# Tag Info

3

As you probably know, elements of $W^{1,2}$ aren't functions but equivalence classes of functions (up to almost everywhere equality). A representative is then just an element of this equivalence class. The Sobolev embedding theorem asserts that every $f\in W^{1,2}(\mathbb{R})$ has a continuous representative. Hopefully, this explains both 1. and 2.

2

A priori, an element $[f] \in W^{1,2}(\mathbb{R}) \subset L^2(\mathbb{R})$ is only an equivalence class of functions in $L^2(\mathbb{R})$ (where two functions are identified if the set on which they differ is of measure zero). However, one can show that if $[f] \in W^{1,2}(\mathbb{R})$, then there is a (unique) $\overline{f} \in [f]$ that is continuous. The ...

2

If $f(x)=|x|$ we have \begin{eqnarray} \sup_{x \in [-1,1]}\left| f_n(x)-f(x) \right|&=&\sup_{x \in [-1,1]}\left| \sqrt{x^2+\frac1n\left( \cos \left( x^n \right) \right)^2} -|x|\right| \\ &\leq& \sup_{x \in [-1,1]}\left| \sqrt{x^2+\frac1n}-\sqrt{x^2} \right|\\ &=& \sup_{x \in [-1,1]}\frac{\left| \sqrt{x^2+\frac1n}-\sqrt{x^2} ...

2

Just use Holder inequality with $p=q=2$ $|F(x) - F(y)|=|\int_y^x f(t) dt|\le \sqrt{\int_y^x |f|^2 dt}\sqrt{\int_y^x 1 dt}\le C|x-y|^{\frac{1}{2}}$ with $C=\sqrt{\int_\mathbb{R} |f|^2 dt}$

2

Take a function $\varphi\in L^2$, we have $\displaystyle \left|\int \varphi f_n -\int \varphi f \right|\leq \int |f_n-f|\varphi$. for every $\varepsilon>0$ we have $\mu(|f_n-f|>\varepsilon)<\varepsilon$ for $n$ large enough, so $\displaystyle \int |f_n-f|\varphi\leq (1-\varepsilon)\varepsilon+\int_{|f_n-f|>\varepsilon}|f_n-f|\varphi$ and we just ...

2

Any continuous function is a uniform limit of polynomials, so pick your favourite non-differentiable continuous function, and that would work! In fact, if $f$ is such a function, say $f(x) := |x- 0.5|$, then take the sequence to be the sequence of Bernstein polynomials $$B_n(f)(x) := \sum_{k=1}^n {n\choose k} f\left(\frac{k}{n}\right) x^k (1-x)^{n-k}$$ ...

1

A function $f$ on $[a,b]$ is Lipschitz continuous with constant $M$ iff there is a constant $A$ and $g \in L^{\infty}[a,b]$ with $\|g\|_{\infty} \le M$ such that $$f(x) = A+\int_{0}^{x}g(t)dt.$$ Can you show that $\{ f \in C[a,b] : |f| \le M \}\subset L^1$ is dense in $\{ f \in L^{\infty}[a,b] : \|f\|_{\infty} \le M \}\subset L^1$ ...

1

The operator $L=\frac{1}{i}\frac{d}{dt}$ defined on the domain $\mathcal{D}(L)$ consisting of periodic absolutely continuous functions $f$ on $[-\pi,\pi]$ with $f' \in L^2[-\pi,\pi]$ is a densely defined linear operator with discrete spectrum. The eigenvalues are the integers, and $\{ \frac{1}{\sqrt{2\pi}}e^{inx}\}$ is an orthonormal basis of eigenvectors. ...

1

The max norm is not the (absolute value of) the function at one specific point, but at a point that depends on the function itself. In a more general setting where functions are not necessarily continuous, it is called the supremum norm. The integral of $|f|$ cannot be higher than the length of the integration interval multiplied by the maximum of $|f|$ ...

1

Hints: For nonexistence of $c$ you could try to construct non-negative continuous functions, which have a small integral, i.e., $\|f\|_1$ is small, but with large $\|f\|_{max}$. The second inequality follows with $c = b - a$.

1

Take a look at the norm of $H^1(\Omega)$: $$\|u\|_{H^1} = \|u\|_{L^2} + \| Du\|_{L^2} \ge \|u\|_{L^2}.$$ Thus if $\|f_n -f\|_{H^1} \to 0$, then $\|f_n - f\|_{L^2} \to 0$. In terms of operator it just says that the inclusion $H^1 (\Omega) \to L^2(\Omega)$ is a bounded linear operator. It is true that $H^1(\Omega)$ is dense in $L^2(\Omega)$, but we are not ...

1

If $\Omega$ is regular enough ($C^1$ or uniformly Lipschitz) then there exists a continuous linear operator $$\operatorname{Tr}\colon W^{1,p}(\Omega) \to L^p(\partial \Omega,\mathcal{H}^{N-1})$$ such that $\operatorname{Tr}(u) = u$ for every $u \in W^{1,p}(\Omega)\cap C^1(\Omega)$, there is a constant $C = C(\Omega,N,p)$ such that ...

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