Let $L:E\rightarrow \mathbb{R}$ be a linear functional and define $E=\{f \in C([a,b]); f(a)=f(b)=0\}$. Let $\|f\|_0 = \max |f(x)|$ and $$\|f\|_1=\int_a^b |f(x)|\,dx$$ be norms on $E$. I must prove that the functional $$L(f)=\int_a^b f(x) \, dx$$ is continuous in $(E,\|\cdot\|_0)$ and $(E,\|\cdot\|_1)$. I know that a functional is continuous if, and only if, it is bounded, but i'm a little confuse on the definition, can i just use something like $$|L(f)|= \left| \int_a^bf(x) \, dx \right| \leq \int_a^b |f(x)| \, dx\leq (b-a) \|f\|_0$$ and a similar argument for $\|f\|_1$ or am I missing something? Thanks!
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I guess you mean the right thing. However, just for clarification, bounded is meant in the sense that the operator norm of $L$ is bounded, i.e. the $\sup_{f \in E, |f| = 1} |L(f)| < \infty$, where $| \cdot|$ is w.r.t. to the norm of the domain space $E$. But yes, in that case you can just do what you wrote. – hal4math Apr 12 '21 at 03:49
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A linear function is continuous when its action on the unit ball is bounded. So let $f$ be some element of $E$ in the unit ball, that is, suppose $\|f\|\leq 1$. For the $\|\cdot\|_0$ case, by maximizing the function $f$ by the constant $\|f\|$, we have
$$ |L(f)| = \Bigg|\int f \Bigg| \leq \Bigg|\int \|f\| \Bigg| =\|f\|(b-a) \leq b-a <\infty$$
and in the $\|\cdot\|_1$ case, by using the triangle inequality for integrals, we have
$$ |L(f)| = \Bigg|\int f \Bigg| \leq \int |f| =\|f\| \leq 1 <\infty. $$
Trevor Fancher
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