definition of weak(*)-convergence We know the standard definition of weak convergence in $L^p$:
$u_n \rightharpoonup u$ in $L^p(U)$ if $\int_U u_nvdx \rightarrow\int_Uuvdx$ for all $v\in L^{p'}(U)$. Why do I always find this definition only for $p\in [1,\infty)$ and not for $p=\infty$?
The same question comes up for weak*-convergence in $L^p$. Why is it only defined for $p=\infty$?
 A: It has to do with compactness. If $1<p<\infty$ then $L^p$ is reflexive (roughly speaking, the bi-dual of $L^p$ coincides with $L^p$) and so if we have a bounded sequence in $L^p$, you can extract a subsequence which converges weakly. 
On the other hand, $L^\infty$ is not reflexive, so even if you can define weak convergence in $L^\infty$, it is pretty useless. However, $L^\infty$ is the dual of $L^1$, which is separable, and so by another compactness theorem  if you have a bounded sequence in $L^\infty$, you can extract a subsequence which converges weakly star. $L^1$ is really the bad case...
A: The definition of weak convergence in a normed vector space $(E,\| \cdot \|_{E})$ relies on its dual space $(E',\| \cdot \|_{E'})$, that is:
$$ x_n \rightharpoonup x \Longleftrightarrow \langle f,x_n \rangle_{E' \times E} \rightarrow \langle f,x \rangle_{E' \times E} \ \forall f \in E' \qquad (1)  $$ 
where the last convergence is in the sense of real numbers as $n \rightarrow +\infty$. The useful characterization you wrote for weak convergence in $L^p$ spaces for $p \in [1,+\infty)$ is possible thanks to Riesz's representation theorem that states the famous indentification:
$$ (L^p)' \equiv L^{p'} $$ 
where $p,p'$ are conjugate Holder exponents ( $p' := \infty$ if $p=1$) . With this in mind, $(1)$ can be simplified to the integral convergence you wrote down. But such identification fails for $p=\infty$, i.e we can't say that $(L^{\infty})' \equiv L^1$. Note that $(1)$ still holds if $E=L^{\infty}$, but we can't precisely characterize $E'$: we can only get the inclusion $L^1 \subset (L^{\infty})'$, but the latter is much bigger.  
The problem with weak * convergence is a little different but has the same motivation. Weak * convergence is defined only on dual spaces, if $(f_n)_n,f \subset E'$:
$$ f_n \rightharpoonup{*} \ \ f \Longleftrightarrow \langle f_n,x \rangle_{E' \times E} \rightarrow \langle f,x \rangle_{E' \times E} \ \forall x \in E \qquad (2)  $$ 
where the last convergence is again in the sense of real numbers. Since the pre-dual space of $L^1$ (i.e, the space whose dual space is $L^1$) is not $L^{\infty}$ (see again Riesz's theorem), we cannot characterize weak * convergence in $L^1$.
A: 
Why do I always find this definition only for $p \in [1,\infty)$ and not for $p = \infty$.

Every normed space has a weak topology. If $X$ is a normed space, then a net $x_n$ converges weakly to $x$ iff $\varphi(x_n) \to \varphi(x)$ for all $\varphi \in X^*$.
Now let $X = L^p$ for some $p \in [1,\infty)$. Then $X^* \simeq L^{p'}$, i.e. for all $\varphi \in X^*$ there exists some $v \in L^{p'}$ such that 
$$
\varphi(u) = \int u v \; dx
$$
for all $u \in L^p$. This duality does not hold for $p = \infty$.

The same question comes up for weak*-convergence in $L^p$. Why is it only defined for $p= \infty?

I don't like the term "defined". In fact the weak* convergence is defined on every normed space that has a predual, i.e. every normed space which is the dual space of another. Using the above identification of $L^p$ spaces, for all $p \in (1,\infty]$ you have that $(L^{p'})^* \simeq (L^p)$. Hence, for every $L^p$ space for $p \in (1,\infty]$ there exists a weak* convergence structure.
