# Continuous Functionals and Norms

In Luenberger Optimization book, pg. 40 upper semicontinuity for a functional is defined as "if given $\epsilon > 0$ there is a $\delta > 0$ s.t. $f(x) - f(x_0) < \epsilon$ for $||x-x_0|| < \delta$". Then it goes on as "a functional is said to be lower semicontinuous at $x_0$ if $-f$ is upper semicontinuous at $x_0$". Functional is continuous if it is both.

However later in the same page Example 1 talks about a functional $f$ defined on $C[0,1]$

$$f(x) = \int_{0}^{1/2}x(t)dt - \int_{1/2}^{1}x(t)dt$$

and says "it is easily verified that f is continuous since, in fact, $|f(x)| \le ||x||$".

How did he make this connection? Where did $x_0$ go? And how come there is a relation between the functional and the norm on $x$?

Thanks,

-

$f$ is linear, so $f(x)-f(x_0)=f(x-x_0)$. From this $|f(x)-f(x_0)|=|f(x-x_0)|\le\Vert x-x_0\Vert$; from which continuity easily follows.

To see why $|f(x)|\le\Vert x\Vert$: $$|f(x)|=\biggl|\int_0^{1/2} x(t)\,dt -\int_{1/2}^1 x(t)\,dt \biggr|\le \biggl|\int_0^{1/2} x(t)\,dt \biggr|+ \biggl| \int_{1/2}^1 x(t)\,dt \biggr|\le \textstyle {1\over2}\Vert x\Vert+ {1\over2}\Vert x\Vert=\Vert x\Vert.$$

-
Hi David, how did you know the inequality $|f(x-x_0)|\le\Vert x-x_0\Vert$ ? This problem doesn't seem to define a norm for $C[0,1]$, or I missed it. –  BB_ML Jan 16 '12 at 13:26
@user6786 It is the supremum norm, defined in Example 1 on page 23. (The inequality $|f(x-x_0)|\le\Vert x-x_0\Vert$ follows directly from $|f(x)|\le \Vert x\Vert$.) –  David Mitra Jan 16 '12 at 13:31
got it, thanks. –  BB_ML Jan 16 '12 at 13:48
One more question, the derivation you've shown shows how to obtain the inequality $|f(x)| \le ||x||$. How do we conclude given this inequality, continuity should be correct as well? Without any use of $\epsilon,\delta$? I guess if $|f(x)| \le ||x||$ than $f(x) - f(x_0) < \epsilon$ for $||x-x_0|| < \delta$ should be automatically true also? –  BB_ML Jan 16 '12 at 13:57
@user6786 Your thoughts on $\varepsilon$-$\delta$ seem right. More precisely, the David's answer shows that the function is $1$-Lipschitz. It is a general fact that all Lipschitz functions are continuous; the proof is an easy application of the $\varepsilon$-$\delta$ definition. It is usually cleaner to prove that a function is Lipschitz: we can do away with $\varepsilon$'s and $\delta$'s. –  Srivatsan Jan 16 '12 at 14:51
You use that if an operator $T: X \to Y$ is linear and continuous at one point in $X$ then $T$ is continuous on all of $X$.
I assume $\|x\| = \|x\|_\infty$.
It's easy to verify that $f$ is linear using the linearity of $\int : C([0,1]) \to \mathbb{R}$. Then you can show $f$ is continuous at $0 \in C([0,1])$:
\begin{align} |f(x)| &= \left | \int_0^{\frac12} x(t) dt - \int_\frac12^1 x(t)dt \right | \\ &\leq \left | \int_0^{\frac12} x(t) dt \right | + \left | \int_\frac12^1 x(t)dt \right | \leq \left | \int_0^{\frac12} \|x\|_\infty dt \right | + \left | \int_{\frac12}^1 \|x\|_\infty dt \right | = \|x\|_\infty \end{align}