# A Connected Graph Theorem

In our early days of calculus, we said that a function is continuous if you could draw its graph without having to lift your pen. This is related to two concepts: that of the graph being closed, and that of it being connected. Indeed, a powerful theorem I've recently learned is that a linear function (and more generally, of any function if the range is compact) between Banach spaces has a closed graph in the product topology if and only if the function which corresponds to that graph is continuous.

Now, because every normed vector space is path connected, so must be its continuous image, and hence its graph, if I'm not mistaken. However, much less clear, is whether or not the converse holds:

If the graph of a function (either, linear between Banach spaces, or perhaps more generally) is path connected, does this imply that the function must have been continuous?

I see similar results for when the domain of the function has the property that bounded sequences contain convergent sub-sequences (example 1 and example 2), but that is of course not going to be true of the domain is an infinite dimensional normed vector space.

Thanks very much.

• No, the graph of any linear function is path connected (any point in the graph is connected to $(0,0)$) but there are linear functions that are not continuous provided the dimension is infinite. – s.harp Dec 22 '20 at 17:34
• I see, let me now try to prove that (that they're connected to $(0,0)$). Thanks for the hint. – Thomas Winckelman Dec 22 '20 at 17:37

Generally this is false. Consider any metric space $$X$$ and let $$f:X\to[0,1]$$ be any discontinuous map. Define

$$h_f:X \times[0,1]\to[0,1]$$ $$h_f(x,t)=tf(x)$$

Clearly $$h_f$$ is discontinuous because $$f$$ is. But the graph of $$h_f$$ is path connected which I leave as an exercise. This shows that the statement is false even in low dimensional and compact case (e.g. $$X=[0,1]$$).

In the case of normed spaces (Banach or not) and linear maps this is even simpler. Let $$f:V\to W$$ be linear and $$(v,f(v))\in Gr(f)$$. Then there is a path

$$\lambda:[0,1]\to Gr(f)$$ $$\lambda(t)=(tv, f(tv))=(tv,tf(v))$$

which connects $$(v,f(v))$$ with $$(0,0)$$. Thus the graph of any linear map is path connected. But in the infinite dimension it is well known that there are discontinuous linear maps.

• Thanks very much for taking the time. What didn't occur to me at first was to try to connect an arbitrary point in the graph to $(0,0)$ (as you might guess, I'm still a bit of a novice at topology). I'll try writing out the proof and check my answer against yours. Still, thanks for putting an answer on the record. – Thomas Winckelman Dec 22 '20 at 17:42