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The functionals $$ \phi_n(x) = \int_{\frac{1}{n} \le |t| \le 1} \frac{x(t)}{t} \mathrm{d} t $$ define a sequence of functionls in $C([-1,1])$ and $C^1([-1,1])$.

a) Show that $(\phi_n)$ converges *-weakly in $C^1([-1,1])'$.

b) Does $(\phi_n)$ converges *-weakly in $C([-1,1])'$?

For me the limit functional $$ \int_{0 \le |t| \le 1} \frac{x(t)}{t} \mathrm{d} t $$ is not well defined so i have trouble evaluating the condition of convergence? Do you have any hints?

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What exactly do you mean by "For me the limit functional ... is not well defined"? Is $x \in C^1$ or $x \in C$? Does that make a difference? (It does.) – Martin Jan 11 '13 at 12:24
Ok, i don't see why it makes a difference, i think the problem lies in the factor $1/t$, which is not defined at $t = 0$? – Stefan Jan 11 '13 at 12:28
Shouldn't it be $\int_{1/n<t<1}\frac{x(t)-x(0)}tdt$? – Davide Giraudo Jan 11 '13 at 12:35
no, in the exercise its written without $x(0)$. – Stefan Jan 11 '13 at 12:50
@DavideGiraudo: In fact it is like you stated, there is an absolute value $1/n < |t| < 1$, so the integral splits into two parts and you can add and substract $x(0)$. – Vobo Jan 11 '13 at 16:56

We can write $$\phi_n(x)=\int_{1/n}^1\frac{x(t)-x(-t)}tdt.$$ When $x$ is in $C^1[-1,1]$, this converge to $\int_0^1\frac{x(t)-x(-t)}tdt$ (and this integral is convergent, as the problem at $0$ is solved by the derivative.

Taking a continuous function $f$ such that $f=1$ on $[n^{-1},1]$ and $-1$ on $[-1,-n^{-1}]$, we can see that $\lVert \phi_n\rVert_{(C[-1,1])'}=2\log n$ so we can't have weak$^*$ convergence in $(C[-1,1])'$.

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could your f choosen such that it is differentiable too, and then contradicting your conclusion that $\phi_n$ is weak* convergent in $C^1([-1,1])$? – Stefan Jan 12 '13 at 16:10
No, because $\lVert\phi_n\rVert_{(C^1[-1,1])'}$ is bounded independently on $n$. – Davide Giraudo Jan 12 '13 at 16:14
but why your $f$ not, think I can choose a function from $C^1([-1,1])$ with your property that $f=1$ on $[n^{-1},1]$ and $-1$ on $[-1, -n^{-1}]$? – Stefan Jan 12 '13 at 16:37

@Davide Giraduo

Regarding your derivation, i came to another result: \begin{align*} \int_{1/n \le |t| \le 1} \frac{x(t)}{t} \mathrm{d} t & = \int_{-1}^{-1/n} \frac{x(t)}{t} \mathrm{d} t + \int_{1/n}^1 \frac{x(t)}{t} \mathrm{d} t \\ & = - \int_{1}^{1/n} \frac{x(-t)}{t} \mathrm{d} t + \int_{1/n}^1 \frac{x(t)}{t} \mathrm{d} t \\ & = \int_{1/n}^{1} \frac{x(-t)}{t} \mathrm{d} t + \int_{1/n}^1 \frac{x(t)}{t} \mathrm{d} t \\ & = \int_{1/n}^{1} \frac{x(-t) + x(t)}{t} \, \mathrm{d} t \end{align*} Which has others sign's?

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The first "-" in the second row is wrong [$t\to -t$ and $dt\to -dt$]. – Vobo Jan 13 '13 at 8:42
i am confused, the rule for linear substitution as it is written in my textbooks reads, for $c \ne 0$: $\int_a^b f(cx) \mathrm{d}x = \frac{1}{c} \int_{ca}^{cb} f(t) \mathrm{d}t$ for $t(x) := cx$. I guess the transformation factor is the $\frac{1}{c}$, which I have put before the integral? – Stefan Jan 13 '13 at 12:59
Yes, equivalently, $\int_a^b f(y) dy = \frac{1}{c} \int_{ca}^{cb} f(\frac{t}{c}) dt$. Now apply this for $f(y)=\frac{x(y)}{y}$. – Vobo Jan 13 '13 at 16:13

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