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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,

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1 Answer 1

up vote 2 down vote accepted

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

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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
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@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
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@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

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