Let $p > 0$ be fixed. For $t > 0$ and $x\in[0,1]$ define $$f(t,x) = \frac{1}{t}x^pe^{-\frac{x}{t}}.$$ For each values of $p$ does each of the following statement hold true?
(a) $f_t \rightarrow 0$ a.e. as $t \rightarrow 0$.
(b) $f_t \rightarrow 0$ uniformly in $[0,1]$ as $t \rightarrow 0$.
(c) $\int_0^1f_t(x)dx \rightarrow 0$ as $t \rightarrow 0$.
Answers:
(a) Notice that $\lim_{t \rightarrow 0}f_t(x)$ yields an indeterminate form $0/0$. Thus using L'Hopital rule we have $$\lim_{t \rightarrow 0}\frac{1}{t}x^pe^{-\frac{x}{t}} = \lim_{t \rightarrow 0}-x^{p+1}e^{-x/t} = -x^{p+1}/\infty = 0.$$And this holds for all $p > 0$ on $[0,1] \setminus \{x = 0\}$ otherwise we keep getting indeterminate forms.
(b) To see uniform convergence we start with $$\left|f_t(x)\right|_\infty = \sup_{x \in [0,1]}\left|\frac{1}{t}x^pe^{-\frac{x}{t}}\right| \leq \left|\frac{1}{t}e^{-\frac{x}{t}}\right|,$$and as $t \rightarrow 0$ we get $\lim_{t \rightarrow 0}\|f_t(x)\|_\infty = 0$. Observe that if $x = 0$ when taking the supremum then we trivially get $0$ as the limit.
(c) Now I am having some troubles satisfying the conditions of the Lebesgue theorem to get the integral over $0$ as to yield the result.
I have seen several variations of the Lebesgue theorem.
(i) Classical; (ii) If $f_t(x)$ is bounded below for each $x$ and $f_t \uparrow f$; (iii) if $|f_t(x)| \leq M$, where $M$ is some constant, for all $x$ and $f_n \rightarrow f$.
I think the main problem is showing that I can bound $|f(x,t)|$ with some non-negative function $g(x)$.
Thank you very much.