot,\cdot)$ shall fulfill a inf-sup condition on $X \times Y$, i.e. there is a $\beta>0$ such that

$$\inf_{x \in X} \sup_{y \in Y} \frac{|a(x,y)|}{\|x\|_X \|y\|_Y}\geq \beta>0.$$

There is a paper, which states in Remark 2.10 that in general one can't follow that an inf-sup condition on $X \times X$ holds i.e. that there is a $\beta_X>0$ such that

$$\inf_{u \in X} \sup_{v \in X}\frac{|a(u,v)|}{\|u\|_X \|v\|_X}\geq\beta_X>0.$$

Now I am looking for a bounded bilinear form, which has inf-sup condition of $\boldsymbol{X \times Y}$, but not on $\boldsymbol{X \times X}$.

My try:

I tried with $X=H^1_0(0,1), Y=L^2(0,1)$. We have an inf-sup condition on $H_0^1 \times L^2$ with the bilinear form $a(u,v):=\int_0^1 u' \cdot v \, dx=(u',v)_{L^2}$ because

$$\sup_{v \in L^2} \frac{|(u',v)_{L^2}|}{ \|v\|_{L^2}}\geq \frac{|(u',u')_{L^2}|}{ \|u'\|_{L^2}} = \|u'\|_{L^2}=|u|_{H_0^1}\geq C \|u\|_{H^1}$$

for all $u \in H_0^1$ by Poincare's inequality (here $|\cdot|_{H_0^1}$ is even a full norm). Dividing by $\|u\|_{H^1}$ and taking the inf over all $u \in H_0^1$ yields the inf-sup condition on $H_0^1 \times L^2$.

Now it would be great if $a(\cdot,\cdot)$ has no inf-sup condition on $H_0^1 \times H_0^1$. So I have to show that there is no $\beta_X>0$ such that

$$\inf_{u \in H_0^1} \sup_{v \in H_0^1} \frac{|(u',v)_{L^2}|}{\|u\|_{H^1} \|v\|_{H^1}}\geq\beta_X>0.$$

Or equivalently I could show that for all $\beta_X>0$ there is a $u \in H_0^1(0,1)$ such that $$\sup_{v \in H_0^1} \frac{|(u',v)_{L^2}|}{\|v\|_{H^1}}<\beta_X \|u\|_{H^1}$$

Does someone have an idea? Up to now I had no success; somehow I have to choose this $u$ in a smart way - likely depending on $\beta_X$. Another example with different Hilbert spaces or a different bilinear form is also fine; since I don't know if my example even works.

As a response to H. H. Rugh's answer:

• If we have $\|v\|_{H^1}\leq 1$, then $|v(x)|=|v(0)+\int_0^x v'(y)dy|\leq \|v'\|_{L^1}\leq \|v'\|_{L^2} \leq \|v\|_{H^1}\leq 1$ hence $|v(x)| \leq 1$
• Let $u_\varepsilon$ be piecewise continuous such that $u_\varepsilon(\tfrac12 \pm \varepsilon)=0$ and $u_\varepsilon(\tfrac12)=\sqrt{\varepsilon}.$ Then $$u_\varepsilon'(x)=\begin{cases} 1/\sqrt{\varepsilon}, &x \in (\tfrac12 -\varepsilon, \tfrac12) \\ -1/\sqrt{\varepsilon}, &x \in (\tfrac12,\tfrac12 +\varepsilon) \end{cases}$$
• Computing the norms we have $\|u_\varepsilon'\|_{L^1}=\tfrac{2}{\sqrt{\varepsilon}}(\tfrac12-\tfrac12+\varepsilon)=2\sqrt{\varepsilon}$ and $\|u_\varepsilon'\|_{L^2}=\sqrt{2}$.
• We have $$\sup_{\|v\|_{H^1} \leq 1} \frac{|\int_0^1 u_\varepsilon' v \, dx|}{\|u_\varepsilon\|_{H^1}} \leq \sup_{\|v\|_{H^1} \leq 1} \frac{|\int_0^1 u_\varepsilon' \, dx|}{\|u_\varepsilon'\|_{L^2}}\leq \frac{\|u_\varepsilon'\|_{L^1}}{\|u_\varepsilon'\|_{L^2}}=\sqrt{2 \varepsilon} \xrightarrow{\varepsilon \to 0} 0$$ and hence for every constant $\beta_X>0$ we have a $u_\varepsilon$ (as chosen above) such that $$\underbrace{\underset{\|v\|_{H^1} \leq 1}{\sup}\frac{|\int_0^1 u_\varepsilon' v \, dx|}{\|u_\varepsilon\|_{H^1}}}_{\to 0} < \beta_X.$$

For fixed $u\in H^1_0$: If $v\in H^1_0$ has norm one, then $|v(x)|\leq 1$ for all $x$ so the sup over such $v$'s is bounded by $$\frac{1}{\|u\|_H} \int_0^1 |u'| dx$$ This can be made arbitrarily small by choosing a suitable spike for $u$, e.g. piecewise linear with $u(\frac12)={\sqrt{\epsilon}}$, $u(\frac12\pm \epsilon)=0$ and letting $\epsilon\rightarrow 0$. The H-norm stays bounded away from zero but the integral goes to zero.

• Thank you very much for your answer. I think I got it, can you please look at my question again? I edited it and wrote down how I understood your answer. – Fritz Sep 3 '17 at 11:50
• @MarvinF Since $v(0)=0$ (when in $H_0^1$) you have just $|v(x)|\leq \int|v'| \times 1 \leq \|v'\|_{L^2}$ (Cauchy-Schwarz). For the second part $u'$ is non-zero only on the interval $[-\epsilon,\epsilon]$ so the square-norm is just 2. A part from this, your interpretation is fine. – H. H. Rugh Sep 3 '17 at 13:53