A vector without minimum norm in a Banach space


Let $E = C[0, 1]$, with sup norm. Let $K$ consist of all $f$ in $E$ such that
Prove that $K$ is a closed convex subset of $E$ which contains no element of
minimum norm.

My proof:

For convexity of $K$
Let $f, g \in K$ and $\alpha\in [0,1]$

$\begin{align}\int_{0}^{\frac{1}{2}}&((1-\alpha)f(s)+\alpha g(s))ds-\int_{\frac{1}{2}}^{1}((1-\alpha)f(s)+\alpha g(s))ds& \\

For closure:

Let $f_{n}$ be a sequence in $K$ such that $f_{n}\to f$ as $n\to \infty$. We need to show that $f\in K$

$\int_{0}^{\frac{1}{2}}f_{n}(s)ds-\int_{\frac{1}{2}}^{1}f_{n}(s)ds=1$ because $f_{n}$ is in $K$.

Taking limit as $n\to \infty$ we have $f\in K$.

Next, is to show $K$ contains no element of minimum norm, but I stuck here, can I go by contradiction? That is supposing K contains at least 1 element of minimum norm, say $f_{0}$ i.e $\|f_{0}\|=\inf_{f\in K}\|f\|$, then what?

Please I need the help of professionals. Thanks

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I can’t think of an “abstract” way to do what you are asking— that is, I think the simplest and perhaps only thing to do is to explicitly evaluate the number
\inf \{\|f\|: f \in K\}
and then prove, using the knowledge of what this number is, that no element of $K$ can have this number as its norm.

If $f$ is in $K$, then (making use of some familiar inequalities)
1 & = \left|\int_0^{\frac{1}{2}} f(s) \, ds – \int_{\frac{1}{2}}^1 f(s) \, ds\right| \\
& \leq \left|\int_0^{\frac{1}{2}} f(s) \, ds\right| + \left| \int_{\frac{1}{2}}^1 f(s) \, ds\right| \\
& \leq \int_0^{\frac{1}{2}} |f(s)| \, ds + \int_{\frac{1}{2}}^1 |f(s)| \, ds \\
& \leq \int_0^{\frac{1}{2}} \|f\| \, ds +\int_{\frac{1}{2}}^1 \|f\| \, ds
= \|f\|.
We conclude that $\inf \{\|f\|: f \in K\} \geq 1$.

Define for $n \geq 2$ the function $f_n: [0,1] \to \mathbb{R}$ by declaring its graph to be the union of the straight line segment from $(0,1)$ to $(\frac{1}{2}, 1)$, the straight line segment from $(\frac{1}{2},1)$ to $(\frac{1}{2} + \frac{1}{n}, \frac{1+n}{1-n})$, and the straight line segment from $(\frac{1}{2} + \frac{1}{n}, \frac{1+n}{1-n})$ to $(1, \frac{1+n}{1-n})$. Clearly $f_n \in C[0,1]$ for all $n$, and short calculations show that
f_n \in K, \qquad \|f_n\| = \frac{n+1}{n-1}, \quad n \geq 2.
Since $\inf \{\|f_n\|: n \geq 2\} = \lim_{n \to \infty} \frac{n+1}{n-1} = 1$, it follows that $\inf \{\|f\|: f \in K\} \leq 1$, and hence that
\inf \{\|f\|: f \in K \} = 1.
I claim that it is impossible for $f \in K$ to satisfy $\|f\| = 1$.

Suppose $f$ in $C[0,1]$ satisfies $\|f\| = 1$. Then each inequality in the above proof that $\|f\| \geq 1$ is an equality. It is easy to see that since $f$ is continuous, the last inequality in that proof is an equality if and only if $|f(s)| = 1$ holds for all $s$. If your $C[0,1]$ is the set of real-valued functions on $[0,1]$ then we already have a contradiction, because a real-valued continuous function satisfying this condition must either be the constant $1$ or the constant $-1$, and a short check shows that neither of these satisfy the condition for membership in $K$. Otherwise we have to work a little harder: the second-to-last inequality being equality forces
\left|\int_0^{\frac{1}{2}} f(s) \, ds\right| = \int_0^{\frac{1}{2}} |f(s)| \, ds
and similarly for the integral from $\frac{1}{2}$ to $1$. But a short examination of your favorite proof of the general inequality $|\int_a^b f(s) \, ds| \leq \int_a^b |f(s)| \, ds$ should make apparent that equality holds in this general inequality if and only if if and only if there is a complex number $c$ with $|c| = 1$ and a nonnegative function $F$ with $f = cF$. Since the $f$ relevant to us is known to have constant modulus, we conclude that there are complex numbers $c_1$ and $c_2$ satisfying $|c_1| = |c_2| = 1$ and $f(x) = c_1$ for all $x \in [0, \frac{1}{2}]$ and $f(x) = c_2$ for all $x \in [\frac{1}{2}, 1]$. Continuity of $f$ forces $c_1 = c_2$; but then $\int_0^{\frac{1}{2}} f(x) \, dx – \int_{\frac{1}{2}}^1 f(x) \, dx = 0$ is not $1$, a contradiction.