From the Fundamental lemma of calculus of variations we know the following: Let $u \in L_{loc}(a,b)$ and for every $\phi \in C_0^\infty$, $$\int_a^b u(x) \phi(x) dx = 0$$ holds, then $u(x) = 0$ almost everywhere. From that I need to prove the following Corollary: Let $u \in L_{loc}(a,b)$ and for every $\phi \in C_0^\infty$, $$\int_a^b […]

I’m tying to understand distributional derivatives. That’s why I’m trying to calculate the distributional derivative of $|x|$, but I got a little confused. I know that a weak derivative would be $\operatorname{sgn}(x)$, but not only I’m not finding that one in my calculations, I ended up with a wrong solution and in my other attempt […]

For a math paper I need to be able to evaluate $\int_{-a}^{a}\delta^{(n)}(x)\ f(x)\ dx$ for differentiable $f$. I know that it is ‘supposed’ to equal $(-1)^nf^{(n)}(0)$: $$\int_{-a}^a\delta^{(n)}(x)\ f(x)\, dx=\int_{-a}^a \frac{d^n}{dx^n}\delta(x) f(x)\, dx =\int_{-a}^a(-1)^n\delta(x) \frac{d^nf}{dx^n}\, dx=(-1)^nf^{(n)}(0).$$ This argument seems like a hack and I have no idea what is actually going on. I want to write […]

Let $d\in\mathbb N$ $\Omega\subseteq\mathbb R^d$ be open $\mathcal D(\Omega):=C_c^\infty(\Omega)$ and $\mathcal D(\Omega,\mathbb R^d):=C_c^\infty(\Omega,\mathbb R^d)$ $H^{-1}(\Omega):=H_0^1(\Omega)’$ and $H^{-1}(\Omega,\mathbb R^d):=H_0^1(\Omega,\mathbb R^d)’$ Let me cite a well-known theorem: Let $u\in\mathcal D'(\Omega,\mathbb R^d)$ $\Rightarrow$ $$\left.u\right|_{\mathfrak D(\Omega)}=0\;\Leftrightarrow\;\exists p\in\mathcal D'(\Omega):u=\nabla p\tag 1$$ where $$\mathfrak D(\Omega):=\left\{\phi\in\mathcal D(\Omega,\mathbb R^d):\nabla\cdot\phi=0\right\}\;.$$ Now, if $F\in H^{-1}(\Omega,\mathbb R^d)$, then $$f:=\left.F\right|_{\mathcal D(\Omega,\:\mathbb R^d)}\in\mathcal D'(\Omega,\mathbb R^d)\;.$$ And since $\mathfrak […]

Show that there exists a unique $v_0 \in H^1(0,1)$ such that $u(0)=\int_0^1(u’v_0’+uv_0), \forall u \in H^1(0,1)$. Further Show that $v_0$ is the solution of some differential equation with appropriate boundary conditions. Compute $v_0$ explicitly. Let $f: H^1(0,1) \to \mathbb{R}$ be defined by $f(u)=u(0)$. Then I showed that $f$ is linear and continuous. Hence there exists […]

I am working in $H^1(S^1)$, the space of absolutely continuous $2\pi$-periodic functions $\mathbb R\to\mathbb R^{2n}$ wih square integrable derivwtives. I have a sequence $z_j$ (for the record, it comes from minimizing a functional, in the middle of Hofer-Zehnder’s proof that a Hamiltonian field has a periodic orbit on a strictly convex compact regular energy surface) […]

Note – I am just starting to learn about theory of distributions, so this may be a trivial question, if so I’d be grateful for a reference, nevertheless the question is the following: suppose I have a function $f$ such that $f \in L^1$ and $f$ is differentiable almost everywhere (in the strong sense) and […]

I’m trying to figure out if convergence in $H^1(a,b)$ implies pointwise convergence (by the way: what is the usual name of this space?). It is defined to be Hilbert space of absolutely continuous functions on some (possibly infinite) interval such that both function and its (weak) derivative are square integrable. Scalar product in this space […]

My quesion involves the weak time derivative. In the book: ‘Partial Differential Equations’ by Evans the time derivative $u’$ of a function $u: [0,T] \rightarrow H^1_0(U)$ is defined by an element $u_t \in L^2(0,T;H^{-1}(U))$ such that $$ \forall \phi \in C_0^{\infty}([0,T]): \int_{[0,T]} u \phi’ dt = -\int_{[0,T]} u’ \phi dt $$ Where $u’\phi$ and $u […]

Let $\Omega \subseteq \mathbb{R}^n$ (open) and $u \in \mathcal{D}’$ be a distribution, that has a distributional derivative which is in $L^p(\Omega)$ (for some $p \geq 1$). Show that $u \in L^p_{loc}(\Omega)$ …where $L^p(\Omega)$ is defined as in https://en.wikipedia.org/wiki/Lp_space#Lp_spaces and $L^p_{loc}(\Omega)$ is defined as in: https://en.wikipedia.org/wiki/Locally_integrable_function#Generalization:_locally_p-integrable_functions My thoughts: 1) For notational simplicity let’s call $Du = […]

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