Prove that $\inf\limits_{f\in X}\,I_a(f)=1-a$. Let $X:=\big\{f\in \mathcal{C}^1[0,1]\,\big|\,f(0)=0\text{ and } f(1)=1\big\}$, $0<a<1$, and $I_a(f):=\displaystyle\int _0 ^1 x^a \left(\frac{\text{d}}{\text{d}x}\,f(x)\right)^2 dx$.
Then prove that $$\inf_{f\in X}\,I_a(f)=1-a\,.$$
I know that $\inf\limits_{f\in X}\, I_a(f)\leq 1-a$ by taking $f(x)=x^{1-a}$ for all $x\in[0,1]$, but I don't know how to prove $\inf\limits_{f\in X}\, I_a(f)\geq 1-a$.
 A: Let $f\in X$.  Using the Cauchy-Schwarz Inequality and the Triangle Inequality, we have
$$\begin{align}\left(\int_0^1\,x^{-a}\,\text{d}x\right)^{\frac12}\,\left(\int_0^1\,x^a\,\big(f'(x)\big)^2\,\text{d}x\right)^{\frac12}&\geq  \int_0^1\,\big|f'(x)\big|\,\text{d}x
\\&\geq\left| \int_0^1\,f'(x)\,\text{d}x\right|=\big|f(1)-f(0)\big|=1\,.\end{align}$$
This shows that
$$I_a(f)\geq \frac{1}{\displaystyle\int_0^1\,x^{-a}\,\text{d}x}=1-a\,.\tag{*}$$
Note that the inequality becomes an equality if and only if $x^a\,\big|f(x)\big|=1$ for almost every $x\in[0,1]$ (in the Lebesgue sense).  Since $f$ is a continuously differentiable function, the only equality case must happen when $x^a\,f'(x)=k$ for some constant $k$, but this cannot happen as $f'(x)$ will not be defined at $x=0$.  Thus, (*) is in fact a strict inequality.  
However, we can take an arbitrary sequence $\left(f_n\right)_{n\in\mathbb{Z}_{>0}}$ of functions $f_n\in\mathcal{C}^1\big([0,1]\to\mathbb{R}\big)$ that converge uniformly on compact sets to $g:[0,1]\to\mathbb{R}$, where $g(x):=x^{1-a}$ for $x\in[0,1]$.  (For example, take $f_n:=\chi_{\left[0,\frac1n\right]}\,g$ for $n=1,2,3,\ldots$.  Here, for a given $\epsilon\in(0,1)$, $\chi_{[0,\epsilon]}:[0,1]\to[0,1]$ is a smooth function equaling $1$ on $[\epsilon,1]$ and vanishing on $\left[0,\frac{\epsilon}{2}\right]$.)  By doing so, we conclude that (*) is sharp, whence
$$\inf_{f\in X}\,I_a(f)=1-a\,.$$
A: By Stone–Weierstrass theorem, it suffices to consider polynomials $f(x)$, by density arguments.

By Euler-Lagrange, it suffices to find $f$ such that:
$$\frac{\mathrm d}{\mathrm dx} [2x^af'] = 0$$
So we solve:
$$\begin{array}{rcl}
\dfrac{\mathrm d}{\mathrm dx} [2x^af'] &=& 0 \\
2x^af' &=& k_1 \\
f' &=& k_2 x^{-a} \\
f &=& k_3 x^{1-a} + k_4
\end{array}$$
The conditions $f(0) = 0$ and $f(1) = 1$ translate to $k_4 = 0$ and $k_3 + k_4 = 1$, so $f(x) = x^{1-a}$ is a stationary point.
It suffices to prove that it is minimal.

The second variation of this functional (i.e. "second derivative") is:
$$\delta^2 I_a = \int_0^1 2x^a (\xi')^2 \ \mathrm dt$$
which is twice the functional itself.
Observe that as long as $\xi \ne 0$ (i.e. $\xi$ is not identically zero), then the second derivative is positive, establishing that $1-a$ is the minimum.

Alternatively, we plug in $f = x$ and observe that $I_a(x) = \dfrac1{1+a} > 1-a$, so $1-a$ corresponds to a minimum.
