About the functionals $J(y)=\int_0^1y'(x)dx$ and $K(y)=\int_0^1y^2(x)dx$ on the space of functions $y$ in $C^1([0,1])$ such that $y(0)=0$ and $y(1)=1$ Let
$$
A=\{y \in{C^1([0,1]):y(0)=0, y(1)=1}\}.
$$
$C^1([0,1])$ is the space of continuous and differentiable functions over the interval $[0,1]$.
Problem 1: Calculate the minimum and maximum, in $A$, of the functional
$$
J[y]=\int_0^1 y'(x) dx.
$$
Problem 2: Consider the functional:
$$
J[y]=\int_0^1 y^2(x) dx
$$
Proof that, in $A$, the infimum of $J$ is $0$, but $J$ does not have minimum.
Sugestion: Consider the sucession of functions $y_n(x)=x^n$.
My answer:
Problem 1:
For $ y\in A$, we have that $y$ is continuous and differentiable over the interval $[0,1]$. By Barrow's formula,
$$
\int_0^1 y'(x) dx=y(1)-y(0)=1-0=1.
$$
This means that $ J[y]=1$ in $A$.We conclude that $1$ is the minumum and the maximum of $J[y]$ in $A$.
Functions that satisfy the conditions: 
$$
y(x)=x^m,m\ge1; y(x)=\sin(\frac{\pi}{2}x).
$$
Problem 2:
We note that:
$$
J[y]=\int_0^1 y^2(x) dx = \Vert y(x)\Vert^2
$$
where $\Vert * \Vert $ represents the 2-norm.
So obviously,
$$
J[y] \ge 0.
$$
However, $J[y]=\Vert y(x)\Vert^2=0$ if and only if $y(x)=0$. But in this case, $y$ does not satisfy $y(1)=1$. So, $J$ does not have a minimum in $A$.
END
In problem 1, I'm having doubts because I didn't got a specific function, and I'm not sure if that was what they were asking...
In problem 2, I think there is something missing so that I can say that "$0$ is the infimum of the functional on $A$". Also, I didn't use the sugestion, so is there something else missing?
 A: You're reasoning on problem 2 is incorrect. Saying that $J[y] \ge 0$ and that $J[y] = 0$ iff $y \equiv 0$ shows that $y \equiv 0$ is the minimizer of $J$ over all functions, but there could be a different minimizer in $A$. As an example, consider minimizing the function $$f(t) = t^2,$$ in the set $A = \{t\in \mathbb R : 2 \le t \le 5\}$. Of course $f(t) \ge 0$ and $f(t) = 0$ iff $t = 0$. But $t = 0$ is not in the set; still, you can't claim that $f(t)$ has no minimizer in $A$. Indeed, the minimizer of $f$ in $A$ is $t =2$ because $2 \in A$ and $f(2)\le f(t)$ for all $t \in A$. 
It is possible that something similar could happen here. I.e., $J$ will never equal zero among functions in $A$, but there may be some $y_* \in A$ such that $J[y_*] \le J[y]$ for all $y \in A$. The problem is asking you to show that no such $y_*$ exists. It is clear that $J[y] > 0$ for all $y \in A$; indeed, if $y$ is continuous and $y(1) = 1$, there is some $\epsilon > 0$ such that $y(x) > 1/2$ for $x \in (1-\epsilon,1]$ and thus $$J[y] \ge \int^1_0 y(x)^2 dx \ge \int^1_{1-\epsilon} \tfrac 1 2 dx = \tfrac \epsilon 2 > 0.$$ Now let $y_n(x) = x^n$, $x \in [0,1]$. We see $$J[y_n] = \int^1_0 x^{2n} dx = \frac{1}{2n+1} \to 0$$ as $n \to \infty$. Consider, for any $y_* \in A$, we have $$J[y_*] > 0.$$ But then for $n$ large enough, we have $$J[y_*] > \frac{1}{2n+1} = J[y_n].$$ This shows that there is no $y_* \in A$ such that $J[y_*] \le J[y]$ for all $y \in A$ (since we can always find $y_n$ which makes $J$ smaller) and so there is no minimizer of $J$ in $A$.
