QEDspace — The practice ground for advanced mathematics.
Real Analysis · Medium
Let $f \in L^1([0,1])$ satisfy $\int_0^1 x^n f(x)\,dx = 0$ for every integer $n \ge 0$. Show that $f = 0$ almost everywhere.
Complex Analysis · Medium
Let $f$ be entire and suppose $|f(z)| \le C\,(1+|z|)^{3/2}$ for every $z \in \mathbb{C}$. Prove that $f$ is a polynomial of degree at most $1$.
Real Analysis · Hard
Suppose $f_n \to f$ in measure on $[0,1]$ and $\sup_n \int_0^1 |f_n|^2\,dx < \infty$. Prove that $\int_0^1 |f_n - f|\,dx \to 0$.
Complex Analysis · Hard
Let $f$ be entire and injective. Prove that $f(z) = az + b$ for some $a \ne 0$.
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QEDspace is a practice ground for advanced mathematics — a growing library of problems in the style of university exams, with the relevant terminology defined in context, escalating hints, graded solutions, and an AI tutor that guides without spoiling.
It’s built for everyone from casual problem solvers looking to strengthen and grow their advanced mathematics knowledge to serious problem solvers preparing for university exams.
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Easy
Let $f \in L^1(\mathbb{R})$. Show that $\lim_{n\to\infty} \int_{\mathbb{R}} f(x)\cos(x/n)\,dx = \int_{\mathbb{R}} f(x)\,dx$.
Your answer
Preview
$f(x)\cos(x/n) \to f(x)$ pointwise, and $|f\cos(x/n)| \le |f| \in L^1$. By Dominated Convergence, $\int f\cos(x/n) \to \int f$. $\blacksquare$
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Grading…
✓ Correct
Minor: justify the pointwise limit — $\cos(x/n) \to 1$ for each fixed $x$.
Formal Proof
∎
Fix $x \in \mathbb{R}$; then $x/n \to 0$, so $\cos(x/n) \to 1$ and $f(x)\cos(x/n) \to f(x)$. Since $|f(x)\cos(x/n)| \le |f(x)|$ with $f \in L^1(\mathbb{R})$, the Dominated Convergence Theorem gives $\int f(x)\cos(x/n)\,dx \to \int f(x)\,dx$. $\blacksquare$
Reference
$L^1$ Space definition
The space of integrable functions: those with $\int |f|\,d\mu < \infty$.
Pointwise Convergence definition
$f_n \to f$ pointwise when $f_n(x) \to f(x)$ for every fixed $x$.
Dominated Convergence Theorem theorem
If $f_n \to f$ a.e. and $|f_n| \le g$ for some $g \in L^1$, then $\int f_n \to \int f$.
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