Let $(n)=\{1, \ldots , n \}$ with $\{0,n\}$ frontier points and $(n) \setminus \{0,n\}$ interior points, we say that a function $f: (n) \rightarrow \mathbb{R}$ is a harmonic function if $f(x)=\frac{f(x-1)+f(x+1)}{2}$ for all interior points of $(n)$
show that
- Every harmonic function reaches its maximum and minimum values at the frontier
- If f and g are harmonic functions and coincide at the frontier then they are equal
First i try to found a recurrence relation $2f(k) - f(k-1)=f(k+2)$ and now i proceded by induction on cardinality of $(n)$ these is correct for 1?
For 2 i have problems i try the induction but i fail
Any hint or help i will be very grateful
Assume the minimum occurs at some point $i$ in the interior, i.e., $f(i) < f(j)$ for all $j \in (n) \setminus \{i\}$. Since $f$ is harmonic, $$ f(i) = \frac{f(i + 1) + f(i - 1)}{2} > \frac{f(i) + f(i)}{2} = f(i),$$ implying a contradiction. Hence, the minimum cannot occur at an interior point. You can use the same logic for maximum.
Consider the function $h = f - g$. Note that $h$ is also harmonic with $h(0) = h(n) = 0$. We claim that for any harmonic function $h$ with $h(0) = 0$, $h(i) = ih(1)$ for all $i \in \{1,2,\dots,n\}$. We prove this claim using induction. The base case for $i = 1$ follows immediately. Assume true for $1, \dots, k$ and we show it for $k + 1$. Using the recurrence relation you derived, we have, $h(k+1) = 2h(k) - h(k-1) = 2kh(1) - (k-1)h(1) = (k+1)h(1)$, as required. Hence, $h(i) = ih(1)$ for all $i \in \{1,2,\dots,n\}$. In particular, $h(n) = nh(1)$. But it is given that $h(n) = 0$ implying $h(1) = 0$ and consequently, $h(i) = 0$ for all $i \in (n)$. Hence, $f - g = 0$, or equivalently, $f = g$.