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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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