Convergence to the Dirac Delta Function Let $h\colon[0,1]\to \mathbb{R}^+$ be any bounded measurable non-negative function with a unique maximum at $a$ and $h$ is continuous at $a$. For $\lambda>0$ define $h_\lambda(x)=C_\lambda h(x)^\lambda$ where $C_\lambda$ normalizes such that $\displaystyle\int_0^1 h_\lambda(x)\,dx=1$. $f$ is any continuous function on $[0,1]$ and $\epsilon>0$. Are the following assertions true? 
1)$\displaystyle\lim\limits_{\lambda\to\infty}\int_{h(x)\le h(a)−\epsilon}h_λ(x)f(x)\,dx=0$ and $\displaystyle\lim\limits_{\lambda\to\infty}\int_0^1 h_λ(x)f(x)\,dx=f(a)$. In words, the limit of $h_\lambda$ is the Dirac delta function $\delta(\cdot-a)$.
2) If $h$ is continuous, $h_\lambda(x)$ converges uniformly to $0$ in $\{x: h(x) \le h(a)-\epsilon\}$.
It seems correct since the exponentiation suppresses the part somewhere below $h(a)$ while heightens the part of $h$ near $a$. The difficulty seems to be giving a partitions of $[0,1]$ where I can utilize the exponentiation to either suppress or heighten $h$ by the exponentiation of $\lambda$ after $h$ is normalized by $C_\lambda$.
This question is a generalization of my more specific case, and was suggested by one commentator whuber. However, he did not supply a proof and I am stumped by the difficulty described the last paragraph.
 A: In this answer I will assume that $h$ is defined on $\mathbb R$ instead of $[0,1]$ and with $\int h <\infty$ (so that I do not need to write $\cap [0,1]$ all the time). Also I will make an extra assumption on $h$ (You will see).
For any $d>0$ we have
\begin{equation}
\begin{split}
\bigg|\int_{\mathbb R} &h_\lambda(x) f(x) dx -f(a) \bigg| =\bigg|\int_{\mathbb R} h_\lambda(x) (f(x) -f(a)) dx \bigg| \\
&\le \bigg|\int_{|x-a|\le d} h_\lambda (x) (f(x)-f(a)) dx \bigg|+ \bigg|\int_{|x-a|> d} h_\lambda (x) (f(x)-f(a)) dx\bigg|\\
&\le \int_{|x-a|\le d} h_\lambda (x) \big|f(x)-f(a)\big| dx + \max \big|f(x)-f(a)\big|\bigg|\int_{|x-a|> d} h_\lambda (x) dx\bigg|\\
\end{split}
\end{equation}
Let now $\epsilon>0$. Then as $f$ is continuous there is $c>0$ so that $|f(x) - f(a)|<\epsilon/2$ whenever $|x-a|<c$. So as long as $c>d$ we have
$$\bigg|\int_{\mathbb R} h_\lambda(x) f(x) dx -f(a) \bigg|  \le \frac{\epsilon}{2} + \max\big|f(x)-f(a)\big|\bigg|\int_{|x-a|>d} h_\lambda (x) dx\bigg|$$
Now the more difficult task is to show that
$$(*)\ \ \ \ \bigg|\int_{|x-a|> d} h_\lambda (x) dx\bigg| \to 0 \ \ \text{as } \lambda \to \infty.$$
For this one I will use that
$$\|h\|_{\infty, d} < \|h\|_\infty$$
for all $d>0$, where $\|h\|_{\infty, d}$ is the $L^\infty$ norm of $h$ when restricted to $\{|x-a|>d\}$ (Note the strict inequality. That is not implied by your assumption).
Now note that for any nonnegative integrable functions $h$ we have
$$\bigg(\int_\mathbb{R} h^\lambda\bigg)^\frac{1}{\lambda} \to \|h\|_\infty\ \ \text{as }\lambda \to \infty$$
So as by definition,
$$C_\lambda = \frac{1}{\int_\mathbb{R} h^\lambda} \Longrightarrow  C_\lambda^\frac{1}{\lambda} \to \frac{1}{\|h\|_\infty}$$
this implies that
$$\bigg(\int_{|x-a|>d} h_\lambda\bigg)^\frac{1}{\lambda} = C_\lambda^\frac{1}{\lambda} \bigg(\int_{|x-a|>d} h^\lambda\bigg)^\frac{1}{\lambda}\to \frac{\|h\|_{\infty, d}}{\|h\|_\infty} <1$$
by the extra assumption I made. Thus there is $\lambda_0$ so that
$$\bigg(\int_{|x-a|>d} h_\lambda\bigg)^\frac{1}{\lambda} <B <1$$
for all $\lambda >\lambda_0$. In particular,
$$\int_{|x-a|>d} h_\lambda < B^\lambda$$
for all $\lambda >\lambda_0$. As $B<1$, $B^\lambda \to 0$. This implies (*) and we are done.
(Note that continuity of $h$ at $a$ is not needed, only some sort of "continuity in measure at $a$" is used).
