# Prove that this functional is continuous and calculate its norm

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?

• Side note: $\Vert$ $\Vert$ renders more nicely than $||$ $||$ for norms. – Martin R Nov 2 '19 at 7:58

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 $$f(\frac 12) = 0$$.

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 $$\|\phi\|=1$$.