functional-analysishilbert-spacesordinary differential equationssobolev-spaces
I consider space $H^{2}(0,a)=\{ f\in L^{2}(0,a): f',f''\in L^{2}(0,a) \}$
I define norm $\Vert w \Vert_{H^{2}}:=b\Vert w''\Vert_{L^{2}}$, where b is positive constant.
I couldn't proof that it is norm equivalent to standard norm in $H^{2}$.
Maybe is easier show that $H^{2}$ with this norm is a Hilbert space?
Could you help me?
First, I am not sure what the author means by Banach's theorem, but it seems like the relevant result from functional analysis which he's using is the bounded inverse theorem: if $T:(X,\left\|\right\|_{X})\rightarrow (Y,\left\|\right\|_{Y})$ is a continuous (i.e. bounded) bijective linear map between two Banach spaces $X$ and $Y$, then $T^{-1}: (Y,\left\|\right\|_{Y})\rightarrow (X,\left\|\right\|_{X})$ is a continuous (i.e. bounded) linear map. Note that the spaces don't need to be finite-dimensional, just Banach; if you don't know this result, you really should consult a text on functional analysis.
This result is used to obtain a contradiction, since if $T=I:(X,\left\|\right\|_{1})\rightarrow (X,\left\|\right\|_{2})$ is the identity map on $X$ and $\left\|\right\|_{1}$ and $\left\|\right\|_{2}$ are two norms on $X$ with respect to which $X$ is complete, then they are equivalent by the preceding result. The author derives a contradiction from this by exhibiting an example $u$ to show that there is no universal constant $C>0$ such that $$\left\|\widehat{u}\right\|_{L^{1}(B(0,1))}\leq C\left\|u\right\|_{\dot{H}^{s}}$$ Here is the example. Let $\mathcal{C}$ be the annulus $\left\{x\in\mathbb{R}^{d} : 1/4\leq\left|x\right|\leq 3/8\right\}$. Note that the annulus $2\mathcal{C}=\left\{1/2\leq\left|x\right|\leq 3/4\right\}$ is disjoitn from $\mathcal{C}$, and $\mathcal{C}\subset B(0,1)$. Define $$\Sigma_{n}:=\left(\sum_{j=1}^{n}\dfrac{2^{n(s+\frac{d}{2})}}{j}\chi_{2^{-j}\mathcal{C}}\right)^{\vee}$$ Since $2^{-j}\mathcal{C}\cap 2^{-i}\mathcal{C}=\emptyset$, $i\neq j$, we have that $$\left\|\widehat{\Sigma}_{n}\right\|_{L^{1}(B(0,1))}=\sum_{j=1}^{n}\dfrac{2^{j(s+\frac{d}{2})}}{j}2^{-jd}C=C\sum_{j=1}^{n}\dfrac{2^{j(s-\frac{d}{2})}}{j},$$ where $C=\left|\mathcal{C}\right|$, and \begin{align*} \left\|\Sigma_{n}\right\|_{\dot{H}^{s}}^{2}&=\int_{\mathbb{R}^{d}}\left|\xi\right|^{2s}\left(\sum_{j=1}^{n}\dfrac{2^{j(s+\frac{d}{2})}}{j}\chi_{2^{-j}\mathcal{C}}(\xi)\right)^{2}d\xi\\ &=\sum_{j=1}^{n}\int_{2^{-j}\mathcal{C}}\left|\xi\right|^{2s}\left(\dfrac{2^{j(s+\frac{d}{2})}}{j}\right)^{2}d\xi\\ &\leq \sum_{j=1}^{n}2^{-2js}2^{-jd}\left(\dfrac{2^{j(s+\frac{d}{2})}}{j}\right)^{2}\int_{\mathcal{C}}d\xi\\ &=C\sum_{j=1}^{n}\dfrac{1}{j^{2}}\leq C_{1}<\infty \end{align*} for all $n$. But since $s\geq d/2$, we see that $\left\|\widehat{\Sigma}_{n}\right\|_{L^{1}(B(0,1))}\uparrow\infty$ as $n\rightarrow\infty$, whence $N(\Sigma_{n})\rightarrow\infty$ as $n\rightarrow\infty$.
