Show that the infimum of $T(v)=\|v-u\|_\infty$ is not attained Let $X=\{u\in C([0,1]):\ u(0)=0\}$ with the norm $\|u\|=\|u\|_\infty$. Let $f\in X^\star$ be defined by $$\langle f,u\rangle=\int_0^1u(x)dx $$
Let $M=\{u\in X:\ \langle f,u\rangle=0\}$ and consider the functional $T_u:M\to\mathbb{R}$ for $u\in X\setminus M$ defined by $$T_u(v)=\|v-u\|$$
Is true that the infimum of $T_u$ is not assumned for any $v\in M$? or equivalently: is there any $v\in M$ such that $$\inf_{w\in M}\|w-u\|=\|v-u\|$$
Thank you
 A: I believe this is correct now.
Let $u \in X$, $U = \int_0^1u$ and $\hat v = u - U$. We first show that $\hat v$ minimizes $T_u$ in the extended domain $\widehat M = \{v \in C[0, 1]: \int_0^1 v = 0\}$. It is obvious that $\hat v \in \widehat M$.
First, observe that $T_u(\hat v) = \|\hat v - u\| = |U|$.
Then, for $w \in \widehat M$,
\begin{align*}
T_u(w) & = \|w - u\| \\
& \ge \left|\int_0^1 (w - u) \right| \\
& = \left|\int_0^1 u\right| = |U|.
\end{align*}
This shows that $\hat v$ minimizes the extended $T_u$.
Next, we show that there exists a sequence $v_1, v_2, \ldots$ in $M$ such that $T_u(v_n) \to |U|$ as $n \to \infty$. For each $n$, define $\tilde v_n \in C[0, 1]$ by $\tilde v_n(x) = x^{1/n}\hat v(x)$. Finally, define $v_n(x) = \tilde v_n(x) - 2x\int_0^1 \tilde v_n(y) dy$.
\begin{align*}
\tilde v_n(x) & = x^{1/n}\hat v_n(x) \\
& = x^{1/n}(u(x) - U) \\
v_n(x) & = x^{1/n}\hat v_n(x) - 2x\int_0^1 y^{1/n}\hat v_n(y)dy \\
& = x^{1/n}\left(u(x) - U\right) - 2x\int_0^1 y^{1/n}(u(y) - U)dy \\
\therefore \quad
|v_n(x) - u(x)|
& \le \left|(x^{1/n} - 1)u(x) - x^{1/n}U\right| + 2x\left|\int_0^1 y^{1/n}(u(y) - U)dy\right| \\
& \le \left|(x^{1/n} - 1)u(x) - x^{1/n}U\right| + 2x\left|\int_0^1 y^{1/n}(u(y) - U)dy\right| \\
\end{align*}
Let us now show that the sequence $I_n = \int_0^1 y^{1/n}(u(y) - U)dy$ converges to $0$.
Suppose $\epsilon > 0$ is given.
We know that the sequence of functions $(x^{1/n})$ converges uniformly on an interval $[x', 1]$ for any $x' > 0$, so we can pick $x' \in (0, \epsilon]$ and $N$ such that for $n > N$, $1 - x^{1/n} < \epsilon$ for $x > x'$. Then,
\begin{eqnarray*}
|I_n| & = &
\left|\int_0^1 y^{1/n}(u(y) - U) dy\right| \\
& \le & \left|\int_0^{x'} y^{1/n}(u(y) - U)dy\right| + \left|\int_{x'}^1 y^{1/n}(u(y) - U) dy\right| \\
& \le & \epsilon \|u - U\| + \left|\int_{0}^1 (y^{1/n} - 1)(u(y) - U) dy\right| + \left|\int_0^1 (u(y) - U) dy\right| \\
& \le & \epsilon \|u - U\| + \epsilon \|u - U\| = 2\epsilon \|\hat v\|.
\end{eqnarray*}
Therefore, $I_n \to 0$. Consequently, the sequence of functions $(2x|I_n|)$ converges uniformly to $0$.
We have dealt with the second part of $|v_n(x) - u(x)|$. We will show that the first part, $J_n(x) = \left|(x^{1/n} - 1)u(x) - x^{1/n}U\right|$, has a converging upper bound $|U|$. Suppose $\epsilon > 0$ is given. Same as before, we use the uniform convergence of the sequence $(x^{1/n})$ on an interval $[x', 1]$ with $x' > 0$ to choose $x' < \epsilon$ and $N$ such that $n > N$ implies $1 - x^{1/n} < \epsilon$ for $x \in [x', 1]$. Then, for $x < x'$, we have
\begin{align*}
J_n & = \left|x^{1/n}(u(x) - U) - u(x)\right| \\
& \le \left|x^{1/n}\right|\cdot\left|u(x) - U\right| + |u(x)| \\
& \le \epsilon \|u\| + \epsilon
\end{align*}and for $x \ge x'$, we have
\begin{align*}
J_n & = \left|(x^{1/n} - 1)u(x) - x^{1/n}U\right| \\
& \le \left|x^{1/n} - 1\right|\cdot|u(x)| + \left|x^{1/n}\right|\cdot|U| \\
& \le \epsilon \|u\| + |U|.
\end{align*}
Therefore, $\|J_n\| \le |U| + \epsilon \|u\| + \epsilon$.
Now we can conclude that $T_u(v_n) \to |U|$ as $n \to \infty$. What is left to investigate is whether $|U|$ is attainable by any $v \in M$.
Let's assume that $v \in M$ and $T_u(v) = |U|$. Then
\begin{align*}
|U| = T_u(v) & = \|v - u\| = \int_0^1 \|v - u\| dx \\
& \ge \int_0^1 \left|v(x) - u(x)\right| dx \ge \left|\int_0^1 (v(x) - u(x)) dx\right| = |U|
\end{align*}
and all inequalities must be equalities.
Hence,
\begin{align*}
\int_0^1\left(\|v - u\| - |v(x) - u(x)|\right) dx & =
\int_0^1\Big|\|v - u\| - |v(x) - u(x)|\Big| dx = 0.
\end{align*}
By continuity, we must have $|v(x) - u(x)| = \|v - u\| = |U|$ for all $x$, and so $v(x) = u(x) \pm |U|$. Since $v(0) = 0$ and $u(0) = 0$, we must have $|U| = 0$, which implies $u \in M$.
Hence,

For $u \in M - X$, the infimum of $T_u$, which is equal to $\left|\int_0^1 u\right|$, is not attainable by functions in $M$.

A: Assuming the infinity norm of $u$ is $\max_{x \in [0, 1]}u(x)$: 
The answer is yes, the infimum is not obtained. I shall give a intuitive answer.
let $u\in X \\ M$. Without loss of generality, let $\int_0^1 u(x)dx$ be $\epsilon$. 
Let w be another function. Under the sole restriction $\int_0^1 w(x)dx =0$, we minimize $\| w-u \|$ by having $w = u-\epsilon$. Indeed, $\|w-u \|$ would be $\epsilon$ in this case.
Now, we want $v$ to be a function with another restriction. $v(0) = 0$. However, we can make a v as close to $w$ as we desire. Therefore, the infimum of $\|v-u\|$ is still $\epsilon$, but it cannot be reached.
