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The answer to your question is negative. Take the smooth function $\chi$ defined by $$ \chi(t)=H(t) e^{-t^{-1}-t^2}, \quad H=\mathbf 1_{(0,+\infty)}. $$ This function is in $L^1(\mathbb R)$, $C^\infty …

Take $k=1$ in your statement. There are two easy cases: the first one is when $f$ is real-valued, then you have only to use the implicit function theorem to get a normal form $t+a(x)$, up to a unit (a …

A most important result is missing in the previous answers, namely the characterization by Lars Hörmander of hypoellipticity in his seminal paper, On the theory of general partial differential operato …

One of the results due to Georges Glaeser is the following: there exists a non-negative $C^\infty$ function $f$ on the real line, flat at its zeroes, such that $\sqrt{f}$ is not $C^2$. On the other ha …

Let $(X,\mathcal M, \mu)$ be a measure space, where $\mu$ is a positive measure and $X$ is topological space. Let $\mathcal B$ the Borel $\sigma$-algebra on $X$. The measure $\mu$ is called a Borel m …

The Hardy-Littlewood-Sobolev Inequality says that $$\text{for $p,q,r\in (1,+\infty)$ such that }\quad 1-\frac1p+1-\frac1q=1-\frac1r,\tag {$\sharp$} $$ $$ \exists C, \forall u\in L^p(\mathbb R^n),\qu …

With $H=\mathbf 1_{\mathbb R_+}$, $t,x$ real, $ H(t)t^{-1/2}e^{-x^2/t} $ is the fundamental solution of the heat equation, $C^\infty$ everywhere except at $(0,0)$, analytic only outside $t=0$.

When the roots are simple, they can be chosen as smooth functions of $\omega$ if the matrix $A$ is smooth of $\omega$ ; both "smooth" above can be replaced by "analytic". This is a consequence of the …

We note that $u$ is Hölder continuous, and we may write for any $M>0$ (to be chosen later), $$ u(x)=\int_{\vert \xi\vert\le M} e^{2π i x\xi}\hat u(\xi)(1+\vert \xi\vert) (1+\vert \xi\vert)^{-1}d\xi + …

Your operator $K$ is a Hilbert-Schmidt operator since its kernel belongs to $L^2$. As a result this is a compact operator whose spectrum contains a sequence of eigenvalues $\\{\lambda_k\not=0\\}$ with …

We have with $\chi\in C^\infty_c(\mathbb R)$, equal to $1$ near 0, $$ \frac{1}{\sinh t}=\frac{\chi(t)}{\sinh t}+\frac{1-\chi(t)}{\sinh t}. $$ The second function belongs to $L^1$ and Young's inequalit …

Let me assume that $a=-\infty, b=+\infty, x_0=0$ and $f$ smooth and compactly supported near 0. Then after a suitable change of variable, you get that $ I(\lambda)=\int g(t) e^{i\lambda t^3/3} dt, $ …

Requiring $C^2$ is too much: you can ask only $C^1$, twice differentiable at the distinguished point with a non-degenerate Hessian matrix. More precisely the following holds true. Theorem. Let $\Omeg …

No. Take for instance the following ODE where $0$ is a regular singular point, $$x y'=\lambda y.$$ The functions $cx_+^{\lambda}$ are solutions and are not analytic. If $\lambda$ is not a non-negative …

Let $f_n:[0,1]\rightarrow \mathbb R$ be a sequence of continuous functions converging pointwise, i.e. such that $\forall x\in [0,1]$, the sequence $(f_n(x))_{n\in \mathbb N}$ converges. We set $f(x)=\ …