GATE Mathematics Real Analysis Syllabus

Give an example showing the equicontinuity hypothesis cannot be dropped in Ascoli–Arzelà.

On $[0,1]$, take $f_n(x) = \sin(nx)$. The family is uniformly bounded by $1$ but not equicontinuous, and no subsequence converges uniformly. So uniform boundedness alone is insufficient.

Give an example showing uniform boundedness cannot be dropped in Ascoli–Arzelà.

On $[0,1]$ take $f_n(x) = n$ (constant functions). They are equicontinuous (in fact constant) but not uniformly bounded, and no subsequence converges. So equicontinuity alone is insufficient.

State the Monotone Convergence Theorem (MCT) for Lebesgue integrals.

If $\{f_n\}$ are measurable with $0 \leq f_1 \leq f_2 \leq \cdots$ and $f_n \to f$ pointwise a.e., then $f$ is measurable and $$\lim_{n\to\infty}\int f_n\,d\mu = \int f\,d\mu = \int \lim_{n\to\infty} f_n\,d\mu.$$

What are the key hypotheses of the Monotone Convergence Theorem?

(1) Each $f_n$ is measurable and nonnegative, and (2) the sequence is monotone increasing: $f_n \leq f_{n+1}$ a.e. No integrability bound is needed; the limit integral may be $+\infty$.

Does MCT hold for a decreasing sequence of nonnegative functions? Give a counterexample.

Not in general. Take $f_n = \chi_{[n,\infty)}$ on $\mathbb{R}$. Then $f_n \downarrow 0$ pointwise but $\int f_n = \infty$ for all $n$, while $\int 0 = 0$. A finite-integral condition (e.g. $\int f_1 < \infty$) is needed for the decreasing version.

State Fatou's Lemma.

If $\{f_n\}$ are measurable and nonnegative, then $$\int \liminf_{n\to\infty} f_n\,d\mu \leq \liminf_{n\to\infty} \int f_n\,d\mu.$$

What is the direction of the inequality in Fatou's Lemma, and can it be strict?

The integral of the liminf is less than or equal to the liminf of the integrals: $\int \liminf f_n \leq \liminf \int f_n$. The inequality can be strict, e.g. $f_n = \chi_{[n,n+1]}$ gives $0 \leq 1$.

Give an example where Fatou's Lemma is a strict inequality.

Let $f_n = n\,\chi_{(0,1/n)}$ on $[0,1]$. Then $\liminf f_n = 0$ a.e., so $\int \liminf f_n = 0$, but $\int f_n = 1$ for all $n$, so $\liminf \int f_n = 1$. Thus $0 < 1$.

State the Dominated Convergence Theorem (DCT).

If $f_n \to f$ pointwise a.e., each $f_n$ is measurable, and there exists an integrable $g$ with $|f_n| \leq g$ a.e. for all $n$, then $f$ is integrable and $$\lim_{n\to\infty}\int f_n\,d\mu = \int f\,d\mu.$$

What is the role of the dominating function $g$ in the Dominated Convergence Theorem?

$g$ is an integrable function ($\int g\,d\mu < \infty$) satisfying $|f_n| \leq g$ a.e. for all $n$. It controls the tails uniformly, preventing mass from escaping and justifying the interchange of limit and integral.

State the conclusion of DCT regarding the $L^1$ norm of $f_n - f$.

Under the DCT hypotheses, $\int |f_n - f|\,d\mu \to 0$, i.e. $f_n \to f$ in $L^1$. This is stronger than mere convergence of the integrals.

Give an example showing the dominating function in DCT is essential.

Let $f_n = n\,\chi_{(0,1/n)}$ on $[0,1]$. Then $f_n \to 0$ pointwise but $\int f_n = 1 \not\to 0$. No integrable $g$ dominates all $f_n$, so DCT does not apply.

How is the Monotone Convergence Theorem typically used to prove Fatou's Lemma?

Define $g_n = \inf_{k \geq n} f_k$, which increases to $\liminf f_n$. By MCT, $\int g_n \to \int \liminf f_n$. Since $g_n \leq f_k$ for $k \geq n$, $\int g_n \leq \inf_{k\geq n}\int f_k$, and taking limits gives Fatou's inequality.

How is the Dominated Convergence Theorem typically proved from Fatou's Lemma?

Apply Fatou to the nonnegative sequences $g + f_n \geq 0$ and $g - f_n \geq 0$. This yields $\int f \leq \liminf \int f_n$ and $\int f \geq \limsup \int f_n$, forcing $\lim \int f_n = \int f$.

Compare the hypotheses of MCT, Fatou's Lemma, and DCT.

MCT: nonnegative and monotone increasing. Fatou: nonnegative (no monotonicity, gives only an inequality). DCT: pointwise convergent and dominated by an integrable $g$ (functions may change sign). DCT and Fatou do not require monotonicity.

State Tonelli-type corollary of MCT for series of nonnegative functions.

