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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?

surfer1311
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  • beware when writing "$p(t) = f(x) + t(f(x) - f(x))$ : you should write $p(t,x)$. – Tlön Uqbar Orbis Tertius Mar 04 '15 at 18:53
  • I think he's fixing $x$ - but agreed, should be made clear. – Sam OT Mar 04 '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 $t \in [0,1]$. – Mnifldz Mar 04 '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? – surfer1311 Mar 04 '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 $g$. – Sam OT Mar 04 '15 at 18:57
  • Thanks for all the help so far. There is one more thing I'm not sure about: how do I show that $|g(x) - f(x)|$ is nonzero and finite? Because if it were either zero or infinite then $\delta$ would not be valid. – surfer1311 Mar 04 '15 at 19:09
  • Edit: My book states that the sup norm is only defined for bounded continuous functions, so even though the question does not explicitly state it, I think it's safe to assume that f and g are bounded. Should I take the case f = g separate? Then $\delta=c$ where $c$ is any constant when $f=g$ and $\delta = \frac{\epsilon}{|g(x)-f(x)|}$ when $f \not= g$. Is this a valid answer for $\delta$? – surfer1311 Mar 04 '15 at 19:19
  • If $f=g$, then $f$ and $g$ are certainly path-connected! Take the trivial path $p_x(t) = f(x) = g(x)$ for all $t$. Then certainly $p_x(0) = f(x)$ and $p_x(1) = g(x)$. But that is a good thought. You should perhaps pracy (however that's spelt!) with "we fix $x$, and wlog $f(x) \neq g(x)$ else trivial. – Sam OT Mar 04 '15 at 21:25
  • I guess a better way to do it really would be to ignore '$x$' completely. You want a path $p$ that has $p(0) = f$ and $p(1) = g$. So $p:[0,1] \to C[0,1]$. Then simply define $p(t) = f + t\cdot(g-f)$, as @Mnifldz. Reading it now, I think that's what he was meaning initially! – Sam OT Mar 04 '15 at 21:27

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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 $fg = {1 \over 2}((f+g)^2 - f^2 - g^2)$.

Anyway, hope this helps! =)

Sam OT
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