# Path-connectedness of continuous functions

I want to prove that the metric space $C[0,1]$ with the metric $d(f,g) = sup_{x \in [0,1]} |f(x) - g(x)|$ is path-connected. I think I've done most of the proof, but I am not too sure about the outcome.

I simply tried the straight line path $p(t) = f(x) + t(g(x) - (f(x))$ so that $p(0) = f(x)$ and $p(1) = g(x)$. To show path-connectedness, $p \in C[0,1]$ must be satisfied, hence I need to prove that $p$ is continuous.

Let $t_0 \in [0,1]$ and $\epsilon > 0$ and assume $|t-t_0| < \delta$. Then $sup_{t \in [0,1]} |p(t) - p(t_0)| = sup_{t \in [0,1]}|t-t_0||(g(x)-f(x)| < \delta|g(x) - f(x)|$.

So if I let $\delta = \frac{\epsilon}{|g(x) - f(x)|}$ is the path p then continuous? Did I forget anything here or is this proof valid?

• beware when writing "$p(t) = f(x) + t(f(x) - f(x))$ : you should write $p(t,x)$. Mar 4 '15 at 18:53
• I think he's fixing $x$ - but agreed, should be made clear. Mar 4 '15 at 18:54
• You can also use the fact that $C[0,1]$ is a vector space to argue that $f + t(g-f)$ is a continuous function for all $t \in [0,1]$. Mar 4 '15 at 18:55
• I forgot to fix x indeed. If I just were to say "let $x \in [0,1]$" before starting the proof on continuity then that should be enough right? Mar 4 '15 at 18:56
• Probably should write $t \to f(t) + t(g(t) - f(t))$, since here '$t$' is a function (identity function, $t \to t$), as are $f$ and $g$. Mar 4 '15 at 18:57

Looks good to me, albeit a bit round-about; see my comment above for more details. As mentioned in the comments above, it's worth saying "let us fix $x$", and then show that it's true for all $x$ ($\in [0,1]$). If writing $p(t,x)$ doesn't seem nice to you, I would personally write it as $p_x(t)$ - *note that these are meaning the same thing just maybe makes it clearer what $x$ and $t$ are 'doing'.
There is something else that it's good to be aware of though. The product of two continuous functions (on $[0,1]$, say) is continuous, as is the sum. It's worth trying to show this for arbitrary functions. When you know this, you can say immediately that $p$ is continuous. Again as mentioned in the comments this makes $C[0,1]$ a vector space.
Hint: If you're struggling to show that $fg$ is continuous when $f$ and $g$ are, then consider this trick. You (hopefully) have shown that $f + g$ is. Show that $f^2$ is (should be easier as a special case). Then use the fact that $fg = {1 \over 2}((f+g)^2 - f^2 - g^2)$.