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Let $X=\{X\in C[0,1]\}$: $f(1/2)=0$, with the induced norm by $C[0,1]$ and the functional $\varphi:X\rightarrow\mathbb{K}$ defined by:

$$ \varphi(f)=\int_0^1f(t)dt\;\;\forall f\in X $$

Prove that $\varphi$ is continuous.

I have made the following: $||\varphi(f)||=|\int_0^1f(t)dt|\leq \int_0^1|f(t)|dt\leq 1 ||f||_\infty$.

Now, that relation is valid for all $f\in C[0,1]$ such as $||f||_\infty=1$. This proves that the operator is bounded in the unit ball so it's continuous.

The problem with this exercise is that I have not made use of $f(1/2)=0$ so I assume my solution is wrong at some point. Any idea?

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    $\begingroup$ Side note: $\Vert$ $\Vert$ renders more nicely than $||$ $||$ for norms. $\endgroup$ – Martin R Nov 2 '19 at 7:58
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Your calculation $$ \Vert \varphi(f)\Vert =|\int_0^1f(t)dt|\leq \int_0^1|f(t)|dt\leq 1 \Vert f\Vert _\infty $$ is correct. It shows that $\varphi$ is continuous on $ C[0,1]$. As a consequence, $\varphi$ is continuous on the subspace $X$ of $ C[0,1]$ and its norm (as a linear functional on $X$) is at most one.

The condition $f(\frac 12) = 0$ must be considered when computing the exact norm. Try to think of functions which are constant apart from a narrow spike to satisfy $f(\frac 12) = 0$.

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Computing the norm of $\phi$: let $f_n(x)=1$ for $|x-\frac 1 2| >\frac 1 n$ and $n|x-\frac 1 2|$ for $|x-\frac 1 2 |\leq \frac 1n$. Then $f_n \in X$, $\|f_n\|=1$ for all $n$ and $\phi (f_n) =1-\frac 1 n \to 1$. Hence $\|\phi\|=1$.

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