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Hi I've got this functional $f(x)=x(0)+x(1)$ where $x(t) $ is from $C[0,1]$.

I've checked it is linear: $f(ax+by)=ax(0)+ax(1)+by(0)+by(1) = af(x)+bf(y)$

How can I check if it is bounded and find its norm?

Here what I've done $|f(x)|=|x(0)+x(1)| \leq |x(0)|+| x(1)| $, what shall I do next?

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The norm of a functional is defined as

$$\sup_{\|x\|_{C[0,1]}=1} |f(x)|.$$

So far, you have shown that

$$\sup_{\|x\|_{C[0,1]}=1} |f(x)| \leq \sup_{\|x\|_{C[0,1]}=1} |x(0)|+|x(1)|.$$

Now, we can bound $|x(0)|\leq \|x\|_{C[0,1]}$ and $|x(1)|\leq \|x\|_{C[0,1]}$. Thus, we have

$$\sup_{\|x\|_{C[0,1]}=1} |x(0)|+|x(1)|\leq \sup_{\|x\|_{C[0,1]}=1} \|x\|_{C[0,1]}+\|x\|_{C[0,1]}=2.$$

We conclude that $f$ is bounded, and that the norm is at most $2$. To complete the proof, see if you can find a function $x\in C[0,1]$ which satisfies $\|x\|=1$ and $f(x)=2.$

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It is important to understand what $f$ is, first of all.

So what does $f$ do? $f$ is a function, that takes a function from $C[0,1]$, and returns a real number. That is, $f$ is a linear functional on $C[0,1]$.

So how do we define the norm of $f$? The norm of $f$ is the smallest constant $c$ such that $||f(x)|| \leq c||x||$ for all $x$. The norm of $x$ is the supremum norm that is $||x||=\sup_{t \in [0,1]} x(t)$.

Now, as you say: $$ ||f(x)|| \leq ||x(0)|| + ||x(1)|| \leq 2\cdot \sup_{t \in [0,1]} x(t) \leq 2||x|| $$

So $f$ is bounded. Hence, $f$ is a continuous linear functional on $C[0,1]$.

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