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3

You can formalize this: Take two Hilbert spaces $H\subset V$ such that The inclusion is dense (i.e. the image of $H$ in $V$ is dense in the topology of $V$) and, The inclusion is continuous (i.e. $H$ has a stronger norm than $V$). Then, if we identify $V$ with its dual (let's assume the spaces are real, though you can do this in general), we have ...


3

You have all the ingredients ready: To show that $\ell$ is bounded, we need $|\ell(f)| \le C||f||_{H^1}$ for some $C$. Now there is $\zeta$ (depending on $f$) so that $$\ell(f) = f(c) = f(\zeta) + \int_\zeta^c f'(x)dx = \int_a^b f(x) dx + \int_\zeta^c f'(x)dx$$ Then $$|\ell (f)| \le \int_a^b |f(x) |dx + \int_\zeta^c |f'(x)| dx \le \cdots$$


3

Your examples are lacking squares (think of homogeneity), but apart from that they are both valid norms for $H^4(I) \cap H^2_0(I)$, as is the normal $L^2$-norm. What you really want in most situations is not just any norm, but a norm with which the space is complete. This is not true for the $H^2$-norm: There holds by definition $$C^\infty_0(I) \subset ...


2

In general, no. Take $U = (0,1) \subset \mathbb{R}^1$, $p=2$ and $F(z,w,x) = z^4$. Then take $$w_n(x) = \begin{cases} n^{1/4} x, & 0 < x < n^{-1} \\ n^{-3/4}, & x \ge n^{-1} \end{cases}$$ so that $w_n' = n^{1/4} 1_{(0, n^{-1})}$. It's easy then to see that $w_n' \to 0$ and $w_n \to 0$ in $L^2$, so $w_n \to 0$ in $W^{1,2}(U)$. Taking $w = ...


2

In the case $q<p^*$ the Rellich–Kondrachov theorem applies, if your domain $\Omega$ has the required properties. If $q=p^*$, the $L^q$ norm is still controlled by the $W^{1,p}$ norm; and since a weakly convergent sequence is norm-bounded, $L^q$ norm is bounded. Moreover, you will have weak convergence in $L^q$, since linear maps preserve weak ...


1

Use integration by parts and the Riesz representation theorem. For the one inclusion, if $u\in H^4(I) \cap H^2_0(I)$, we can integrate by parts twice to obtain $$a(u,v) = \int\limits_I u''(x)v''(x)\,dx = \int\limits_I u^{(4)}(x)v(x)\,dx$$ and see that $$\lvert a(u,v)\rvert \leqslant \lVert u^{(4)}\rVert_{L^2(I)}\cdot \lVert v\rVert_{L^2(I)}.$$ For the ...


1

The following claim is not true in general. Let $X$ and $Y$ be Banach spaces and $X_1 \subset X$ a subspace. If $T: X_1 \to Y$ is a continuous bounded operator, then there exists a continuous linear extension to the whole space $X$, i.e. there exists a bounded linear operator $\hat T: X \to Y$ such that $ \hat{T}\restriction_{X_1} = T$. Take for ...


1

The argument is the following: Let $v_n$ be a sequence in $W^{1, p}_0(\Omega)$ for some $\Omega \subset \mathbb R^N$ and $p >N$. If $v_n \to v$ weakly in $W^{1,p}_0(\Omega)$, then $||v_n||_{1, p}$ is uniformly bounded. By the Sobolev Embedding (Theorem 7.17) and the fact that $$C^{0, \alpha}(\overline \Omega) \to C(\overline \Omega)$$ is compact, there ...


1

Okay so (strong) measurability of $f$ is equivalent to weak measurability and $f$ being a.e. separably valued. Since $H^{-1}(\Omega)$ is separable we get the equivalence you mentioned. Now to show weak measurability we have to show that for every $L\in H^{-1}(\Omega)^*$ the map $t\mapsto \langle f(t), L\rangle$ is measurable. So far this is exactly what you ...


1

For $H^1$- functions, this does not hold in general. It is basically the same argument that $L^2$- functions generally do not vanish at $\infty$, you just have to take such a function and integrate it, for example a bump function where the bumps get thinner when you go outside. For $H^2$-functions however, it does hold: Take $$\int_0^a ...


1

An "abstract" construction can be done as follows: Take $u$ the solution of the problem $-\Delta u= 1$ in $\Omega$ and $u=0$ on $\partial \Omega$. If $\Omega$ is smooth enough then $u\in H^2(\Omega)$ by elliptic regularity. On the other hand, each partial derivative is harmonic so if $u\in H_0^2(\Omega)$, this would mean that the partial derivatives have ...


1

$C_c^\infty$, the space of $C^\infty$ functions with compact support is dense in $W^{1,2}$. Let $\{\phi_n\}$, $\{\psi_n\}$ be sequences in $C_c^\infty$ converging to $\phi$ and $\psi$ respectively in $W^{1,2}$. We have $$ \int_{\mathbb{R}}\phi_n\,\psi'_n=-\int_{\mathbb{R}}\phi'_n\,\psi_n. $$ Taking limits as $n\to\infty$ we get $$ ...


1

Since $u' \in H_0^1(I)$, you can apply Poincaré to $u'$. This gives $$\|u'\|_{H^1} \le C \, \|u''\|_{L^2}.$$ Can you conclude?


1

I think you have the inclusion backwards: For instance the function $f(t)=|t|^{1/2}$ is Holder continuous in, say, $\Omega=(-1,1)$ but $f'(t)=t^{-1/2}\notin L^2(\Omega)$, therefore the inclusion $C^{0,1/2} \to W^{1,2}$ fails. To prove that every $u\in W^{1,p}$ has a (locally absolutely) continuous representative see here.


1

Suppose that $f_n$ is a Cauchy sequence in $W^{1,2}$ and write $$f_n(b)-f_n(a)=\int_a^b f_n'(t)dt.\tag{1}$$ By one hand, there is $f\in L^2$ such that $f_n\to f$. On the other hand, there is $g\in L^2$, such that $f'_n\to g$. From the estimate $\|f_n\|_\infty\le K\|f_n\|_{1,2}$, we must conclude that $f_n(x)\to f(x)$ for all $x\in\mathbb{R}$ and by using ...


1

Assume $c=0$. The desired result follows easily when we consider the Gagliardo norm on $H^{\frac 12}$ and from the identity $|u^+(x)-u^+(y)| \leq |u(x)-u(y)|$ for any $u \in H^{\frac 12}$. Am I right?



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