# Tag Info

2

Yes, zero derivative implies constant distribution. One way to see this is by mollification: if $T'=0$, then for any smooth mollifier $\phi$ we have $$(T*\phi)' = T'*\phi = 0$$ Since $T*\phi$ is a smooth function, it's identically constant. Then the distributional limit is constant of $T*\phi_\epsilon$ as $\epsilon\to 0$ is constant too. (Here ...

2

Take any compact $K$ not containing zero, then take any test function $\phi$ with support in $K$, then $$\langle pv(f),\phi\rangle = \lim_{\epsilon\to 0}\int_{\|x\|\ge \epsilon}f(x)\phi(x)dx.$$ If $$\epsilon< dist(K,0),$$then $$\int_{\|x\|\ge \epsilon}f(x)\phi(x)dx = \int_Kf(x)\phi(x)dx,$$hence $$\langle pv(f),\phi\rangle = \int_{K}f(x)\phi(x)dx$$ and ...

1

As you did the first part, I do not address it. Concerning the examples, $c(\sin z)/z$ is the first example that you need. For the second example, take $\phi(t)$ infinitely differentiable, with support on $(-1,1)$, and consider the function $$f(z)=\int_{-\infty}^\infty\phi(t)e^{izt}dt.$$ This is evidently bounded: $|f(x)|\leq \|\phi\|_1$. Now, ...

1

Let $g$ be smooth and compactly supported. Write $u=nx$ so that $$\int_{-\infty}^\infty n^2 f(nx) g(x) dx = \int_{-\infty}^\infty n f(u) g(u/n) du.$$ Now by the hypotheses, $g_n(u) := g(u/n)$ converges uniformly to $g(0)$. (We needed the compact support for this.) This motivates splitting the integral by adding and subtracting $g(0)$ as follows: ...

1

Joelafrite made a good suggestion: consider the first distributional derivative of a function $f$ that is not absolutely continuous. By definition, this distribution acts as $$\phi\to -\int f\phi'$$ If $f$ is increasing (like Cantor staircase and Minkowski's ?-function), then the distribution $f'$ is a measure. If $f$ has bounded variation, then $f'$ is ...

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Yes, they are norms. My guess is that the book calls them semi-norms because the topology in a locally convex topological vector space is defined through a family of semi-norms (which may or may not be norms.)

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This is similar to the fact that pointwise convergence does not imply $L^1$ convergence. Construct a sequence $f_n$, each with integral $1$, that converges to $0$ pointwise.

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$f_n \to f$ in $D'$ if $$\forall \phi \in D, \ \lim_n \int_U (f-f_n)\phi = 0$$ If the convergence is strong in $L^p$, Hölder inequality gives us the result : $$\left|\int_U (f-f_n)\phi \right| \leq \int_U \left|(f-f_n)\phi \right| \leq \|f_n-f\|_p\|\phi\|_q \to 0$$ If the convergence is weak, as $D\subset L_q$, the answer is immediate

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You say you get to $$-\int_{-l}^l e^{ikx} \phi''(x) \, \mathrm{d} x$$ but then apply the Riemann-Lebesgue lemma.

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