If $f_n \geq 0$ are measurable, then $$\int \sum_{n=1}^{\infty} f_n\,d\mu = \sum_{n=1}^{\infty} \int f_n\,d\mu.$$ This term-by-term integration follows from applying MCT to the partial sums.

Under DCT, state the condition allowing differentiation under the integral sign.

If $\partial_t f(x,t)$ exists and $|\partial_t f(x,t)| \leq g(x)$ with $g$ integrable, then $$\frac{d}{dt}\int f(x,t)\,d\mu(x) = \int \frac{\partial f}{\partial t}(x,t)\,d\mu(x).$$

Does pointwise a.e. convergence alone guarantee $\int f_n \to \int f$? Explain.

No. Pointwise convergence is insufficient; one needs extra control such as monotonicity (MCT), nonnegativity for an inequality (Fatou), or domination (DCT). The example $f_n = n\chi_{(0,1/n)}$ shows failure without such control.

State the bounded convergence theorem as a special case of DCT.

If $\mu(E) < \infty$, $f_n \to f$ pointwise a.e. on $E$, and $|f_n| \leq M$ (a constant) for all $n$, then $\lim_n \int_E f_n\,d\mu = \int_E f\,d\mu$. Here the dominating function is the constant $M$, integrable because $\mu(E)<\infty$.

What does 'almost everywhere (a.e.)' mean in the convergence theorems?

A property holds almost everywhere if it holds for all points except a set of measure zero. In MCT/Fatou/DCT, hypotheses like $f_n \to f$ or $|f_n| \leq g$ need only hold outside a $\mu$-null set.

State the relationship $\liminf \int f_n \leq \limsup \int f_n$ and when equality forces convergence.

Always $\liminf \int f_n \leq \limsup \int f_n$. When DCT applies, both equal $\int f$, so $\lim \int f_n$ exists and equals $\int \lim f_n$. Fatou bounds only the liminf from below by $\int \liminf f_n$.

Can uniform convergence on an unbounded domain allow interchange of limit and integral?

Not necessarily. On $\mathbb{R}$, $f_n = \frac{1}{n}\chi_{[0,n]}$ converges uniformly to $0$, but $\int_{\mathbb{R}} f_n = 1 \not\to 0$. Uniform convergence guarantees interchange only on finite-measure (bounded) domains.

State the term-by-term integration theorem for a uniformly convergent series.

If $\sum f_n$ converges uniformly to $S$ on $[a,b]$ with each $f_n$ continuous, then $$\int_a^b S\,dx = \sum_{n=1}^{\infty} \int_a^b f_n\,dx.$$

What is the pointwise limit and uniform-convergence status of $f_n(x) = \frac{nx}{1+n^2x^2}$ on $[0,1]$?

Pointwise $f_n(x) \to 0$ for all $x \in [0,1]$. But $f_n(1/n) = \tfrac{1}{2}$, so $\sup |f_n| \geq \tfrac12$ and convergence is not uniform on $[0,1]$.

State Fatou's Lemma in its $\limsup$ form (reverse Fatou).

If $f_n \leq g$ with $g$ integrable, then $$\limsup_{n\to\infty} \int f_n\,d\mu \leq \int \limsup_{n\to\infty} f_n\,d\mu.$$ This requires an integrable upper bound, dual to the standard nonnegative lower bound.

What general property of $C(K)$ makes Ascoli–Arzelà a compactness criterion (vs. finite-dimensional spaces)?

$C(K)$ is infinite-dimensional, so closed and bounded sets need not be compact (unlike $\mathbb{R}^n$ by Heine–Borel). Equicontinuity is the extra condition that restores compactness in this function space.

In MCT, may the common limit $\int f\,d\mu$ be infinite?

Yes. MCT holds in $[0, +\infty]$; if $\int f_n \to +\infty$ then $\int f = +\infty$ as well. The theorem asserts equality of the (possibly infinite) limits, requiring no finiteness assumption.

Classify these as 'inequality' vs 'equality' results: Fatou's Lemma, MCT, DCT.

Fatou's Lemma gives an inequality ($\int \liminf f_n \leq \liminf \int f_n$). MCT and DCT give equalities ($\lim \int f_n = \int \lim f_n$). Fatou is the weakest, needing only nonnegativity.

For a sequence converging uniformly on $E$, relate $\|f_n - f\|_\infty$ to convergence in the supremum metric.

Uniform convergence is exactly convergence in the sup-norm metric $d(f,g) = \sup_{x\in E}|f(x)-g(x)|$. Thus $f_n \to f$ uniformly $\iff \|f_n - f\|_\infty \to 0$, i.e. convergence in $(C_b(E), \|\cdot\|_\infty)$.

Summarize the logical chain MCT $\Rightarrow$ Fatou $\Rightarrow$ DCT.

MCT (monotone nonnegative) is proved first; applying MCT to $\inf_{k\geq n} f_k$ yields Fatou's Lemma; applying Fatou to $g \pm f_n$ (with dominating $g$) yields the Dominated Convergence Theorem. Each builds on the previous.