# A vector without minimum norm in a Banach space

Question:

Let $E = C[0, 1]$, with sup norm. Let $K$ consist of all $f$ in $E$ such that $$\int_{0}^{\frac{1}{2}}f(s)ds-\int_{\frac{1}{2}}^{1}f(s)ds=1$$ 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 $\alpha\in [0,1]$

Then \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& \\ &=(1-\alpha)\left(\int_{0}^{\frac{1}{2}}f(s)ds-\int_{\frac{1}{2}}^{1}f(s)ds\right)+\alpha\left(\int_{0}^{\frac{1}{2}}g(s)ds-\int_{\frac{1}{2}}^{1}g(s)ds\right)&\\ &=(1-\alpha)+\alpha=1 \end{align}

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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First construct functions in $K$ with norm $1+\epsilon$. Then show no function of norm 1 is in $K$. –  David Mitra Nov 10 '11 at 8:35
Implication arrows ("$\implies$") are for implications between statements; you're using them to denote equality, which is more usually denoted by equals signs. Also, note that you can put \left and \right before parentheses to get them to adapt their size to their content. –  joriki Nov 10 '11 at 8:46
How is the step "Taking [the] limit as $n\to\infty$ we have $f\in K$" justified? Note that this wouldn't hold for pointwise convergence, so you have to use the uniform convergence of the $f_n$. –  joriki Nov 10 '11 at 8:56

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) \begin{align} 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\|. \end{align} 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.
That last argument is wrong -- equality holds as long as $f$ doesn't change phase; any phase factor in $f$ drops out of the inequality. –  joriki Nov 10 '11 at 8:51