A discontinuous function weakly converging. This is an exercise from a course.
Suppose that $f \in C^{\infty}_{C}(R^3),$ and consider the function for t $\in [0,1)$, $$v(t,x)= \frac{1}{1-t}f(\frac{t}{1-t}).$$
Show that $v \in L^{\infty}([0,1),L^3), v\not \in C([0,1),L^3)$, and $v(t,x) \rightharpoonup 0$ as $t \rightarrow 1$.
I'm pretty lost on each of these.  First off, shouldn't this be $f\in C^{\infty}_{C}(R)$? Also, I don't see what $L^3$ has to do with the first one.  If $f$ has compact support, then as $t \rightarrow 1$, we are going to get out of the support of $f$ eventually.  But then we are integrating $(\frac{1}{1-t})^3 \times 0$ off the support, and $f$ is bounded on its support.  So this holds for any $L^p$ space.
For the second one, we should estimate $||\frac{1}{1-s}f(\frac{s}{1-s}) - \frac{1}{1-r}f(\frac{r}{1-r})||_{L^3}$.  So we need to get a contradiction to continuity.  That means given any $\delta$, we can find points $r,s$ with $|r-s|<\delta$ but $||\frac{1}{1-s}f(\frac{s}{1-s}) - \frac{1}{1-r}f(\frac{r}{1-r})||_{L^3} > \epsilon$.  It seems that the only thing to do is make sequences going to $1$ for $r$ and $s$.  But the support of $f$ may be very small, and then how can we get a contradiction there?
I have no idea on the third one.  What does the dual of $L^{\infty}(0,T,X)$ look like? I know that if $1\leq p<\infty$, and $X$ is reflexive, we have the dual is isomorphic to $L^{p'}(0,T,X')$, but here we don't have it.  So... we somehow have to use some more abstract thing? 
 A: It's probably 
$$v(t,x) = \frac{1}{1-t} f\left( \frac{x}{1-t} \right)$$
Then for the first question,
$$\| v(t,\cdot) \|_{L^3}^3 = \left( \frac{1}{1-t} \right)^3 \int_{\mathbb{R}^3} \left| f\left( \frac{x}{1-t} \right) \right|^3 dx = \left( \frac{1}{1-t} \right)^3 \int_{\mathbb{R}^3} (1-t)^3 |f\left( x \right)|^3 dx = \|f\|_{L^3}^3$$
So $v\in L^\infty( [0,1), L^3)$
Note that $v\not\in L^\infty( [0,1), L^4)$, as $\| v(t,\cdot) \|_{L^4}^4 = \frac{1}{1-t}$ is unbounded (this answer why $L^3$).
For the second question, I find that it's continuous:
Let $t \in [0,1)$. Now for $t_n \to t$, with $t_n \in [t-\epsilon, t+\epsilon] \subset [0,1)$ (or $t_n \in [0, \epsilon]$ if $t=0$), we define
$$\phi_n(x) = \left| \frac{1}{1-t} f(\frac{x}{1-t}) - \frac{1}{1-t_n} f( \frac{x}{1-t_n} ) \right|^3$$
It's clear that $\phi_n$ converge pointwise to $0$.
And if the support of $f$ is included in $B(0,R)$ we have
$$| \phi_n(x) | \leq \left( \left| \frac{1}{1-t} M \right| \mathbb{1}_{B(0,(1-t)R)} (x) + \left| \frac{1}{1-t_n} f( \frac{x}{1-t_n} ) \right| \mathbb{1}_{B(0,(1-t_n)R)} (x)\right)^3$$
$$ \leq M^3\left( \frac{1}{1-t}+\frac{1}{1-t_n}\right)^3\mathbb{1}_{B(0,R)} (x)$$
And for $t_n$ close enough of $t$, we have 
$$ \leq M^3\left( \frac{3}{1-t}\right)^3\mathbb{1}_{B(0,R)} (x)$$
And this is an integrable function, so we get by Dominated convergence theorem that
$$\| \phi_n \|_1 \to 0$$
And as this is true for all sequences $t_n$, we get the continuity 
So either my function is not the right one, or I made a mistake, or there is another mistake in the exercice
For the third question, the idea is that you're looking for the weak convergence of the familly of functions :
$$\phi_t(x) = v(t,x)$$
The problem is what space are we considering for $\phi_t(x)$? 
If we're considering $C^\infty_c$, then clearly, it doesn't converge to $0$ if $f(0) \neq 0$. Indeed, for the Dirac distribution,
$$\langle \phi_t, \delta \rangle = \phi_t(0) = \frac{1}{1-t} f(0) \to \pm \infty$$ 
If we're considering $L^3$ (a more interesting choice, as $\| \phi_t\|_3$ is constant) , now it converge to 0 : 
Let's take a $g\in L^q$, $q = \frac{3}{2}$.
Then
$$\langle \phi_t, g \rangle = \int_{\mathbb{R}^3} \frac{1}{1-t} f(\frac{x}{1-t})   g(x) dx $$
Now, if the support of $f$ is in $B(0,R)$, the support of $f(\frac{x}{1-t})$ is in  $B(0,(1-t)R)$
$$ = \frac{1}{1-t} \int_{B(0,(1-t)R)} f(\frac{x}{1-t})   g(x) dx $$
$$ \leq \frac{M}{1-t} \int_{B(0,(1-t)R)} g(x) dx $$
But
$$ \left| \int_{B(0,(1-t)R)} g(x) dx  \right| \leq \left(\int_{B(0,(1-t)R)} |g(x)|^{\frac{3}{2}} dx  \right)^{\frac{2}{3}} \left(\int_{B(0,(1-t)R)} 1 dx  \right)^{\frac{1}{3}}$$
$$ \leq C(1-t) \left( \int_{B(0,(1-t)R)} |g(x)|^{\frac{3}{2}} dx  \right)^{\frac{2}{3}} $$
So 
$$\frac{M}{1-t} \int_{B(0,(1-t)R)} g(x) dx \leq C' \left(\int_{B(0,(1-t)R)} |g(x)|^{\frac{3}{2}} dx  \right)^{\frac{2}{3}}$$
And by dominated convergence theorem, 
$$\lim_{t\to 1} \int_{B(0,(1-t)R)} |g(x)|^{\frac{3}{2}} dx = 0$$
Hence 
$$\forall g \in L^{\frac{3}{2}}( \mathbb{R}^3 ),\  \langle v(t,x) , g \rangle \to 0$$
