Bound of functional $g(0) - 7g(\frac12)$ Let's consider functional
$$f\colon (C[0,1], \Vert\cdot \Vert_\infty) \ni g\rightarrow g(0) - 7g(\frac12)$$
I'm trying to judge whether functional $f$ is bounded or not, however I'm dealing with problems of interpretation of condition to consider.
We want to check whether there is some ball that $f$ will be inside it.
But I'm not sure how exactly this ball will look in this space
In my opinion it should look like following:
$$B(C[0,1], r) = \{h \in C[0,1] : \lVert f-h \rVert < r\}$$
But I'm not sure if it's true. Could you please tell me if I'm thinking correctly, or maybe this ball looks differently in this space?
 A: For $g \in V = (\mathcal C([0,1], \mathbb R)$ you have
$$\vert f(g) \vert = \left\vert g(0) - 7 g\left(\frac{1}{2}\right) \right\vert \le \Vert g \Vert_\infty + 7 \Vert g \Vert_\infty = 8 \Vert g \Vert_\infty.$$
This proves that $f$ is bounded: for $\Vert g \Vert_\infty \le 1$, you have $\vert f(g) \vert \le 8$. As $f$ is linear, you immediately get for $g,h \in V$
$$\vert f(h) - f(g) \vert \le 8 \Vert h - g \Vert_\infty.$$
Even more, the norm of $f$ is equal to $8$ as for the map $h(x)=\vert 4x - 2 \vert -1$ we have $\Vert h \Vert_\infty = 1$ and $f(h)=8$.
A: Let $g \in C([0,1])$ be arbitrary and let $\epsilon>0$ be given. Then we have:
$$
|f(g)-f(h)|=|g(0)-7g(0.5)+h(0)-7h(0.5)| \leq |g(0)-h(0)|+|7g(0.5)-7h(0.5)| \leq \\
||g-h||_{\infty}+7||g-h||_{\infty}=8||g-h||_{\infty}<^! \epsilon
$$
So if we choose $||g-h||_{\infty} < \frac{\epsilon}{8}=\delta$, we have checked the epsilon-delta definition of continuity using open balls. You then have
$$
f(B_{\delta}(g)) \subset B_{\epsilon}(f(g))
$$
A short note on your notation for balls:
$$
B_r(f)=\{g \in C([0,1]) | \; ||f-g||_{\infty}<r   \}
$$
You can also use $B_r(f)=B(f,r)$ as a notation. You usually just mention the centre and the radius, not the space you are taking this ball in. However, you can or even have to denote it somewhere in your text/question/whatever.
