K(subset of C[0,1]) does not attain the minimal norm

Question

Let $V=C[0,1]$ and $K\subset V$ be defined by $K= \{f \in V | \int_{0}^{1/2} f(t) dt - \int_{1/2}^{1} f(t) dt = 1\}$.

Show that $K$ does not admit an element with minimal norm.

Answer 1

Hints:

Show that if $\|f\| < 1$ then $f \notin K$. Hence the distance is at least one.

Let $f_n$ be the function whose graph is given by joining the points $(0,{1 \over 1-{1 \over n}}), ({1 \over 2}-{1 \over n},{1 \over 1-{1 \over n}}), ({1 \over 2}+{1 \over n},-{1 \over 1-{1 \over n}}), 1,-{1 \over 1-{1 \over n}})$ and show that $f_n \in K$ and $\|f_n\| \to 1$.

Hence $\inf_{f \in K} \|f\| = 1$.

Now suppose $\|f\| = 1$ and $f \in K$ and show that this contradicts continuity of $f$.

To see the latter, let $a=\int_{0}^{1/2} f(t) dt, b=\int_{1/2}^{1} f(t) dt$, note that $|a|,|b| \le {1 \over 2}$ and $a-b=1$. Hence $a=1+b \ge {1 \over 2}$ and so $a={1 \over 2}, b = -{1 \over 2}$. Now note that $\int_{0}^{1/2} (1-f(t)) dt = 0$ and since $f(t) \le 1$ we see that $f(t) = 1$ for $t \in [0, {1 \over2}]$. Repeating for the second side, we see that $f(t) = -1$ for $t \in [{1 \over2},1]$. This gives a contradiction since $f({1 \over 2})$ cannot be both $\pm1$.