To see that $(\dot{H}^{s}(\mathbb{R}^{d}),N)$ is a Banach space, let $(u_{n})_{n}$ be a sequence in $\dot{H}^{s}(\mathbb{R}^{d})$ which is Cauchy with respect to $N$. The definition of $N$ implies that $\widehat{u}_{n}$ is a Cauchy sequence in $L^{2}(\mathbb{R}^{d};\left|\xi\right|^{2s}d\xi)$, which is complete, whence there exists an element $f\in L^{2}(\mathbb{R}^{d};\left|\xi\right|^{2s}d\xi)$ such that $(\widehat{u}_{n})$ converges to $f\in L^{2}(\mathbb{R}^{d};\left|\xi\right|^{2s}d\xi)$. Similarly, since $L^{1}(B(0,1))$ is complete, there exists an element $g\in L^{1}(B(0,1))$ such that $\widehat{u}_{n}$ converges to $g$ in $L^{1}(B(0,1))$.
I claim that $f=g$ a.e. on $B(0,1)$. To show this, we need the following simple lemma.
Lemma. $\left|\xi\right|^{2s}d\xi$ and Lebesgue measure $d\xi$ (we'll also use notation $\left|\cdot\right|$) are mutually absolutely continuous.
Proof. Since $\left|\xi\right|^{2s}$ is nonnegative, it is evident that $\left|\xi\right|^{2s}d\xi$ is absolutely continuous with respect to Lebesgue measure. Conversely, suppose $\int_{A}\left|\xi\right|^{2s}d\xi=0$. Then for any $\epsilon>0$, $$\int_{A}d\xi=\int_{A\cap B(0,\epsilon)}d\xi+\int_{A\setminus B(0,\epsilon)}d\xi\leq\int_{B(0,\epsilon)}d\xi+\epsilon^{-2s}\underbrace{\int_{A}\left|\xi\right|^{2s}d\xi}_{=0}\leq \left|B(0,1)\right|\epsilon^{d}$$ Letting $\epsilon\downarrow 0$ completes the proof. $\Box$
Since $\widehat{u}_{n}$ convereges to $f$ in $L^{2}(\mathbb{R}^{d};\left|\xi\right|^{2s}d\xi)$, passing to a subsequence if necessary, we may assume that $\widehat{u}_{n}\rightarrow f$ a.e. with respect to the measure $\left|\xi\right|^{2s}d\xi$. Similarly, since $\widehat{u}_{n}\rightarrow g$ in $L^{1}(B(0,1))$, passing to a further subsequence if necessary, we may assume that $\widehat{u}_{n}\rightarrow g$ (Lebesgue) a.e. on $B(0,1)$. The desired conclusion follows immediately from the above lemma.
We now use this result to show that the $L_{loc}^{1}(\mathbb{R}^{d})$ function $f$ is a tempered distribution. Then we can define $u$ to be the inverse Fourier transform of $f$, and we will have shown that $u\in\dot{H}^{s}$ and $u_{n}\rightarrow u$ with respect to the norm $N$.
For any Schwartz function $\varphi\in\mathcal{S}(\mathbb{R}^{d})$, we have \begin{align*} \int_{\mathbb{R}^{d}}\left|f(\xi)\varphi(\xi)\right|d\xi&\leq\int_{B(0,1)}\left|g(\xi)\varphi(\xi)\right|d\xi+\int_{\mathbb{R}^{d}}\left|\xi\right|^{s}\left|f(\xi)\right|\left|\varphi(\xi)\right|d\xi\\ &\leq\left\|g\right\|_{L^{1}(B(0,1))}\left\|\varphi\right\|_{L^{\infty}}+\left\|f\right\|_{L^{2}(\left|\xi\right|^{2s}d\xi)}\left\|\varphi\right\|_{L^{2}}, \end{align*} where we use Holder's inequality in the last step. Both $\left\|\varphi\right\|_{L^{\infty}}$ and $\left\|\varphi\right\|_{L^{2}}$ are controlled by a finite sum of Schwartz seminorms, from which the conclusion follows.
First problem: typically by $C^1(\mathbb R^n\times[0,\infty))$ we mean the space of continuously differentiable functions on $\mathbb R^n\times[0,\infty)$, with no assumption of boundedness. Clearly what you call "the standard norm" is not a norm on this space.
