**Moderator's edit**:

Given that \(f(x) \) is a function continuous at \(x=1\) and satisfy the functional equation \(f(xy) = f(x) f(y) \) for all \(x\) and \(y\). Prove that \(f(x) \) is continuous at all non-zero \(x\).

**Moderator's edit**:

Given that \(f(x) \) is a function continuous at \(x=1\) and satisfy the functional equation \(f(xy) = f(x) f(y) \) for all \(x\) and \(y\). Prove that \(f(x) \) is continuous at all non-zero \(x\).

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TopNewest\( f(1)^2 = f(1) \Rightarrow f(1) = 1, \text {or } f(1) = 0. \\ f(1) = 0 \Rightarrow f(x) = 0 \text{ } \forall \text{ } x \in R \Rightarrow \text{f is continuous over R. Otherwise, } f(1) = 1. \\~\\ \text{Suppose we had a sequence of real numbers, } (a_n), \text{ such that, } \lim_{n\to\infty} (a_n) = 1. \\ \text{The existence of such a sequence is guaranteed by the density of real numbers.} \\ \text{Now f is continuous at 1, so, by the definition of continuity, } \\ \lim_{n\to\infty} f((a_n)) = f(\lim_{n\to\infty} (a_n)) = f(1) = 1. \\~\\ \text{Take an arbitrary real number, } c \text{. Now, } \\ \lim_{n\to\infty} (c \cdot a_n) = c \cdot \lim_{n\to\infty} (a_n) = c. \\ \text{So, } f(c) = f(c) \cdot 1 = f(c) \cdot (\lim_{n\to\infty} f(a_n)) = \lim_{n\to\infty} f(c) \cdot f(a_n) = \lim_{n\to\infty} f(c \cdot (a_n)). \\ \text{But, } \lim_{n\to\infty} c\cdot (a_n) = c. \\ \text{So, as } n \to\infty, c \cdot a_n \to c \text{. Set } c \cdot a_n = x. \text{ Then, } \lim_{x \to c} f(x) = f(c)\text{, which is what we needed to show. } \\~\\ \text{This finishes the proof! I like your questions, Akhilesh. } \) – Ameya Daigavane · 10 months, 3 weeks ago

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@parv morCan you please post a solution to this one too – Akhilesh Prasad · 11 months, 1 week ago

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e raised to x – Asif Mujawar · 11 months, 1 week ago

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– Akhilesh Prasad · 11 months, 1 week ago

I dont think \({ e }^{ x }\) can be the answer as it is continuous at \(x=0\) too. Please a provide a solution if you have arrived at this answer.Log in to reply

Can you please provide a solution for this one too.@Rishabh Cool – Akhilesh Prasad · 11 months, 1 week ago

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