Proving Constancy of a Function

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[slow elbow]

putting aside the case a = 0 which is trivial, you can suppose without loss of generality that |a| < 1 for starters

first question for you: why do you not lose generality?

second question: use the property, but instead of x, plug ax. What happens?

[acoustic vortex]

The main point is that the property f(x) = f(ax) holds for all x in R, and a is not equal to plus or minus 1. If the absolute value of a is greater than 1, I can redefine the function as g(x) = f(x/a). This transforms the equation into g(x) = g(ax), where the absolute value of a in the new function is now less than 1. Since the reasoning works the same way for both cases (whether the absolute value of a is less than 1 or greater than 1), I don't lose any generality by assuming the absolute value of a is less than 1.

.

lim x->0 f(x) = lim x->0 f(ax) = f(0) ?

Or did you mean something else?

[slow elbow]

Very good

Good job

[slow elbow]

Here's another bone: I want to show that for any x, f(x) = f(0)

in particular, for x fixed, f(x) = f(ax) = f(a²x)

f(x) = f(ax) is given by the property immediately

but by replacing "x" with y = ax, you get: f(y) = f(ay), that is, f(ax) = f(a²x)

[acoustic vortex]

I use the given property that f(x) = f(ax) for all x. To show that f(x) = f(0) for any fixed x, I first note that f(x) = f(ax) by the given property. Then, I replace x with y = ax, which gives me f(ax) = f(a²x), so I get f(x) = f(a²x).

By repeatedly applying this process, I get f(x) = f(ax) = f(a²x) = f(a³x), and so on. Since aⁿx approaches 0 for |a| < 1 as n becomes large, I conclude that f(x) = f(0) for any x. This shows that f is constant.

[slow elbow]

Since aⁿx approaches 0 for |a| < 1 as n becomes large,
yes, as n tends to infinity

I conclude that f(x) = f(0) for any x
The conclusion is true, but you're jumping the gun: why?

@acoustic vortex

you're missing an argument to conclude that f(x) = f(0)

[acoustic vortex]

So I have to prove that from lim (n → ∞) f(a^n x) = f(0) follows f(x) = f(0) ?

[slow elbow]

lim (n → ∞) f(a^n x) = f(0)
this is what you need to show

f(a^n x) converges as n tends to infinity, and the limit is f(0)

this is the unproven claim

you do know that a^n x tends to 0, which is a good start

but you need a specific property on f as well

hint: read the text again

[acoustic vortex]

It is because of |a| < 1 . Because if you often multiply a number less than 1 by itself, it always becomes smaller.

[slow elbow]

yes, but that's for a^n x

a^n x tends to 0

but that doesn't necessarily mean f(a^n x) has a limit

and even if it did, it might not be f(0)

for example, consider: $f = 1_{{0}}$

[glad linden]

Rion

[slow elbow]

the indicator function of the singleton {0}

so f(x) = 1 if x = 0, and otherwise f(x) = 0

it does verify f(x) = f(ax)

[acoustic vortex]

Because f is continuous at 0, the continuity guarantees that whenever the input a^n x approaches 0, the corresponding outputs f(a^n x) must approach f(0). ?

[slow elbow]

good job

yes, continuity at 0 guarantees that for any sequence $(x_n)_{n \in \bN}$ that converges to 0, you have $f(x_n) \xrightarrow[n \to \infty]{} f(0)$

[glad linden]

Rion

[slow elbow]

so you take $x_n = a^n x$

[glad linden]

Rion

[acoustic vortex]

Thank you 🫶

[slow elbow]

👍