Even if we assume that the space consists of bounded functions with bounded derivatives, it's not a Banach space. Intuitively this is because the $y$-derivative is not involved in the norm. For an explicit counterexample, define
$$\psi_n(t):=\begin{cases} |t|,&|t|>\frac1n,\\ \frac n2t^2+\frac1{2n},&|t|\le\frac1n \end{cases}$$ and let $f_n(x,y)=\varphi(y)\psi_n(1-y)$, where $\varphi$ is a smooth function such that $\varphi=1$ on $[0,2]$ and $\varphi=0$ on $[3,\infty)$. Note that $\psi_n$ is continuously differentiable with $\psi_n(t)\to|t|$ uniformly, so $f_n(x,y)\to f(x,y):=\varphi(y)|1-y|$ uniformly and $\nabla f_n\equiv0$, which implies $\{f_n\}$ is Cauchy. But if $f_n\to g$ in $\|\cdot\|_{C^1}$ then also $f_n\to g$ uniformly, so $g=f\notin C^1$.
So then it seems like we are interested in the space of functions $f(x,y)$ which are bounded, continuous, and have bounded and continuous $x$-derivative $\nabla f(x,y)$. This space looks nothing like what anyone would call $C^1$, so I'm going to call it $\mathcal X$.
Next problem: if $c>0$, then $\|\cdot\|$ is not a norm on $\mathcal X$. Indeed, the constant function $1\in\mathcal X$, and yet $\|1\|=\infty$. If $c\le0$ then $\|\cdot\|$ is indeed a norm - this is easy to prove. However, unless $c=0$ (in which case $\|\cdot\|=\|\cdot\|_{C^1}$), $\mathcal X$ is not a Banach space under this norm. To see this, let $g_n$ be a smooth, non-decreasing function such that $g_n(t)=t$ for $t\le n-\frac1n$ and $g_n(t)=n$ for $t\ge n$, then define $f_n(x,y)=g_n(y)$. Observe that $f_n(x,y)e^{cy}\to ye^{cy}$ uniformly and again $\nabla f_n(c,y)\equiv0$, so $\{f_n\}$ is Cauchy in $(\mathcal X,\|\cdot\|)$. But if $\|f_n-f\|\to0$ then $f_n(x,y)e^{cy}\to f(x,y)e^{cy}$ uniformly, so $f(x,y)=y\notin\mathcal X$.
Basically, the answer to all your questions under any reasonable interpretation is "this is not true". There are two workarounds I can think of. One is to consider the space $\mathcal Y$ of continuous function $f(x,y)$ with continuous $x$-derivative $\nabla f(x,y)$ such that $\|f\|<\infty$. Then indeed we have that $\|\cdot\|$ is a norm on $\mathcal Y$ and $(\mathcal Y,\|\cdot\|)$ is a Banach space. It is a different space from $\mathcal X$, so it doesn't make a lot of sense to talk about the norm being "equivalent" to $\|\cdot\|_{C^1}$. What we have instead is an isometric isomorphism $T:(\mathcal Y,\|\cdot\|)\to(\mathcal X,\|\cdot\|_{C^1})$ given by $Tf(x,y)=f(x,y)e^{cy}$ (this works for any $c\in\mathbb R$, or indeed if you replacing $e^{cy}$ with any positive continuous function). The proof of this is trivial - we basically constructed the space $\mathcal Y$ so that $\|Tf\|_{C^1}=\|f\|$.
The other workaround is to replace $[0,\infty)$ with $[0,T]$. This perhaps makes the most sense, because then $\|\cdot\|$ is a norm on $\mathcal X$. Moreover, since $e^{cy}$ is bounded above and below by positive constants on $[0,T]$, it is equivalent to $\|\cdot\|_{C^1}$ and hence $(\mathcal X,\|\cdot\|)$ is a Banach space. This is again rather obvious, and you can replace $e^{cy}$ with any positive continuous function which is bounded above and below by positive constants.
Best Answer
This is false; consider $w = 1$ (a constant function on the interval). Then clearly $w$ has non-zero $H^2$-norm, but if you just take the $L^2$ norm of its derivative, it'll of course be zero.
However, it is true if you're considering the space $H^2_0(0,a)$, the space of $H^2$ functions with zero trace.