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It’s been about a year since I had started teaching math and English to high (and junior) school students. I really started liking this job for its uniquely variant aspects of facing kids; in particular, when I am to guide them, although I first started it for my own career fortification in a way that “explaining”, “observing”, and “making myself understood” are the foundation to become a keen researcher in a field of math.

These days, some high school kids and even co-workers (which of course are instructors) have called me “math geek teacher” in a friendly manner, and I’m kind of proud of it for them giving me a large trust as the last stronghold in math at the private school (I’m not that good in math at all for studying for quite a few hours everyday).

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From here on, I’ll introduce a few propositions and problems from a past grad school exam that I’ve solved today. If there you find my misunderstanding or false proof, please do not be hesitant to help me.

【1】

(1) Let f(x) be a function defined in a closed interval [-1,1]. Explain the definition of f(x) being continuous at the point x=c (-1＜c＜1).

Proof)

Kind of a trivial question. With logic symbol, it’s written either as:

(∀ε＞0)(∃δ＞0)(|x-c|＜δ⇒|f(x)-f(c)|＜ε)

or

lim[x→c]f(x)=f(c) □

(2) A continuous sequencing function {f_n(x)} on an interval [-1,1] uniformly converges to a function f(x) when n→∞. Denote the continuity of f(x) on the inteval [-1,1].

Proof)

Assumption leads for all ε＞0, there exists n_0∈N, for all n＞n_0 (n∈N) it follows that |f-f_n|＜ε/3 (∀x∈[-1,1])〔i〕. Against the same arbitrary ε, taken a proper δ＞0, it leads:

|x-y|＜δ ⇒ |f_n(x)-f_n(y)|＜ε/3〔ii〕.

At the given points of x and y, 〔i〕 also indicates that:

|f(x)-f_n(x)|＜ε/3〔iii〕,

|f_n(y)-f(y)|＜ε/3〔iv〕

on the assumption that n＞n_0.

Taking a sum of〔ii〕,〔iii〕and〔iv〕, it follows that:

|f(x)-f(y)|≦|f_n(x)-f_n(y)| + |f(x)-f_n(x)| + |f_n(y)-f(y)|＜ε

is denoted with the above conditions underneath. This indicates f(x) is continuous for all x in the interval [-1,1] □

(3) At the previous question, prove following equation.

lim[n→∞]∫[-1,1]f_n(x)dx=∫[-1,1]f(x)dx.

Proof)

For f’s uniformly continuous, it follows that:

|∫[-1,1]f_n(x)dx – ∫[-1,1]f(x)dx|

≦∫[-1,1]|f_n(x)-f(x)|dx

≦2||f_n-f||＜2ε(n→∞) □

(4) Let ρ(x) be a continuous function on R, assume ρ(x)=0 when |x|≧1 and it meets following condition,

∫[-1,1]ρ(x)dx=1.

Now let φ(x) be another continuous function on [-1,1], prove following equation:

lim[n→∞]n∫[-1,1]ρ(nx)φ(x)dx=φ(0).

Proof)

Changing variable nx→x, the given equation results in

n∫[-1,1]ρ(nx)φ(x)dx

=∫[-n,n]ρ(x)φ(x/n)dx

=∫[-1,1]ρ(x)φ(x/n)dx.

For the third equation, the assumption applies to zero volume set out of interval [-1,1] and it stays an integral between [-1,1].

Now that the 1st integral mean theorem meets the condition to the integratable functions ρ(x) and φ(x/n), which both are continuous. And it follows that:

∃θ∈[-1,1],

∫[-1,1]ρ(x)φ(x/n)dx

=φ(θ/n)∫[-1,1]ρ(x)dx,

This implies the given equation converges to φ(0) for any constant number θ□